## Going The Distance

The main way to get a handle on your ship definition is to decide what kinds of missions it will be capable of. Let's decide that the Solar Guard cruiser ** Polaris** will be capable of taking off from Terra, travelling to Mars, landing on Mars, taking off from Mars, travelling to Terra, and landing on Terra. All without re-fuelling.

Keep in mind that this is an incredibly silly sort of ship to design. Any real spacecraft designer would design two craft: one surface to orbit shuttle, and one orbit to orbit vehicle.

## Delta-V

The main number of interest is **deltaV**. This means "change of velocity" and is usually measured in meters per second *(m/s)* or kilometers per second *(km/s)*. A spacecraft's maximum deltaV can be though of as how fast it will wind up traveling at if it keeps thrusting until the propellant tanks are dry.

If that means nothing to you, don't worry. The important thing is that a "mission" can be rated according to how much deltaV is required. For instance: lift off from Terra, Hohmann orbit to Mars, and Mars landing, is a mission which would take a deltaV of about 18,290 m/s. **If the spacecraft
has equal or more deltaV capacity than the mission, it is capable of performing that mission.**
The sum of all the deltaV requirements in a mission is called the deltaV budget.

This is why it makes sense to describe a ship's performance in terms of its total deltaV capacity, instead of its "range" or some other factor equally silly and meaningless. In Michael McCollum's classic **Antares Dawn**, when the captain asks the helmsman how much propellant they have, the helmsman replies that they have only 2200 kps *(kilometers per second)* left in the tanks.

The basic deltaV cost for liftoff and landing is what is needed to achieve orbital *(or circular)* velocity.

For a back-of-the-envelope calculation, figure boosting from Terra's surface into LEO will require about 9,400 m/s of deltaV. For other planets use the equation:

**Δ _{vo} = sqrt[ (G * P_{m}) / P_{r} ]**

where:

- Δ
_{vo}= deltaV to lift off into orbit or land on a planet from orbit*(m/s)* - G = 0.00000000006673 or 6.673e-11
*(gravitational constant, don't ask)* - P
_{m}= planet's mass*(kg)* - P
_{r}= planet's radius*(m)* **sqrt[**x**]**= square root of x

Δ_{vo} is what you will use for missions like the Space Shuttle, where you just climb into orbit, deliver or pick up something, then land from orbit. However, if the mission involved travelling to other planets, you will have to use Δ_{esc} instead. This is "escape velocity", and is also the delta V required to land from deep space instead of landing from orbit.

**Δ _{esc} = sqrt[ (2 * G * P_{m}) / P_{r} ]**

**Δ _{esc} = sqrt[ (1.3346e-10 * P_{m}) / P_{r} ]**

where:

- Δ
_{esc}= deltaV for escape velocity from a planet*(m/s)* - G = 0.00000000006673 or 6.673e-11
*(gravitational constant)* - 1.3346e-10 = 2 * G
- P
_{m}= planet's mass*(kg)* - P
_{r}= planet's radius*(m)*

So for our ** Polaris** mission, basic deltaV for Terra escape or capture: 11,180 m/s, basic deltaV for Mars escape or capture: 5030 m/s

Please note that Δ_{esc} already includes the deltaV for Δ_{vo}. In other words, when figuring the total deltaV for a given mission, you will add in either Δ_{esc} or Δ_{vo}, but not both. Use Δ_{vo} for surface-to-orbit missions and use Δ_{esc} for planet-to-planet missions

## Drag

The above equation does not take into account gravitational drag or atmospheric drag. Both are very difficult to estimate.

For a back-of-the-envelope calculation, figure boosting from Terra's surface into LEO will require an extra 1,500 m/s to 2,000 m/s to compensate for the combined effects of atmospheric drag and gravity drag.

Gravitational drag (aka "gravity tax") depends on the planet's gravity, the angle of the flight path, and the acceleration of the spacecraft. For Terra, the first approximation is 762 m/s *(acceleration of ten gees)*. You won't use this equation,
but the actual first approximation is

**Δ _{vd} = g_{p} * t_{L}**

where:

- Δ
_{vd}= deltaV to counteract gravitational drag*(m/s)* - g
_{p}= acceleration due to gravity on planet's surface*(m/s*^{2})*(this assumes that the majority of the burn is close to the ground)* - t
_{L}= duration of liftoff or duration of liftoff burn*(seconds)*

Arthur Harrill has made a nifty Excel Spreadsheet that calculates the liftoff deltaV for any given planet.

Gravitational drag grows worse with each second of burn, so one wants to reduce the burn time. Unfortunately reducing the burn time is the same as increasing the acceleration, and there is a limit to what the human frame can stand. Thorarinn Gunnarsson noted that the eyes are very vulnerable to high-gravity acceleration, second only to bad hearts and full bladders.

You won't use this equation either but

**t _{L} = Δ_{vo} / A**

where:

- A = spacecraft's acceleration
*(m/s*^{2})

The spacecraft's acceleration will be discussed on the page about blast-off.

The equation you will use is this:

**A _{pg} = A / g_{p}**

**Δ _{vd} = Δ_{esc} / A_{pg}**

where:

- A
_{pg}= acceleration of spacecraft in terms of planetary gravities

For our ** Polaris** mission, with an acceleration of 10 g, gravitational drag during Terra lift off will be 11,180 m/s / 10 = 1,118 m/s.

Atmospheric drag only occurs on planets with atmospheres (Δ_{va}). There ain't many planets in the solar system with atmospheres. At least none that you'd care to land on. Landing on Jupiter is a quick way to convert your spacecraft into a tiny ball of crumpled metal. The same holds true for Venus, except that the tiny ball will be acid-etched. So for a planet with no atmosphere, Δ_{va} will be zero.

For Terra, the first approximation is 610 m/s. It is not possible to give a general equation for atmospheric drag due to the large number of factors and variables. You can probably get away with proportional scaling, comparing atmospheric density, assuming you can find data on planetary atmospheric density *(translation: I don't know how to do it)*.

## Total Delta-V

The total lift-off or landing deltaV is the basic deltaV plus the extra deltaV due to atmospheric drag (if any) and gravitational drag.

**Δ _{tvo} = Δ_{vo} + Δ_{vd} + Δ_{va}**

**Δ _{tesc} = Δ_{esc} + Δ_{vd} + Δ_{va}**

where:

- Δ
_{tvo}= total orbital deltaV*(m/s)* - Δ
_{tesc}= total escape deltaV*(m/s)* - Δ
_{vo}= basic deltaV cost for liftoff and orbital landing*(m/s)* - Δ
_{esc}= basic deltaV cost for escape and deep space landing*(m/s)* - Δ
_{vd}= deltaV to counteract gravitational drag*(m/s)* - Δ
_{va}= deltaV to counteract atmospheric drag*(m/s)*

So the total deltaV to lift off from Terra for our ** Polaris** mission is 11,180 + 1118 + 610 = 12,908 m/s. Maybe 13,058 if you add in about 150 m/s for course corrections and as a safety margin.

## Lift-off Acceleration Profile

So you want to keep the acceleration at a maximum of 4g or so otherwise the astronauts cannot manipulate the controls (max of 30g to avoid causing serious injury). But you want to spend as little time as possible getting into orbit in order to minimize gravitational drag. Therefore you want to maintain a steady 4g *(throttling back the thrust as the mass of the propellant drops)* until you get into orbit, right?

Well, I found that it was not that simple. You see, if you are lifting off from a planet with an atmosphere, you have to have to keep your spacecraft's speed such that the maximum dynamic pressure *(or "Max Q")* is too low to shred the ship into titanium confetti. The Space Shuttle's acceleration profile keeps Max Q below about 700 pounds per square foot, but a more sturdy spacecraft could probably survive 800 pounds per square foot.

On the NASA Spaceflight forum I asked what the optimal "acceleration profile" would be for an atomic rocket with a thrust-to-weight ratio above 1, an unreasonable specific impulse of 20,000 (a NSWR), single-stage surface to orbit.

A gentleman who goes by the Internet handle of "Strangequark" was kind enough to answer me.

The acceleration profile says that the spacecraft takes off and accelerates at 4g for about five seconds. From second 5 to second 8 it drastically throttles back to an acceleration of about 0.25g. From second 8 to second 50 it gradually increases acceleration until it is back to 4g. It then stays at 4g until second 215, where it achieves orbit and the engine is shut off.

## General Orbits

If you jump out a window of a skyscraper, or out of an aircraft at high altitude, you will fall down. You will be in a wonderful state of free-fall…until you hit the ground.

An orbit is a clever way to constantly fall towards a planet **but never hit the ground**.

There are certain preferred orbits.

An equatorial orbit is a non-inclined orbit with respect to Terra's equator (*i.e.,* the orbit has zero inclination to the equator, 180° inclination if retrograde). Most civilian satellites use such orbits. The United States uses Cape Canaveral Air Force Station and the Kennedy Space Center to launch into equatorial orbits.

An ecliptic orbit is a non-inclined orbit with respect to the solar system ecliptic.

An inclined orbit is any orbit that does not have zero inclination to the plane or reference (usually the equator).

A polar orbit is a special inclined orbit that goes over each pole of the planet in turn, as the planet spins below (*i.e.,* the orbit is inclined 90° to the equator). Heinlein calls it a "ball of yarn" orbit since the path of the station resembles winding yarn around a yarn ball. The advantage is that the orbit will eventually pass over every part of the planet, unlike other orbits. Such an orbit is generally used for military spy satellites, weather satellites, orbital bombardment weapons, and Google Earth. The United States uses Vandenberg Air Force Base to launch into polar orbits. Google Earth uses data from the Landsat program, whose satellites are launched from Vandenberg.

How long it takes a space station or ship to make one orbit depends upon how massive the planet is and the altitude of the orbit. The mass of the station doesn't matter. The equations below are for a circular or near-circular orbit (low eccentricity) and where the mass of the planet is much larger than the mass of the space station (which is always the case unless the station is built out of stellar black holes or something). The equations for elliptical orbits are a bit more complicated.

An orbit with an orbital period exactly equal to the planetary rotation (one planetary "day") is highly prized for communication satellites.

**OrbitalRadius = OrbitalAltitude + PlanetRadius**

**OrbitalAltitude = OrbitalRadius - PlanetRadius**

**OrbitalPeriod = (2 * π) * sqrt[ OrbitalRadius ^{3} / (G * PlanetMass) ]**

**OrbitalVelocity = sqrt[ (G * PlanetMass) / OrbitalRadius ]**

**OrbitalRadius = cubeRoot[ (G * PlanetMass * OrbitalPeriod ^{2}) / (4 * π^{2}) ]**

**μ = G * PlanetMass**

**μ _{Terra} = 3.99×10^{14}**

**OrbitalPeriodTerra = 6.28318 * sqrt[ (6.37×10 ^{6} + OrbitalAltitude)^{3} / 3.99×10^{14} ]**

**OrbitalVelocityTerra = sqrt[ 3.99×10 ^{14} / OrbitalRadius ]**

**OrbitalRadiusTerra = cubeRoot[ (3.99×10 ^{14} * OrbitalPeriodTerra^{2}) / 39.478 ]**

where:

OrbitalRadius= distance from station to center of planet (m)

OrbitalAltitude= distance from station to surface of planet (m)

PlanetRadius= distance from center of planet to planet's surface (m) (Terra = 6.37×10^{6}m)

OrbitalPeriod= time it takes station to make one orbit around the planet (sec)

OrbitalVelocity= mean velocity of station in its orbit (m/s)

π= pi = 3.14159...

G= Newton's gravitational constant = 6.673×10^{-11}(N m^{2}kg^{-2})

PlanetMass= mass of planet (kg) (Terra = 5.98×10^{24}kg)

μ= standard gravitational parameter

sqrt[= square root ofx]x

cubeRoot[= cube root ofx]x

Just so you know, when it comes to planetary orbits and spacecraft trajectories, none of them are perfectly circular. It is just that so many of them are close enough to being a circle that a science fiction author can get away with using the above equations. We call them **Kepler's** laws of planetary motion because Kepler found that the equations worked if you assumed the planet orbits were ellipses (which are eccentric circles). Kepler's boss Tycho Brahe was dumped in the dust-bin of history because he stubbornly insisted that planet orbits were perfect circles.

And when you get to things like spacecraft transfer orbits, some are not even close to being circular.

What you have to do is use the orbiting object's **Semi-Major Axis** instead of OrbitalRadius.

Don't panic, it is easy to calculate. As long as you have the object's Periapsis and Apoapsis (in meters), which means the object's closest approach and farthest retreat from the planet it is orbiting. Those numbers are easy to find, for example see Wikipedia's entry for the Moon. Periapsis (called perigee) of 362,600 kilometers and Apoapsis (called apogee) of 405,400 kilometers, right in the data bar on the right. Don't forget to convert the values to meters for the equations, *e.g.,* multiply 362,600 km by 1,000 to convert to 362,600,000 m.

Sometimes an orbit will actually be specified by the periapsis and apoapsis. For instance the Orion bomber's patrol orbit is described as a 190,000-410,000 km Terran orbit.

Given periapsis and apoapsis in meters, the Semi-Major Axis is:

**SemiMajorAxis = (Periapsis + Apoapsis) / 2**

Take the equation for OrbitalPeriod, replace OrbitalRadius with SemiMajorAxis, and you are good to go.

OrbitalVelocity unfortunately is a major pain. You see, the orbiting object moves at different speeds at different parts of its orbit. It moves fastest at periapsis and slowest at apoapsis. Only if the orbit is perfectly circular does the orbiting object always move at the same speed.

If you use the OrbitalVelocity equation replacing OrbitalRadius with SemiMajorAxis, you will get the Mean or Average orbital velocity.

If you want the orbital velocity at a specific point in the orbit, you will specify said point by its distance from the primary. The distance will be somewhere between periapsis and apoapsis, inclusive. Again it will be fastest at periapsis and slowest at apoapsis. The equation is:

**OrbitalVelocity = sqrt[ (G * PlanetMass) * ( (2 / CurrentOrbitalRadius) - (1 / SemiMajorAxis) ) ]**

This is the famous Vis Viva Equation, which comes in real handy to calculate delta-V requirements for various missions.

If for some reason you want to draw the orbit, it isn't too hard. As long as have a drawing program that can create an ellipse given a bounding box (The Gimp, Inkscape, Adobe Photoshop, Adobe Illustrator). First you calculate the semi-major axis and the semi-minor axis.

**SemiMajorAxis = (Periapsis + Apoapsis) / 2**

**SemiMinorAxis = sqrt[ SemiMajorAxis ^{2} - ( SemiMajorAxis - Periapsis )^{2} ]**

Chose a convenient scale for your drawing program, like 1,000,000 meters equals one pixel. Draw the upper and lower sides of the box (red lines in diagram above) which are twice the length of the SemiMajorAxis in scale. Draw the left and right sides of the box (green lines) which are twice the length of the SemiMinorAxis in scale. That is the bounding box.

Draw a horizonal center line so it is equidistant from the top and bottom edges of the box. Draw a vertical line (blue in diagram above), and move it so it is one Periapsis scale length away from the right edge. Where these two lines cross is the location of the planet.

Move the box so the cross-hairs are on the planet image. Use the "draw ellipse" function of the drawing program such that the ellipse fits in the bounding box. That is the orbit. You can erase the bounding box now, you don't need it any more.

It is greatly desired that satellites and space stations stay in stable orbits, because corporations and insurance companies become quite angry if hundred million dollar satellites or expensive space stations with lots of people are gravitationally booted into The Big Dark.

A good first approximation is ensuring that the orbiting object stays inside the parent's Hill Sphere. This is an imaginary sphere centered on the parent planet (the planet or moon the satellite is orbiting). Within the sphere, the planet's gravity dominates any satellites.

For first approximation you have three players: the space station (*e.g.,* Supra-New York), the planet or moon it is orbiting (*e.g.,* Terra), and the object the planet is orbiting (*e.g.,* Sol) otherwise known as the planet's "primary".

The point is that **Sol cannot gravitationally capture Supra-New York as long as all of the space station's orbit is inside Terra's Hill Sphere**.

You can calculate the approximate radius of a planet's Hill Sphere with the following equation:

**r ≈ a * cbrt( m / (3 * M) )**

where:

r = Radius of Hill Sphere (kilometers)

a = Distance between the planet and its primary (kilometers)

m = mass of the planet (kilograms)

M = mass of the primary (kilograms)

cbrt(x) = cube root of x (the ∛x key on your calculator)

Actually you can use any desired unit of distance for *r* and *a* **as long as you use the same for both**. The same goes for units of mass for *m* and *M*.

This equation assumes that the planet is in a near-circular orbit. If it has some weird eccentric orbit the Hill Sphere link has the more complicated equation. The above equation also assumes that the mass of the station or sattelite is miniscule compared to the object it is orbiting. It further assumes that the mass of the primary is quite a bit bigger than the mass of the planet.

In practice, for long term stability, the station should not orbit its planet further than one-half the Hill sphere radius. No further than one-third the Hill sphere radius if you are ultra-cautious.

What is the Hill sphere radius of Terra?

In this case, the planet is Terra, the primary is Sol, the mass of Terra (m) is 5.97×10

^{24}kg, the mass of Sol (M) is 1.99×10^{30}kg, the distance between Terra and Sol (a) is 149.6 million kilometers (149,600,000 km).r ≈ a * cubrt( m / (3 * M) )

r ≈ 149,600,000 * cubrt( 5.97×10^{24}/ (3 * 1.99×10^{30}) )

r ≈ 149,600,000 * cubrt( 5.97×10^{24}/ 5.97×10^{30})

r ≈ 149,600,000 * cubrt( 0.000001 )

r ≈ 149,600,000 * 0.01

r ≈ 1,496,000 kilometersSo the ultimate Hill sphere radius is 1.496 million kilometers from Terra's center. Luna is at 0.384 million kilometers, safely inside the Hill sphere. The implication is that all of Terra's stable satellites have an orbital period of less than seven months. Supra-New York would do well to stay inside.

The long term stable radius is 0.748 million kilometers from Terra's center (1/2 Hill sphere). The ultra-cautious radius is 0.499 million kilometers (1/3 Hill sphere).

If you were interested in Lunar satellites, the planet would be Luna, the primary would be Terra, and *a* would be the distance between Terra and Luna.

Any object (like a spaceship) which enters a planet's Hill sphere but does not have enough energy to escape, will tend to start orbiting the planet. The surface of the Hill sphere is sometimes called the "zero-velocity surface" for complicated reasons.

(the section about launch site inclinations has been moved here)

*LEO: light blue*

MEO: yellow

HEO: orange

Hill Sphere is about four times the radius of the orange sphere

Orbits around Terra (geocentric) are sometimes classified by altitude above Terra's surface (which is 6.37×10^{3} km from Terra's center):

**Low Earth Orbit (LEO)**: 160 kilometers to 2,000 kilometers. At 160 km one revolution takes about 90 minutes and circular orbital speed is 8 km/s. Affected by inner Van Allen radiation belt.**Medium Earth Orbit (MEO)**: 2,000 kilometers to 35,786 kilometers. Also known as "intermediate circular orbit." Commonly used by satellites that are for navigation (such as Global Positioning System aka GPS), communication, and geodetic/space environment science. The most common altitude is 20,200 km which gives an orbital period of 12 hours.**Geosynchronous Orbit (GEO)**: exactly 35,786 kilometers from surface of Terra (42,164 km from center of Terra). One revolution takes one sidereal day, coinciding with the rotational period of Terra. Circular orbital speed is about 3 km/s. It is jam-packed with communication satellites like sardines in a can. This orbit is affected by the outer Van Allen radiation belt.**High Earth Orbit (HEO)**: anything with an apogee higher than 35,786 kilometers. If the perigee is less than 2,000 km it is called a "highly elliptical orbit."**Lunar Orbit**: Luna's orbit around Terra has a pericenter of 363,300 kilometers and a apocenter of 405,500 kilometers.**Ultra-Cautious Hill Sphere**: 496,540 kilometers from surface of Terra (498,670 km from center)**Long Term Stable Hill Sphere**: 744,820 kilometers from surface of Terra (748,000 km from center)**Ultimate Hill Sphere**: exactly 1,489,630 kilometers from surface of Terra (1,496,000 km from center)

*Diagram from Arthur Clarke's 1945 paper on extra-terrestrial relays*

It shows how three space stations in geostationary orbit spaced at 120° can provide coverage over all of Terra

Geosynchronous Orbits (aka "Clarke orbits", named after Sir Arthur C. Clarke) are desirable orbits for communication and spy satellites because they return to the same position over the planet after a period of one sidereal day (for Terra that is about four minutes short of one ordinary day).

A Geostationary Orbit is a special kind of geosynchronous orbit that is even more desirable for such satellites. In those orbits, the satellite always stays put over one spot on Terra like it was atop a 35,786 kilometer pole (remember: 42,164 km from *center* of Terra). For complicated reasons all geostationary orbits have to be over the equator of the planet. In theory only three communication satellites in geostationary orbit and separated by 120° can provide coverage over all of Terra.

All telecommunication companies want their satellites in geostationary orbit, but there are a limited number of "slots" available do to radio frequency interference. Things get ugly when you have, for instance, two nations at the same longitude but at different latitudes: both want the same slot. the International Telecommunication Union does its best to fairly divide up the slots.

The collection of artificial satellites in geostationary orbit is called the **Clarke Belt**.

Note that geostationary communication satellites are marvelous for talking to positions on Terra at latitude zero (equator) to latitude plus or minus 70°. For latitudes from ±70° to ±90° (north and south pole) you will need a communication satellite in a polar orbit, a highly elliptical orbit , or a statite. Russia uses highly eccentric orbits since those latitudes more or less define Russia. Russian communication satellites commonly use Molniya orbits and Tundra orbits.

About 300 kilometers above geosynchronous orbit is the "graveyard orbit" (aka "disposal orbit" and "junk orbit"). This is where geosynchronous satellites are moved at the end of their operational life, in order to free up a slot. It would take about 1,500 m/s of delta V to de-orbit an old satellite, but only 11 m/s to move it into graveyard orbit. Most satellites have nowhere near enough propellant to deorbit.

"Okay, T.K., look at it this way. Those three hundred people in LEO Base can get back to Earth in less than an hour if necessary; we'll have lifeboats, so to speak, in case of an emergency. But out there at GEO Base, it's a long way home. Takes eight hours or more just to get back to LEO, where you have to transfer from the deep-space passenger ship to a StarPacket that can enter the atmosphere and land. It takes maybe as long as a day to get back to Earth from GEO Base— and there's a lot of stress involved in the trip."

Hocksmith paused, and seeing no response from the doctor, added gently, "We can get by with a simple first-aid dispensary at LEO Base, T.K., but not at GEO Base. I'm required by my license from the Department of Energy as well as by the regulations of the Industrial Safety and Health Administration, ISHA, to set up a hospital at GEO Base."

He finished off his drink and set the glass down. "If building this powersat and the system of powersats that follow is the biggest engineering job of this century, T.K., then the GEO Base hospital's going to be the biggest medical challenge of our time. It'll be in weightlessness; it'll have to handle construction accidents of an entirely new type; it'll have to handle emergencies resulting from a totally alien environment; it'll require the development of a totally new area of medicine— true space medicine. The job requires a doctor who's worked with people in isolated places—like the Southwest or aboard a tramp steamer. It's the sort of medicine you've specialized in. In short, T.K., you're the only man I know who could do the job . . . and I need you."

Stan and Fred discovered that it took almost nineteen minutes just to get to Charlie Victor, Mod Four Seven. There were a lot of hatches to go through and a lot of modules to traverse. "Fred, if we don't find some faster way to move around this rabbit warren, a lot of people are going to be dead before we reach them," Stan pointed out, finally opening the hatch to Mod Four Seven.

Fred was right behind him through the hatch. "I'll ask Doc to see Pratt about getting us an Eff-Mu."

"What's that?"

"Extra Facility Maneuvering Unit. A scooter to anybody but these acronym-happy engineers."

Transporting was easy in zero-g, but getting through all the hatches while continuing to monitor his condition and maintain the positive-displacement IVs was difficult. It required almost a half hour to bring the man back to the med module.

**SPACE DOCTOR**by Lee Correy (G. Harry Stine) 1981

Lagrangian points are special points were a space station can sit in a sort-of orbit. Lagrange point 1, 2, and 3 are sort of worthless, since objects there are only in a semi-stable position. The ones you always hear about are L4 and L5, because they have been popularized as the ideal spots to locate giant space colonies. Especially since the plan was to construct such colonies from Lunar materials to save on boost delta V costs. The important thing to remember is that the distance between L4 — Terra, L4 — Luna, and Terra — Luna are all the same (about 384,400 kilometers). Meaning it will take just as long to travel from Terra to L4 as to travel from Terra to Luna.

For a more exhaustive list of possible Terran orbits refer to NASA.

It is also possible for a satellite to stay in a place where gravity will not allow it. All it needs is to be under thrust. Which is rather expensive in terms of propellant. Dr. Robert L. Forward noted that solar sails use no propellant, so they can hold a satellite in place forever (or at least as long as the sun shines and the sail is undamaged). This is called a Statite.

If the planet has an atmosphere and the station orbits too low, it will gradually slow down due to atmospheric drag. "Gradually" up to a point, past the tipping point it will rapidly start slowing down, then burn up in re-entry. Some fragments might survive to hit the ground.

The "safe" altitude varies, depending upon the solar sunspot cycle. When the solar activity is high, the Earth's atmosphere expands, so what was a safe altitude is suddenly not so safe anymore.

NASA found this out the hard way with the Skylab mission. In 1974 it was parked at an altitude of 433 km pericenter by 455 km apocenter. This should have been high enough to be safe until the early 1980's. Unfortunately "should" meant "according to the estimates of the 11-year sunspot cycle that began in 1976". Alas, the solar activity turned out to be greater than usual, so Skylab made an uncontrolled reentry in July 1979. NASA had plans to upgrade and expand Skylab, but those plans died in a smoking crater in Western Australia. And a NOAA scientist gave NASA a savage I Told You So.

The International Space Station (ISS) orbited at an even lower at 330 km by 410 km during the Space Shuttle era, but the orbit was carefully monitored and given a reboost with each Shuttle resupply mission. The low orbit was due to the Shuttle carrying up massive components to the station.

After the Shuttle was retired and no more massive components were scheduled to be delivered, the ISS was given a big boost into a much higher 381 km by 384 km orbit. This means the resupply rockets can carry less station reboost propellant and more cargo payload.

If the planet the station orbits has a magnetic field, it probably has a radiation belt. Needless to say this is a very bad place to have your orbit located, unless you don't mind little things like a radiation dosage of 25 Severts per year.

There are known radiation belts around Terra, Jupiter, Saturn, Uranus and Neptune.

## Hohmann Transfer Orbits

*From***We Claim These Stars**by Poul Anderson, 1959

In 1925 Walter Hohmann made your life incredibly easier when he discovered Hohmann transfer orbits. Be grateful.

*Hohmann orbit to a superior planet (Terra to Mars)**Hohmann orbit to an inferior planet (Terra to Venus)**Three phases of a Hohmann*

(Planets orbit counter-clockwise)*Polaris launches from Terra when Mars is 44.36° ahead. This happens every 26 months**8.6 months later both Polaris and Mars arrives at Point X*

Terra is now 75.14° ahead of MarssmAxis =

(OrbitRadius_{s}+ OrbitRadius_{d}) / 2

Now we need to figure the deltaV for Terra-Mars transits.

Current spacecraft propulsion systems are so feeble that they cannot manage much more than the lowest deltaV missions. So they tend to use a lot of *"Hohmann transfer orbits"*.

A Hohmann orbit between two planets is guaranteed to take the smallest amount of deltaV possible. For the Terra-Mars Hohmann, deltaV is 5,596 m/s.

Notice that the deltaV required to get into orbit is 11,180 m/s while the Terra-Mars deltaV is only 5,596 m/s. As Robert Heinlein noted, once one gets into Earth orbit, you are "halfway to anywhere."

And note that it is not strictly necessary that the destination be a physical planet. It can be a virtual point in space, like a reserved slot in geostationary orbit for your communication satellite obtained at great expense and prolonged negotiation with the International Telecommunication Union. Communication satellites are generally delivered via Hohmann transfer, the equations still work even though there is not a planet at the destination. The virtual point still mathematically moves and acts like a planet, even though there ain't nuttin' there.

**Drawbacks of Hohmann Transfers**

Unfortunately a Hohmann orbit also takes the **maximum** amount of transit time. For the Terra-Mars Hohmann mission, transit time is about 8.6 months.

The other drawback is that there are only certain times that one can depart for a given mission, the so-called *"Synodic period"* or Hohmann launch window. The start and destination planets have to be in the correct positions. For the Terra-Mars Hohmann mission, the Hohmann launch windows occur only every 26 months! If you do not launch at the proper time, when you get to the destination planet's orbit the planet won't be there. And then your life span is the same as your rapidly dwindling oxygen supply.

Hop David has computed the cosmic train schedule for Hohmann railroad towns in the Asteroid belt.

### CALCULATING HOHMANN TRANSFERS

If you are in a hurry and just want the transfer parameters between major planets, you can use Erik Max Francis' Mission Tables. These provide the Hohmann delta-V requirements, the transit time, and the delay between Hohmann launch windows.

If the planets you want are not in the tables (because you've made your own solar system or something), the equations are below:

**Hohmann Components**

A Hohmann transfer consists of three phases:

**Insertion Burn:**A large burn to leave circular orbit around starting planet and enter the Hohmann transfer- A long
**Coasting Phase**where the spacecraft travels on an elliptical orbit with engines off **Arrival Burn:**A large burn to leave the Hohmann transfer and enter into a circular orbit around the destination planet (otherwise you are doing a flyby mission)

So the total delta V required is the Insertion Burn plus the Arrival Burn.

Note that when launching only an idiot or somebody absolutely desperate will have their Hohmann going contrary to the planet's native orbital motion. Launching in the same direction as the orbital motion means your spacecraft starts out will that motion as free delta V. The Terra-Mars insertion burn requires 32,731 m/s of delta V. Launching with Terra's orbital motion means the ship starts out with 29,785 m/s for free, and only has to burn for an additional 2,946 m/s. And in the same way the Mars arrival burn in theory requires 21,476 m/s but by using Mars orbital velocity the ship only needs 2,650 m/s. The total delta V required is only 5,596 m/s, not the outrageous 54,207 m/s it needs in theory.

Also note that with a Hohmann, the starting point and the ending point will be 180° from each other. That is, if you draw a line from the start point, the center point, and the end point, you will make a straight line.

* Warning:* the following technique is a simplification. It assumes that the planet orbits are perfectly circular, and the two orbits are coplanar. Neither of these are true in reality, but they are close enough for goverment work. The following technique will give you figures that are in the ballpark, but please do not use them for real astrogation. The perfect technique that gives perfect results is a nightmare of mathematical calculation. If you really want to know, find a copy of Fundamentals of Astrodynamics or Introduction to Space Flight and I salute you.

At the start, you have to chose the starting planet and destination planet (or moon, or asteroids, or whatever). Both have to be orbiting the same primary object, the sun or central planet.

First you need "**μ**_{primary}" ("mu") the gravitational parameter for the sun or planet at the center. If you are calculating Hohmann transfers between planets orbiting Sol, I've precalculated the value of μ for you:

**μ _{SolPrimary} = 1.32715×10^{20} m^{3}/s^{2}**

If you are doing something fancy like transfers between the moons of Jupiter, you have to calculate μ_{primary} for yourself, using the mass of the central body:

**μ _{primary} = 6.674×10^{-11} * M_{primary}**

where M_{primary} = mass of central planet or moon (in kilograms). 6.674×10^{-11} is Newton's gravitational constant expressed in units such that the resulting delta V will be in meters per second, instead of something worthless like furlongs per fortnight. So for Jupiter, Planetary Fact Sheets tell you it has a mass of 1,898.3×10^{24} kilograms, therefore its μ_{primary} is 1.2669×10^{17}

For both the starting and destination planets you'll need:

- The mean orbital radius in meters, i.e., the distance between the planet and the primary. Remember 1 AU = 1.496×10
^{11}meters, since very few astronomical books are silly enough to give orbital radii in meters. - The planet's mass in kilograms
- The planet's mean radius in meters, i.e., distance from the center of the planet and the surface
- The altitude of the parking orbit in meters, i.e., the distance between the planet's surface and the orbiting spacecraft. The orbital altitude at the start planet and destination planet can be totally different from each other. To make life easier on you the parking orbits are assumed to be circular.

Now for the Hohmann delta V calculation. This will give the delta V required to leave low orbit around the starting planet and brake into low orbit around the destination planet. For a crewed mission presumably the crew want to return home again, so you'll have to do the calculations over again with the start and destination data swapped. This will give the delta V for the homeward trip. Add these together to find the minimal delta V rating for the spacecraft.

Yes, there certainly are a lot of equations. That's why they call it rocket science. You probably should make a spreadsheet or something to do the work for you. I tried to encode the following into a spreadsheet (download Microsoft Excel 97-2003 XLS, download Libre Office Calc ODS). It may contains mistakes, use at your own risk.

The "s" subscript means "starting planet" and the "d" subscript means "destination planet". Note that this symbol "∞" should be an infinity symbol, a figure 8 lying on its side. Apologies if your browser cannot render it. In some textbooks they use instead the subscript "inf".

SemimajorAxis = (OrbitRadius_{s}+ OrbitRadius_{d}) / 2

μ_{primary}= 6.674×10^{-11}* M_{primary}OrbitVelocity_{s}= sqrt[ μ_{primary}/ OrbitRadius_{s}]

Velocity_{s}= sqrt[ μ_{primary}* ((2 / OrbitRadius_{s}) - (1 / SemimajorAxis)) ]

Velocity_{∞s}= abs[ Velocity_{s}- OrbitVelocity_{s}]

μ_{s}= 6.674×10^{-11}* M_{s}

ParkingOrbitRadius_{s}= PlanetRadius_{s}+ ParkingOrbitAltitude_{s}

ParkingOrbitCircularVel_{s}= sqrt[ μ_{s}/ ParkingOrbitRadius_{s}]Velocity_{escS}= sqrt[ (2 * μ_{s}) / ParkingOrbitRadius_{s}]

Velocity_{hyperS}= sqrt[ Velocity_{∞s}^{2}+ Velocity_{escS}^{2}]

DeltaV_{s}= Velocity_{hyperS}- ParkingOrbitCircularVel_{s}OrbitVelocity_{d}= sqrt[ μ_{primary}/ OrbitRadius_{d}]

Velocity_{d}= sqrt[ μ_{primary}* ((2 / OrbitRadius_{d}) - (1 / SemimajorAxis)) ]

Velocity_{∞d}= abs[ Velocity_{d}- OrbitVelocity_{d}]

μ_{d}= 6.674×10^{-11}* M_{d}

ParkingOrbitRadius_{d}= PlanetRadius_{d}+ ParkingOrbitAltitude_{d}

ParkingOrbitCircularVel_{d}= sqrt[ μ_{d}/ ParkingOrbitRadius_{d}]

Velocity_{escD}= sqrt[(2 * μ_{d}) / ParkingOrbitRadius_{d}]

Velocity_{hyperD}= sqrt[Velocity_{∞d}^{2}+ Velocity_{escD}^{2}]

DeltaV_{d}= Velocity_{hyperD}- ParkingOrbitCircularVel_{d}DeltaV = abs[DeltaV_{s}] + abs[DeltaV_{d}]

where:

= square of*x*^{2}*x***sqrt[**= square root of*x*]*x***abs[**= absolute value of*x*]*x*, that is, remove any negative sign**SemimajorAxis**= Semi-major axis of Hohmann Transfer orbit (meters)**μ**= mass of primary star (kg) (or whatever) that starting and destination planets are orbiting, multiplied by gravitational constant_{primary}**μ**= mass of starting planet (kg), multiplied by gravitational constant_{s}**OrbitVelocity**= orbital velocity of the starting planet (m/s),_{s}*i.e.,*free delta V**Velocity**= velocity of Insertion Burn (m/s)_{s}**Velocity**= actual velocity needed for Insertion Burn after taking advantage of the free delta V. Called_{∞s}*"hyperbolic velocity at infinity"*(m/s)**PlanetRadius**= mean radius of starting planet (m)_{s}**ParkingOrbitAltitude**= altitude of ship's parking orbit above starting planet's surface (m)_{s}**ParkingOrbitRadius**= radius of ship's parking orbit at starting planet (m)_{s}**Velocity**= local escape velocity from starting planet (m/s)_{escS}**DeltaV**= delta V required to insert spacecraft in parking orbit around starting planet into Hohmann transfer (m/s)_{s}**μ**= mass of destination planet (kg), multiplied by gravitational constant_{d}**OrbitVelocity**= orbital velocity of the destination planet (m/s),_{d}*i.e.,*free delta V**Velocity**= velocity of Arrival Burn (m/s)_{d}**Velocity**= actual deta V needed for Arrival Burn after taking advantage of the free delta V. Called_{∞d}*"hyperbolic velocity at infinity"*(m/s)**PlanetRadius**= mean radius of destination planet (m)_{d}**ParkingOrbitAltitude**= altitude of ship's parking orbit above destination planet's surface (m)_{d}**ParkingOrbitRadius**= radius of ship's parking orbit at destination planet (m)_{d}**Velocity**= local escape velocity from destination planet (m/s)_{escD}**DeltaV**= delta V required for spacecraft to leave Hohmann transfer and enter parking orbit around destination (m/s)_{d}**DeltaV**= actual total delta V needed for the entire Hohmann transfer, which is what you were doing all these calculations for in the first place

**NOMENCLATURE NOTE:**

Depending upon which NASA document you are reading, **Velocity _{∞s}** is also called

**Departure V-infinity**or

**C3**. In missions it is sometimes called

**Trans-**,

*{destination planet}*-Injection*e.g.,*

**TMI**=

**Trans-Mars Injection**.

**Velocity _{∞d}** is also called

**Arrival V-infinity**or

**V**. In missions it is sometimes called

_{∞}**,**

*{destination planet}*-Orbit Insertion*e.g.,*

**MOI**=

**Mars Orbit Insertion**.

For a Terra-Mars Hohmann with central body being Sol, Terra is starting planet, Mars is destination planet.

- Mass of Sol = 1.9885×10
^{30}kg = M_{primary}- Terra's orbital radius = 1.000 AU = 1.496×10
^{11}meters = OrbitRadius_{s}- Mars orbital radius = 1.524 AU = 2.280×10
^{11}meters = OrbitRadius_{d}- Mass of Terra = 5.9720×10
^{24}kg = M_{s}- Mass of Mars = 6.4171×10
^{23}kg = M_{d}- Mean Radius of Terra = 6.3710×10
^{6}m = PlanetRadius_{s}- Mean Radius of Mars = 3.3895×10
^{6}m = PlanetRadius_{d}- Terra Parking Orbit Altitude = 300 km = 300,000 m = ParkingOrbitAltitude
_{s}- Mars Parking Orbit Altitude = 300 km = 300,000 m = ParkingOrbitAltitude
_{d}Doing the math:

- SemimajorAxis = (OrbitRadius
_{s}+ OrbitRadius_{d}) / 2- SemimajorAxis = (1.496×10
^{11}+ 2.280×10^{11}) / 2- SemimajorAxis = 1.888×10
^{11}meters- μ
_{primary}= 6.674×10^{-11}* M_{primary}- μ
_{primary}= 6.674×10^{-11}* 1.9885×10^{30}kg- μ
_{primary}= 1.32715×10^{20}m^{3}/s^{2}- OrbitVelocity
_{s}= Sqrt[ μ_{primary}/ OrbitRadius_{s}]- OrbitVelocity
_{s}= Sqrt[1.32715×10^{20}/ 1.496×10^{11}]- OrbitVelocity
_{s}= 29,785 meters/sec- Velocity
_{s}= sqrt[ μ_{primary}* ((2 / OrbitRadius_{s}) - (1 / SemimajorAxis))]- Velocity
_{s}= sqrt[1.32715×10^{20}* ((2 / 1.496×10^{11}) - (1 / 1.888×10^{11}))]- Velocity
_{s}= 32,731 meters/sec- Velocity
_{∞s}= abs[ Velocity_{s}- OrbitVelocity_{s}]- Velocity
_{∞s}= abs[ 32,731 - 29,785 ]- Velocity
_{∞s}= 2,946 meters/sec- μ
_{s}= 6.674×10^{-11}* M_{s}- μ
_{s}= 6.674×10^{-11}* 5.9720×10^{24}- μ
_{s}= 3.9857×10^{14}- ParkingOrbitRadius
_{s}= PlanetRadius_{s}+ ParkingOrbitAltitude_{s}- ParkingOrbitRadius
_{s}= 6.3710×10^{6}+ 300,000- ParkingOrbitRadius
_{s}= 6.671×10^{6}m- ParkingOrbitCircularVel
_{s}= sqrt[ μ_{s}/ ParkingOrbitRadius_{s}]- ParkingOrbitCircularVel
_{s}= sqrt[ 3.9857×10^{14}/ 6.671×10^{6}]- ParkingOrbitCircularVel
_{s}= 7,730 m/s- Velocity
_{escS}= sqrt[ (2 * μ_{s}) / ParkingOrbitRadius_{s}]- Velocity
_{escS}= sqrt[ (2 * 3.9857×10^{14}) / 6.671×10^{6}]- Velocity
_{escS}=10,931 m/s- Velocity
_{hyperS}= sqrt[ Velocity_{s∞}^{2}+ Velocity_{sesc}^{2}]- Velocity
_{hyperS}= sqrt[ 2,946^{2}+ 10,931^{2}]- Velocity
_{hyperS}= 11,321 m/s- DeltaV
_{s}= Velocity_{hyperS}- ParkingOrbitCircularVel_{s}- DeltaV
_{s}= 11,321 - 7,730- DeltaV
_{s}= 3,592 m/s- OrbitVelocity
_{d}= sqrt[ μ_{primary}/ OrbitRadius_{d}]- OrbitVelocity
_{d}= sqrt[1.32715×10^{20}/ 2.280×10^{11}]- OrbitVelocity
_{d}= 24,126 meters/sec- Velocity
_{d}= sqrt[ μ_{primary}* ((2 / OrbitRadius_{d}) - (1 / SemimajorAxis))]- Velocity
_{d}= sqrt[1.32715×10^{20}* ((2 / 2.280×10^{11}) - (1 / 1.888×10^{11}))]- Velocity
_{d}= 21,476 meters/sec- Velocity
_{∞d}= abs[ Velocity_{d}- OrbitVelocity_{d}]- Velocity
_{∞d}= abs[ 21,476 - 24,126 ]- Velocity
_{∞d}= 2,650 m/s- μ
_{d}= 6.674×10^{-11}* M_{d}- μ
_{d}= 6.674×10^{-11}* 6.4171×10^{23}- μ
_{d}= 4.2828×10^{13}- ParkingOrbitRadius
_{d}= PlanetRadius_{d}+ ParkingOrbitAltitude_{d}- ParkingOrbitRadius
_{d}= 3.3895×10^{6}+ 300,000- ParkingOrbitRadius
_{d}= 3.6895×10^{6}m- ParkingOrbitCircularVel
_{d}= sqrt[ μ_{d}/ ParkingOrbitRadius_{d}]- ParkingOrbitCircularVel
_{d}= sqrt[ 4.2828×10^{13}/ 3.6895×10^{6}]- ParkingOrbitCircularVel
_{d}= 3,407 m/s- Velocity
_{escD}= sqrt[(2 * μ_{d}) / ParkingOrbitRadius_{d}]- Velocity
_{escD}= sqrt[(2 * 4.2828×10^{13}) / 3.6895×10^{6}]- Velocity
_{escD}= 4,818 m/s- Velocity
_{hyperD}= sqrt[Velocity_{∞d}^{2}+ Velocity_{escD}^{2}]- Velocity
_{hyperD}= sqrt[2,650^{2}+ 4,818^{2}]- Velocity
_{hyperD}= 5,499 m/s- DeltaV
_{d}= Velocity_{hyperD}- ParkingOrbitCircularVel_{d}- DeltaV
_{d}= 5,499 - 3,407- DeltaV
_{d}= 2,092- DeltaV = abs[DeltaV
_{s}] + abs[DeltaV_{d}]- DeltaV = abs[2,946] + abs[2,092]
- DeltaV = 3,592 + 2,650
DeltaV = 5,684 meters/secSo the

Polarishas to be capable of 5,684 m/s of delta V in order to do the Terra-Mars Hohmann transfer from Low Earth Orbit to Low Mars Orbit.

How does the above mess of equations work? By the power of the Vis Viva Equation aka *"orbital-energy-invariance law"*. It is used multiple times.

If you don't give a rat's heinie about how this works, please skip ahead to the next section.

If a planet, moon, spacecraft, or whatever is in an elliptical (non-circular) orbit around a primary object (sun or moon), the Vis Viva equation is:

**μ _{primary} = G * M_{primary}**

**V = sqrt[ μ _{primary} * ((2/r) - (1/a)) ]**

where

M= mass of primary object (kg)_{primary}

G= Newton's constant of gravitation = 6.674×10^{-11}N⋅kg^{-1}⋅m^{2}

μ= standard gravitational parameter of the primary object_{primary}

V= orbital velocity at a given point along the elliptical orbit (m/s)

r= distance from primary of the given point along the elliptical orbit (m)

a= semi-major axis of elliptical orbit (m)

sqrt[= square root ofx]x

According to Kepler's Third Law, a planet in an elliptical orbit around a primary has a different orbital velocity at different points in the orbit. The closer that orbital point is to the primary, the faster the orbital velocity is.

If you have a circular orbit, **r = a** so the equation reduces to:

**V = sqrt[ μ _{primary} / r ]**

and the orbital velocity is the same at all points in the circular orbit.

The elliptical Vis Viva equation is used to calculate **Velocity _{s}** and

**Velocity**.

_{d}The circular Vis Viva equation is used to calculate **OrbitalVelocity _{s}**,

**OrbitalVelocity**,

_{d}**ParkingOrbitCircularVel**, and

_{s}**ParkingOrbitCircularVel**

_{d} **Calculating Hohmann Travel Time**

T_{h}= 0.5 * sqrt[ (4 * π^{2}* SemimajorAxis^{3}) / μ_{primary}]

where:

**T**= Hohmann travel time (seconds)_{h}**SemimajorAxis**= Semi-major axis of Hohmann Transfer orbit (meters) from above calculation**μ**= given above, depends on mass of central body_{primary}**sqrt[**= square root of*x*]*x*= square of*x*^{2}*x*= raise*x*^{3}*x*to the third power**π**= 3.14159...

Remember:

- seconds / 2,592,000 = months
- seconds / 31,536,000 = years

The "0.5" factor is because in a Hohmann, the spacecraft only travels over half the Hohmann orbit before it reaches the destination.

For travel time of a Terra-Mars Hohmann with central body being Sol:

- μ
_{primary}= 1.32715×10^{20}- SemimajorAxis = 1.888×10
^{11}metersDoing the math:

- T
_{h}= 0.5 * sqrt[(4 * π^{2}* SemimajorAxis^{3}) / μ_{primary}]- T
_{h}= 0.5 * sqrt[(4 * 3.14159^{2}* (1.888×10^{11})^{3}) / 1.32715×10^{20}]- T
_{h}= 0.5 * sqrt[2.65684×10^{35}/ 1.32715×10^{20}]- T
_{h}= 0.5 * sqrt[2,001,915,283,599,362]- T
_{h}= 0.5 * 44,742,768T_{h}= 22,371,384 seconds = 8.6 months

**Calculating Hohmann Launch Windows**

Hohmann launch windows occur at each synodic period between the two planets.

OrbitPeriod_{i}= 2 * π * sqrt[OrbitRadius_{i}^{3}/ μ_{primary}]

OrbitPeriod_{s}= 2 * π * sqrt[OrbitRadius_{s}^{3}/ μ_{primary}]

SynodicPeriod = 1 / ( (1/OrbitPeriod_{i}) - (1/OrbitPeriod_{s}))

where:

**SynodicPeriod**= time delay between Hohmann launch windows (seconds)**OrbitRadius**= orbital radius of planet closer to central object (meters)_{i}**OrbitRadius**= orbital radius of planet further away from central object (meters)_{s}**OrbitPeriod**= one planetary year for the inferior planet (seconds)_{i}**OrbitPeriod**= one planetary year for the superior planet (seconds)_{s}**μ**= given above, depends on mass of central body_{primary}**x**= raise x to the third power^{3}**π**≅ 3.14159...

Remember:

- seconds / 2,592,000 = months
- seconds / 31,536,000 = years

Delay between Hohmann launch windows for Terra-Mars Hohmann, central body is Sol, Terra is the inferior planet.

- μ
_{primary}= 1.32715×10^{20}- OrbitRadius
_{i}= 1.496×10^{11}meters- OrbitRadius
_{s}= 2.280×10^{11}metersDoing the math:

- OrbitPeriod
_{i}= 2 * π * sqrt[OrbitRadius_{i}^{3}/ μ_{primary}]- OrbitPeriod
_{i}= 2 * 3.14159 * sqrt[(1.496×10^{11})^{3}/ 1.32715×10^{20}]- OrbitPeriod
_{i}= 2 * 3.14159 * sqrt[3.348071936×10^{33}/ 1.32715×10^{20}]- OrbitPeriod
_{i}= 2 * 3.14159 * sqrt[25,227,532,200,580]- OrbitPeriod
_{i}= 2 * 3.14159 * 5,022,702- OrbitPeriod
_{i}= 31,558,565 seconds- OrbitPeriod
_{s}= 2 * π * sqrt[OrbitRadius_{s}^{3}/ μ_{primary}]- OrbitPeriod
_{s}= 2 * 3.14159 * sqrt[(2.280×10^{11})^{3}/ 1.32715×10^{20}]- OrbitPeriod
_{s}= 2 * 3.14159 * sqrt[1.1852352×10^{34}/ 1.32715×10^{20}]- OrbitPeriod
_{s}= 2 * 3.14159 * sqrt[89,306,800,286,328]- OrbitPeriod
_{s}= 2 * 3.14159 * 9,450,228- OrbitPeriod
_{s}= 59,377,531 seconds- SynodicPeriod = 1 / ( (1 / OrbitPeriod
_{i}) - (1 / OrbitPeriod_{s}))- SynodicPeriod = 1 / ( (1 / 31,558,565) - (1 / 59,377,531))
SynodicPeriod = 67,359,430 seconds = 26 months = 2.14 years

**Calculating Launch Timing**

This is for calculating two things:

- What is the configuration of the two planets indicating it is time to launch?
- If you do a Hohmann from planet A to planet B, how long do you have to wait on planet B before the launch window to planet A opens?

For the first question, the best I can do is indicate the angular separation between the two planets when the Hohmann window opens. For example: with the Terra-Mars Hohmann, when the launch window opens, what is angle Terra-Sol-Mars? Note that 0° is where the start planet is at. And at the end of the Hohmann journey, both the spacecraft and the destination planet will be at 180° from the the location of the start planet at the beginning of the journey.

α = π * (1 - ( (1/(2*sqrt[2])) * sqrt[(r1/r2 + 1)^{3}]))

or

α = π * (1 - (0.35355* sqrt[(r1/r2 + 1)^{3}]))

where:

**α**= Phase Angle, or*angle StartPlanet-CenterObject-DestPlanet*(radians). If negative, DestPlanet is behind StarPlanet, otherwise it is ahead.**x**= raise x to the third power^{3}**π**≅ 3.14159...**r1**= OrbitRadius_{s}= orbital radius of start planet (meters)**r2**= OrbitRadius_{d}= orbital radius of destination planet (meters)**0.35355**≅ 1 / (2 * sqrt[2])

Convert radians into decimal degrees by muliplying by (180/π), which is approximately 57.29578...

Spacecraft and Mars will both arrive at spot marked "X" in 8.6 months. Planets are rotating counter-clockwise.Angle between planets for Terra-Mars Hohmann.

- r1 = 1.496×10
^{11}meters- r2 = 2.280×10
^{11}metersDoing the math:

- α = π * (1 - ( (1/(2*sqrt[2])) * sqrt[(r1/r2 + 1)
^{3}]))- α = 3.14159 * (1 - (0.35355 * sqrt[(1.496×10
^{11}/2.280×10^{11}+ 1)^{3}]))- α = 3.14159 * (1 - (0.35355 * sqrt[(0.65617 + 1)
^{3}]))- α = 3.14159 * (1 - (0.35355 * sqrt[1.65617
^{3}]))- α = 3.14159 * (1 - (0.35355 * sqrt[4.54271]))
- α = 3.14159 * (1 - (0.35355 * 2.13136))
- α = 3.14159 * (1 - 0.75355)
- α = 3.14159 * 0.24645
α = +0.77424 radians = +44.36°Since Phase Angle α is positive, Mars is ahead of Terra.

For the Mars-Terra Hohmann, Phase Angle α = -1.31229 radians, or -75.19° behind Mars.

**Calculating Stayover Time Before Return Trip**

For details about how long the ship will have to delay at Mars before the return trip Hohmann window opens, refer here

In a typical round trip, you start at a planet (say, Earth), then execute a Hohmann transfer to another planet (say, Mars). However, you cannot return immediately, since Earth and Mars are then not in the right places. You must wait a certain amount of time before taking another Hohmann transfer back.

How long will that be?

Working out the problem turns out to be relatively straightforward. (For maintaining intuition, I'll continue with the example of visiting Mars, but note that the analysis remains generally applicable.)

First, we define four times:

: the time of departure from Earth.t_{0}: the time of arrival at Mars.t_{1}: the time of departure from Mars.t_{2}: the time of arrival at Earth.t_{3}Next, we define the (heliocentric) angular position of these planets at given times. As time progresses over the planet's orbital period, these angles sweep out

2πradians of passage over a single planetary year (a period of timeandP_{E}for Earth and Mars, respectively). We are only interested in their values at the four points above, though. E.g.P_{M}is the angle of Mars at timeθ_{M,2}.t_{2}We need to define how these angles relate to each other. Let's call

the duration of the Hohmann transfer; notice that it is the same going out as coming back. For the time periodt_{H}:= t_{1}-t_{0}tot_{0}, the spacecraft is on a Hohmann transfer from Earth to Mars. The angles relate as:t_{1}In English, this just says that the (angular) positions of both Earth and Mars advance over the time of the outgoing transfer.

At Mars, we'll wait for some unknown time

(waiting time). Again, the planets' angular positions advance:t_{W}Finally, coming back, we do another Hohmann:

We still need some more information to solve this problem, though. The first two pieces are that a Hohmann transfer always takes you an angle

around the center:πThe first equation says that the spacecraft departing Earth must arrive at Mars after going halfway around the orbit, while the second says that the spacecraft leaving Mars must arrive back at Earth, again after a half-orbit. We need to be more-careful, though. Although the angles for the Earth and Mars can be taken to increase monotonically, one of them might do a full orbit while the other has not. We therefore need to be able to add/subtract multiples of

of the angles to get them to agree:2πWe need just one more equation. The entire problem can be shifted by a constant amount around the Sun. To constrain it, without loss of generality we can simply set:

(I'll also rename the remaining

to justk_{1}.)kWe can rewrite this huge mess of equations as a matrix equation to impose some semblance of sanity on the complexity:

Solving this is not too difficult (I did it by hand first before checking my result with sympy). The only part of it we care about is the value for

, which works out to be:t_{W}Because

is just some integer, we can remove thatk. Generalizing the notation a little, so that the home planet has period-1and the visited planet has periodP_{0}, we get:P_{1}For

, any value can be chosen so long as the resultingkis nonnegative.t_{W}This was the result I gave here (the Google+ question which nerd-sniped me into doing this whole thing). I'm fairly convinced it's correct, but, it's still a little unsatisfactory. Choosing

from the integers is obnoxious. It would be nice to choose it from the natural numbers (i.e.,kℕ := {1,2,3,…}), and thereby get an increasing sequence of possible departure dates, starting from a value that is the earliest.Requiring that

t_{W}≥ 0, some algebra shows that, if, then we must haveP>_{0}P_{1}). Similarly, ifk≥ (-2t)/(_{H}P_{0}, thenP<_{0}P_{1}). And, of course, ifk≤ (-2t)/(_{H}P_{0}, then the planets are co-orbital and you cannot travel between them with Hohmann transfers (so we shall assume this is not the case forthwith).P=_{0}P_{1}We now want to generate a sequence of

values from natural-valuedkvalues that produces the correct result either way:nSome ad-hoc finagling with the intersection of these lines in the middle produces:

By substituting back into our answer, we can get an equation that tells you every possible waiting time, starting from the first, indexed by

:n∈ ℕ = {1,2,3,…}I'm less-confident of this result than the previous but, as we'll see below, it seems to work—at-least for the

case I tested.P<_{0}P_{1}Let's check and demonstrate our work by returning to the Earth/Mars example we started with.

The actual values are (note sidereal value for

whereas Gregorian calendar used to convert from months):P_{i}P= 365.256363004_{0}

P= 686.980_{1}

t=“8.5 months” ≈ 259 days_{H}

t=“14.9 months” ≈ 454 days_{W}According to our first formula, we get:

kt_{W}-3 ≈ 1235 days -2 ≈ 455 days -1 ≈ -325 days 0 ≈ -1105 days 1 ≈ -1885 days 2 ≈ -2665 days 3 ≈ -3445 days Therefore,

must bek-2or less. Notice also the separation between the possible waiting times. They should come in multiples of the transfer window times (because if you miss your departure date for the aligned planets, you'll have to wait until the next transfer window). For Earth/Mars, this is“26 months” ≈ 791 days. Indeed, we see that successive launch times differ by roughly this many days. The agreement is to within1.5%or so, which is pretty good given the imprecision ofand the expectedt_{H}, as well as the astronomical fact that the orbits are imperfect.t_{W}For our second formula, we have:

nt_{W}1 ≈ 455 days 2 ≈ 1235 days 3 ≈ 2015 days This confirms that the formula works when

, but the case forP<_{0}P_{1}remains untested.P>_{0}P_{1}

**HOHMANN WAITING TIME**by Ian Mallett

*(2019)*

### Hohmanns In More Detail

**Planetary Transfer Calculator** is an on-line calculator for various types of transfers (including Hohmanns and torchship brachistochrone transfers). It can calculate ballistic transfers between planets and moons, and powered (constant acceleration) transfers between stars (including effects of relativity). It can also calculate propagation delay due to the absolute speed of light between planets and moons.

For a more in-depth look at the equations for the deltaV of a given Hohmann mission, go here

- Discussion of Hohmann tranfer and travel time
- Discussion of Hohmann delta V
- Discussion of Hohmann launch windows

There is a more in-depth example of calculating both Hohmann and more energetic orbits using Fundamentals of Astrodynamics at the incomparable Voyage to Arcturus. The entries in question are here, here, and here. The discussion is about the superiority of Nuclear-Ion propulsion as compared to Nuclear-Thermal propulsion.

There are good basic tutorials on orbital mechanics and trajectory here, here and here.

There is a simple listing of the appropriate equations at http://scienceworld.wolfram.com/physics/HohmannTransferOrbit.html and at http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

Here is an Excel spreadsheet called "Pesky Belter" which will calculate Hohmann deltaV, transit times, and synodic periods.

Erik Max Francis has written a freeware Hohmann orbit calculator in Python, available here. Be warned that the documentation is rudimentary, and operating the calculator requires a beginners knowledge of the Python language.

Information about the mass and semi-major axes of various planets can be found here: http://nssdc.gsfc.nasa.gov/planetary/planetfact.html

~~The Windows utility program Swing-by calculator can be found at http://www.jaqar.com~~

There is a freeware Windows program called Orbiter that allows one to fly around the solar system using real physics. A gentleman named Steven Ouellette has created an Orbiter add-on that re-creates the Rolling Stone from the Heinlein novel of the same name, along with the
mission it flew *(follow the above link)*.

Terra Space Station and the school ship

Randolphare in a circular orbit 22,300 miles above the surface of the Earth, where they circle the Earth in exactly twenty-four hours, the natural period of a body at that distance.Since the Earth's rotation exactly matches their period, they face always one side of the Earth — the ninetieth western meridian, to be exact. Their orbit lies in the ecliptic, the plane of the Earth's orbit around the Sun, rather than in the plane of the Earth's equator. This results in them swinging north and south each day as seen from the earth. When it is noon in the Middle West, Terra Station and the

Randolphlie over the Gulf of Mexico; at midnight they lie over the South Pacific.The state of Colorado moves eastward about 830 miles per hour. Terra Station and the

Randolphalso move eastward nearly 7000 miles per hour — 1.93 miles per second, to be finicky. The pilot of the Bolivar had to arrive at theRandolphprecisely matched in course and speed. To do this he must break his ship away from our heavy planet, throw her into an elliptical orbit just tangent to the circular orbit of theRandolphand with that tangency so exactly placed that, when he matched speeds, the two ships would lie relatively motionless although plunging ahead at two miles per second. This last maneuver was no easy matter like jockeying a copter over a landing platform, as the two speeds, unadjusted, would differ by 3000 miles an hour.Getting the Bolivar from Colorado to the

Randolph, and all other problems of journeying between the planets, are subject to precise and elegant mathematical solution under four laws formulated by the saintly, absent-minded Sir Isaac Newton nearly four centuries earlier than this flight of the Bolivar — the three Laws of Motion and the Law of Gravitation. These laws are simple; their application in space to get from where you are to where you want to be, at the correct time with the correct course and speed, is a nightmare of complicated, fussy computation.

**SPACE CADET**by Robert Heinlein

*(1948)*

### Rocket Railroad

You are probably using Hohmann transfer orbits because your rocket ain't a torchship. That is the spacecraft has such a pathetically small amount of delta-V that it is forced to use bargain-basement bin cheap Hohmanns instead of fast but hideously expensive Brachistochrone tranfers.

Since the ship is on such a tight delta-V budget it cannot afford to leave the pre-plotted Hohmann trajectory. If you do, you'll run out of the propellant you need to reach your destination, the ship will sail off into the Big Dark, and everybody will die when the oxygen runs out. This was highlighted in a famous story called The Cold Equations by Tom Godwin.

The net result is that when it comes to side trips, rockets are about as capable of that as is a railroad locomotive. The rocket has to stick to its planned Hohmann like it was a choo-choo train on solid steel girders. Much like a locomotive, leaving the tracks for an off-road excursion is a disaster (yes I know that some cargo spacecraft are arranged like train with the engines in the front dragging the cargo behind, but that's another matter).

In spite of the title this post is about spaceships, not trains. It is inspired by commenter Ferrell's remark, in discussion of the aesthetics of space travel, that

'cycler stations' seem more like railroads than ships. To expand on my own response there, this is largely true of spacecraft in general, at least those without magitech drives.

Trains, it has been observed, differ from other common terrestrial vehicles in that they have no steering wheel. Once they leave the station platform they go not where 'the governor [helmsman] listeth,' but where the tracks take them. Spaceships may have a joystick for attitude control, but once they light up their main drive they go where the laws of physics take them.As I noted last year in Space Warfare IV: Mobility, the way they actually get around resembles... self-propelled artillery shells. Once they fire themselves into a particular orbit they can change that orbit only by another burst of power, expending more propellant in the process.Regular readers here are probably geeky enough that you already know this, and in particular you likely appreciate the tactical military implications — what space wargamers call vector movement, AKA why spaceships don't maneuver the way Hollywood usually portrays.So why am I beating you over the head with it? Because it is so easy to forget that this applies not only to tactical maneuvers but to strategic or 'operational' movements, and to commercial traffic.

If a spaceship in Earth orbit is fueled up and ready to go to Mars, once you punch the 'go' button you are on your way to Mars. Yes, in the early stages of your departure burn you can abort back to Earth orbit (or, very occasionally, to lunar orbit). But once past that initial abort window any subsequent change of orbit will, in nearly all cases, take you only on a long, slow trip to nowhere.This applies most rigidly to economical Hohmann (or near-Hohmann) transfer orbits, but it applies with nearly as much force even to fast ships taking steep orbits. Unless provided for in your mission plan, the chances that your fuel allowance permits you to change orbit to one that will get you somewhere else is slim to none.

Military missions may — and certainly should, if possible — provide an abort option that will get you to some friendly base before life support runs out. Commercial missions, probably not: These trips will be costly enough without carrying along extra fuel and life support for a change of destination. And for most space emergencies such an abort would be useless anyway — whatever keeps you from safely reaching Mars would make it even harder to reach anywhere else.

Thus space operations will 'run on rails,' with the route and destination fixed not just by the space line's policy but by constraints of time, motion, and propellant supply.All of which has some interesting secondary implications, ranging from space rescue to command structure.

Rescue is plausible between ships on similar orbits, as in Heinlein's Rolling Stones, where Dr. Stone transfers to a nearby liner, black bag in hand, to fight a disease outbreak on board. But if two ships are passing on different orbits, don't expect one to be able to assist the other. Similarly, 'lifeboats' are pretty much useless in deep space — if you take to the boats you're still on the same orbit as the stricken ship, and unless the lifeboats have delta v and life support comparable to the ship itself they won't help. (Two hab structures with independent life support are a much better bet.)The constraints of space motion also raise a question about who should be in command. In the movie Casablanca, Rick Blaine suggests to Ilsa that they get married on a train. "The captain of a ship can perform marriages; why not the engineer on a train?" But the 'captain' of a train is not the engineer; it is the conductor. (In British railway usage, not the driver but the guard.)

At sea and in the air a pilot/navigator traditionally has command, because they are the most skilled at handling the vehicle under abnormal conditions, to change course and reach sheltered waters or a safe landing. But in space, especially deep space, brilliant shiphandling is probably not an option. Survival, if possible, will generally depend on the crew's ability to function as a social unit, and on the life support system holding out.

In human dramatic terms a spaceship is more like an isolated outpost than any terrestrial vehicle.

Finally, a way that spaceships differ even from trains is that nearly all travel is nonstop, from point of origin to final destination.Terrestrial vehicles can and often do make intermediate stops along the way, each time letting off some passengers and cargo, and taking on others. This trip pattern lends itself well to RPGs, picaresque scenarios in general, and especially episodic television, with each waypoint an Adventure Town.This is practical because ships and trains (or caravans, etc.) lose little time and expend insignificant fuel in making intermediate stops. Planes need extra fuel to climb back to cruise altitude, but they can top off their tanks, and by not carrying fuel for a nonstop trip they can usually carry more payload.

Alas, it does not work that way in space. Spaceships don't burn their fuel while cruising; they burn it to speed up and slow down. So even if several planets were neatly lined up, each intermediate stop would involve major burns. Carrying passengers or cargo to Saturn, with intermediate stops at Mars and Jupiter, means accelerating and decelerating your Saturn-bound manifest three times — a much better way to reach the poorhouse than Saturn. Ships may make several passages before returning to their home base, but nearly all passengers and cargo will turn over at each port of call. (Cargo may not travel by 'ship' at all.)

There are some specialized exceptions to most or all of these rules.

And, of course, with a suitable magitech drive all bets are off.But that is a topic for a different discussion.Related post: In Space Warfare IV: Mobility, I discussed military aspects of space motion.

**RUNNING ON RAILS**by Rick Robinson

*(2010)*

THE LINER

Pegasus, with three hundred passengers arid a crew of sixty, was only four days out from Earth when the war began and ended. For some hours there had been a great confusion and alarm on board, as the radio messages from Earth and Federation were intercepted. Captain Halstead had been forced to take firm measures with some of the passengers, who wished to turn back rather than go on to Mars and an uncertain future as prisoners of war. It was not easy to blame them; Earth was still so close that it was a beautiful silver crescent, with the Moon a fainter and smaller echo beside it. Even from here, more than a million kilometers away, the energies that had just flamed across the face of the Moon had been clearly visible, and had done little to restore the morale of the passengers.They could not understand that the law of celestial mechanics admit of no appeal. The

Pegasuswas barely clear of Earth, and still weeks from her intended goal. But she had reached her orbiting speed, and had launched herself like a giant projectile on the path that would lead inevitably to Mars, under the guidance of the sun's all-pervading gravity.There could be no turning back: that would be a maneuver involving an impossible amount of propellant.The Pegasus carried enough dust in her tanks to match velocity with Mars at the end of her orbit, and to allow for reasonable course corrections en route. Her nuclear reactors could provide energy for a dozen voyages—but sheer energy was useless if there was no propellant mass to eject. Whether she wanted to or not, the Pegasus was headed for Mars with the inevitability of a runaway streetcar. Captain Halstead did not anticipate a pleasant trip.The words MAYDAY, MAYDAY came crashing out of the radio and banished all other preoccupations of the

Pegasusand her crew. For three hundred years, in air and sea and space, these words had alerted rescue organizations, had made captains change their course and race to the aid of stricken comrades. But there was so little that the commander of a spaceship could do; in the whole history of astronautics, there have been only three cases of a successful rescue operation in space.There are two main reasons for this, only one of which is widely advertised by the shipping lines. Any serious disaster in space is extremely rare; almost all accidents occur during planetfall or departure. Once a ship has reached space, and has swung into the orbit that will lead it effortlessly to its destination, it is safe from all hazards except internal, mechanical troubles. Such troubles occur more often than the passengers ever know, but are usually trivial and are quietly dealt with by the crew. All spaceships, by law, are built in several independent sections, any one of which can serve as a refuge in an emergency. So the worst that ever happens is that some uncomfortable hours are spent by all while an irate captain breathes heavily down the neck of his engineering officer.

The second reason why space rescues are so rare is that they are almost impossible, from the nature of things. Spaceships travel at enormous velocities on exactly calculated paths, which do not permit of major alterations—as the passengers of the

Pegasuswere now beginning to appreciate.The orbit any ship follows from one planet to another is unique; no other vessel will ever follow the same path again, among the changing patterns of the planets. There are no "shipping lanes" in space and it is rare indeed for one ship to pass within a million kilometers of another.Even when this does happen, the difference of speed is almost always so great that contact is impossible.

**EARTHLIGHT**by Arthur C. Clarke (1955)

Holden leaned back in his chair and listened to the creaks of the

Canterbury'sfinal maneuvers, the steel and ceramics as loud and ominous as the wood planks of a sailing ship. Or an Earther's joints after high g. For a moment, Holden felt sympathy for the ship.They weren't really stopping, of course. Nothing in space ever actually stopped; it only came into a matching orbit with some other object. They were now following CA-2216862 on its merry millennium-long trip around the sun.

**LEVIATHAN WAKES**by "James S.A. Corey"

*(2011)*. First novel of The Expanse

## Mid-Course corrections

When a probe or spacecraft performs a maneuver, the idea is to enter into a pre-calculated trajectory (hopefully arriving at your destination). But nobody and nothing is perfect. The performance of the maneuver might be a hair off, though not enough to be immediately noticeable. Mission Control or the spacecraft's astrogator has the job of monitoring the spacecraft's current position and vector at this specific point in time, to see if the spacecraft is still on track for the specified trajectory. If it is not in the groove, the astrogator will calculate a mid-course corrections (Trajectory Correction Maneuver or TCM). This is a tiny maneuver to put the spacecraft back on track.

Currently I have no idea how to calculate such a thing. In Proceeding of the Symposium on Manned Planetary Missions 1963/1964 they suggested that with then-current navigation gear the delta V required for TCM on the Terra-Mars trajectory was about 105 m/s and 92 m/s for the Mars-Terra trajectory.

## Other Transfer Orbits

CONJUNCTION CLASS

artwork by Edward BellFor high-thrust rockets, the most fuel-efficient way to get to Mars is called a

Hohmann transfer. It is an ellipse that just grazes the orbits of both Earth and Mars, thereby making the most use of the planets’ own orbital motion. The spacecraft blasts off when Mars is ahead of Earth by an angle of about 45 degrees (which happens every 26 months). It glides outward and catches up with Mars on exactly the opposite side of the sun from Earth’s original position. Such a planetary configuration is known to astronomers as a conjunction. To return, the astronauts wait until Mars is about 75 degrees ahead of Earth, launch onto an inward arc and let Earth catch up with them.Each leg requires two bursts of acceleration. From Earth’s surface, a velocity boost of about 11.5 kilometers per second breaks free of the planet’s pull and enters the transfer orbit. Alternatively, starting from low Earth orbit, where the ship is already moving rapidly, the engines must impart about 3.5 kilometers per second. (From lunar orbit the impulse would be even smaller, which is one reason that the moon featured in earlier mission plans. But most current proposals skip it as an unnecessary and costly detour.) At Mars, retrorockets or aerobraking must slow the ship by about 2 kilometers per second to enter orbit or 5.5 kilometers per second to land. The return leg reverses the sequence. The whole trip typically takes just over two and a half years: 260 days for each leg (increasing astronaut exposure to galactic cosmic radiation) and 460 days on Mars. In practice, because the planetary orbits are elliptical and inclined, the optimal trajectory can be somewhat shorter or longer. Leading plans, such as Mars Direct and NASA’s reference mission, favor conjunction-class missions but quicken the journey by burning modest amounts of extra fuel. Careful planning can also ensure that the ship will circle back to Earth naturally if the engines fail (a strategy similar to that used by

Apollo 13).

OPPOSITION CLASS

artwork by Edward Bell

OPPOSITION CLASSTo keep the trip short (reducing astronaut exposure to galactic cosmic radiation), NASA planners traditionally considered opposition-class trajectories, so called because Earth makes its closest approach to Mars—a configuration known to astronomers as an opposition—at some point in the mission choreography. These trajectories involve an extra burst of acceleration, administered en route. A typical trip takes one and a half years: 220 days getting there, 30 days on Mars and 290 days coming back. The return swoops toward the sun, perhaps swinging by Venus, and approaches Earth from behind. The sequence can be flipped so that the outbound leg is the longer one. Although such trajectories have fallen into disfavor—it seems a long trip for such a short stay—they could be adapted for ultrapowerful nuclear rockets or “cycler” schemes in which the ship shuttles back and forth between the planets without stopping.

LOW THRUST

artwork by Edward Bell

LOW THRUSTLow-thrust rockets such as ion drive save fuel but are too weak to pull free of Earth’s gravity in one go (high specfic impulse but very low thrust). They must slowly expand their orbits, spiraling outward like a car switchbacking up a mountain. Reaching escape velocity could take up to a year, which is a long time to expose the crew to the Van Allen radiation belts that surround Earth. One idea is to use low-thrust rockets only for hauling freight. Another is to move a vacant ship to the point of escape, ferry astronauts up on a “space taxi” akin to the shuttle and then fire another rocket for the final push to Mars. The second rocket could either be high or low thrust. In one analysis of the latter possibility, a pulsed inductive thruster fires for 40 days, coasts for 85 days and fires for another 20 days or so on arrival at the Red Planet.

A VASIMR engine opens up other options. Staying in low gear (moderate thrust but low efficiency), it can spiral low efficiency), it can spiral out of Earth orbit in 30 days. Spare propellant shields the astronauts from radiation. The interplanetary cruise takes another 85 days. For the first half, the rocket upshifts; at the midpoint it begins to brake by downshifting. On arrival at Mars, part of the ship detaches and lands while the rest—including the module for the return flight—flies past the planet, continues braking and enters orbit 131 days later.

**HOW TO GO TO MARS**by George Musser and Mark Alpert

*(Scientific American March 2000)*

Circumlunar free return trajectory (not to scale)

A

free-return trajectoryis a trajectory of a spacecraft traveling away from a primary body (for example, the Earth) where gravity due to a secondary body (for example, the Moon) causes the spacecraft to return to the primary body without propulsion (hence the termfree).(ed note:

Translation, if an Apollo lunar mission had an engine malfunction in the first half of the journey, the spacecraft would go sailing off into an eccentric Terran orbit and all the astronauts would die. But if it was using a free-return trajectory, the Lunar gravity would automatically sling the spacecraft back towards Terra with no engines needed and the astronauts could land on Terra Firma.)## Earth–Moon

The first spacecraft traveled using free-return trajectory on October 4, 1959 was Russian «Луна-3» (Moon-3). Then, free-return trajectories were introduced by Arthur Schwaniger of NASA in 1963 with reference to the Earth–Moon system. Limiting the discussion to the case of the Earth and the Moon, if the trajectory at some point crosses the line going through the centre of the Earth and the centre of the moon, then we can distinguish between:

- A circumlunar free-return trajectory around the Moon. The spacecraft passes behind the Moon. It moves there in a direction opposite to that of the Moon. If the craft's orbit begins in a normal (west to east) direction near Earth, then it makes a figure 8 around the Earth and Moon.
- A cislunar free-return trajectory. The spacecraft goes beyond the orbit of the Moon, returns to inside the Moon's orbit, moves in front of the Moon while being diverted by the Moon's gravity to a path away from the Earth to beyond the orbit of the Moon again, and is drawn back to Earth by Earth's gravity. (There is no real distinction between these trajectories and similar ones that never go beyond the Moon's orbit, but the latter may not get very close to the Moon, so are not considered as relevant.)
For trajectories in the plane of the Moon's orbit with small periselenum radius (close approach of the Moon), the flight time for a cislunar free-return trajectory is longer than for the circumlunar free-return trajectory with the same periselenum radius. Flight time for a cislunar free-return trajectory decreases with increasing periselenum radius, while flight time for a circumlunar free-return trajectory increases with periselenum radius.

Using the simplified model where the orbit of the Moon around the Earth is circular, Schwaniger found that there exists a free-return trajectory in the plane of the orbit of the Moon which is periodic: after returning to low altitude above the Earth (the perigee radius is a parameter, typically 6555 km) the spacecraft would return to the Moon, etc. This periodic trajectory is counter-rotational (it goes from east to west when near the Earth). It has a period of about 650 hours (compare with a sidereal month, which is 655.7 hours, or 27.3 days). Considering the trajectory in an inertial (non-rotating) frame of reference, the perigee occurs directly under the Moon when the Moon is on one side of the Earth. Speed at perigee is about 10.91 km/s. After 3 days it reaches the Moon's orbit, but now more or less on the opposite side of the Earth from the Moon. After a few more days, the craft reaches its (first) apogee and begins to fall back toward the Earth, but as it approaches the Moon's orbit, the Moon arrives, and there is a gravitational interaction. The craft passes on the near side of the Moon at a radius of 2150 km (410 km above the surface) and is thrown back outwards, where it reaches a second apogee. It then falls back toward the Earth, goes around to the other side, and goes through another perigee close to where the first perigee had taken place. By this time the Moon has moved almost half an orbit and is again directly over the craft at perigee.

There will of course be similar trajectories with periods of about two sidereal months, three sidereal months, and so on. In each case, the two apogees will be further and further away from Earth. These were not considered by Schwaniger.

This kind of trajectory can occur of course for similar three-body problems; this problem is an example of a circular restricted three-body problem.

While in a true free-return trajectory no propulsion is applied, in practice there may be small mid-course corrections or other maneuvers.

A free-return trajectory may be the initial trajectory to allow a safe return in the event of a systems failure; this was applied in the Apollo 8, Apollo 10, and Apollo 11 lunar missions. In such a case a free return to a suitable reentry situation is more useful than returning to near the Earth, but then needing propulsion anyway to prevent moving away from it again. Since all went well, these Apollo missions did not have to take advantage of the free return and inserted into orbit upon arrival at the Moon.

Due to the landing-site restrictions that resulted from constraining the launch to a free return that flew by the Moon, subsequent Apollo missions, starting with Apollo 12 and including the ill-fated Apollo 13, used a hybrid trajectory that launched to a highly elliptical Earth orbit that fell short of the Moon with effectively a free return to the atmospheric entry corridor. They then performed a mid-course maneuver to change to a trans-Lunar trajectory that was not a free return. This retained the safety characteristics of being on a free return upon launch and only departed from free return once the systems were checked out and the lunar module was docked with the command module, providing back-up maneuver capabilities. In fact, within hours after the accident, Apollo 13 used the lunar module to maneuver from its planned trajectory to a free-return trajectory. Apollo 13 was the only Apollo mission to actually turn around the Moon in a free-return trajectory (however, two hours after perilune, propulsion was applied to speed the return to Earth by 10 hours and move the landing spot from the Indian Ocean to the Pacific Ocean).

## Earth–Mars

A free-return transfer orbit to Mars is also possible. As with the Moon, this option is mostly considered for manned missions. Robert Zubrin, in his book

The Case for Mars, discusses various trajectories to Mars for his mission design Mars Direct. The Hohmann transfer orbit can be made free-return. It takes 250 days in the transit to Mars, and in the case of a free-return style abort without the use of propulsion at Mars, 1.5 years to get back to Earth, at a total delta-v requirement of 3.34 km/s. Zubrin advocates a slightly faster transfer, that takes only 180 days to Mars, but 2 years back to Earth in case of an abort. This route comes also at the cost of a higher delta-v of 5.08 km/s. Zubrin claims that even faster routes have a significantly higher delta-v cost and free-return duration (e.g. transfer to Mars in 130 days takes 7.93 km/s delta-v and 4 years on the free return), and thus advocates for the 180-day transfer even if more efficient propulsion systems, that are claimed to enable faster transfers, should materialize. A free return is also the part of various other mission designs, such as Mars Semi-Direct and Inspiration Mars.However it should be noted that, travel duration (to Mars or back to Earth) and delta-v requirement depend on the departure year (eg. 2020 or 2022 or so on). 2-year-free-return means from Earth to Mars (aborted there) and then back to earth all combine total is 2 years (0.5 yrs + 1.5 yrs). If entry corridor to Mars is limited (eg. +/- 0.5 deg entry with <9km/s speed as in the reference), 2-year-return is not possible for some years and for some years, delta-v kick of 0.6km/s to 2.7km/s at Mars may be needed to reach back to Earth.

NASA published the Design Reference Architecture 5.0 for Mars in 2009, advocating a 174-day transfer to Mars, which is close to Zubrin's proposed trajectory. It cites a delta-v requirement of approximately 4 km/s for the trans-Mars injection, but does not mention to the duration of a free return to Earth.

**FREE-RETURN TRAJECTORY**

And if you have a Torchship with an outrageous amount of delta V, you can do a Brachistochrone transfer. This is kind of the opposite of a Hohmann, it is a **maximum** delta V cost / **minimum** transit time trajectory.

You launch whenever you want, none of this "launch window" nonsense. Point the nose of your spacecraft at Mars, burn the engine for 1 gee of acceleration for 1.75 days, do a skew flip to aim your tail at Mars, and burn for 1.75 days of 1 gee deceleration. You get to Mars in 3.5 days flat...

...provided your spacecraft is a torchship that can manage a whopping **2,990,000 meters per second** of delta V!

## Pork-Chop Plots

Expending more deltaV than a Hohmann requires can also allow a ship to depart more often than the Hohmann's limit of one per synodic period, but this is hideously complicated to calculate *(no, I don't know how to do this either, it is called Lambert's problem)*.

Instead of calculating this, you can look it up in graphs called a "pork-chop" plot for a given Hohmann trajectory (so-called because some rocket engineer with an odd sense of humor thought the contour lines looked vaguely like a pork chop).

There is an example of how to use a pork-chop plot here.

If you are lucky you can find them in various NASA documents, though almost all of them are for the Terra-Mars mission. Failing that, there are some on-line calculatiors and stand alone applications that will plot them for you.

Luckily *(for Windows users at least)* there is a Windows program called Swing-by Calculator by http://www.jaqar.com which can calculate all the orbits over a series of dates and export a datafile, which can be imported into Excel, which can then draw the pork chop plot. Full instructions on how to do this is included with the software, which currently is free so long as it is not used for commercial purposes. Unfortunately Swing-By calculator seems to have vanished.

For missions to asteroids you can use NASA's Jet Propulsion Laboratory's online Small-Body Mission-Design Tool.

There is an online calculator called EasyPorkshop. It draws two separate types of plot, Trajectory injection and Orbit Insertion. It cannot draw Total delta-v plots.

Windows users can use Trajectory Optimization Tool by Adam Harden. It also draws two separate types of plot, Trajectory injection and Orbit Insertion. It cannot draw Total delta-v plots.

There is an old-school Windows command-line program with no graphic user interface that only does Earth-Mars porkchop plots here. Go to section **A Computer Program for Creating Pork Chop Plots of Ballistic Earth-to-Mars Trajectories**, and download PDF document, Zipped file of executable program, and JPL DE421 ephemeris binary data file. It is actually written in Fortran, so good luck with that.

**Planetary Transfer Calculator** is an on-line calculator that can create Total delta-v plots plots. Alas, they are somewhat tiny and the scales are not labeled.

Now you have to understand that there are **four** types of pork-chop plots. Remember that every Hohmann trajectory has two propulsion burns:

**TRAJECTORY INJECTION BURN:**burn that pays the delta-V cost to inject the spacecraft into the Hohmann trajectory to the destination planet. The amount of delta-V is called**Velocity**,_{∞s}**Departure V-infinity**or**C3**. In missions it is sometimes called**Trans-**,*{destination planet}*-Injection*e.g.,***TMI**=**Trans-Mars Injection**.**ORBIT INSERTION BURN:**burn that pays the delta-V cost to take the spacecraft out of the Hohmann trajectory and insert it into a circular orbit around the destination planet. The amount of delta-V is called**Velocity**,_{∞d}**Arrival V-infinity**or**V**. In missions it is sometimes called_{∞},*{destination planet}*-Orbit Insertion*e.g.,***MOI**=**Mars Orbit Insertion**.

The total mission delta-V is the sum of the trajectory injection and orbit insertion delta-Vs. This is sometimes called **V _{total}**.

So the four types of pork-chop plots are:

- Plot shows Trajectory Injection delta-V contour lines
- Plot shows Orbit Insertion delta-V contour lines
- Plot shows both Trajectory Injection and Orbit Insertion as two sets of overlapping contour lines
- Plot shows Total Mission delta-V contour lines (i.e, the sum of Trajectory Injection and Orbit Insertion)

**TERRA-MARS PORK CHOP PLOT**

Y-axis is arrival date

Diagonal red lines are transit time scale lines

LEFT: Trajectory Injection delta-V (C3)

MIDDLE: Orbit Insertion delta-V (V_{∞})

RIGHT: Total delta-V (V_{total}), the sum of Trajectory Injection and Orbit Insertion

*click for larger image*

A pork-chop plot shows the delta-V for a given Hohmann mission: starting in an orbit around Planet A and ending up in an orbit around Planet B. Usually the x-axis shows departure date of the spacecraft. The y-axis is either the arrival date of the spacecraft OR it is the *duration* of the trajectory (the transit time). If the y-axis is arrival date, there are sometimes diagonal scale lines displaying transit time.

The delta-V (**C3**, **V _{∞}**, or

**V**) is displayed as a series of contour lines, often colored. This shows as two bullseyes, with the center of one of the bullseyes being the lowest delta-V and the most economical Hohmann transfer.

_{total}For every Departure Date and Arrival Date, you find the intersection of the corresponding x and y axis values, and see what delta-V contour it lies in. This is how much delta-V your spacecraft will need. The center of the bullseye with the lowest delta-V defines the allowed departure and arrival dates if you are on a tight budget. By varying your departure time you can see the deltaV cost of launching at other than the proper synodic period. By varying your arrival time you can see the deltaV cost of shortening the duration of the trip.

For an easy to read explanation of pork-chop plots, check out Hollister Davis' Deboning the Porkchop Plot.

For more than you want to know about pork-chop plots, read On the nature of Earth-Mars porkchop plots.

**TERRA-MARS PORK CHOP PLOT**

Y-axis is arrival date

Red contours are Trajectory Injection delta-V (C3)

Blue contours are Orbit Insertion delta-V (V_{∞})

Diagonal green lines are transit time scale lines

*click for larger image***TERRA-MARS PORK CHOP PLOT**

Y-axis is arrival date

Departures and arrivals are in Modified Julian Day, instead of ordinary dates

Contours are Trajectory Injection delta-V (C3)

Orbit Insertion delta-V (V_{∞}) are not shown, they are in a separate plot

Created by Trajectory Optimization Tool**TERRA-MARS PORK CHOP PLOT**

Y-axis is arrival date

Departures and arrivals are in Modified Julian Day, instead of ordinary dates

Trajectory Injection delta-V (C3) are not shown, they are in a separate plot

Contours are Orbit Insertion delta-V (V_{∞})

Created by Trajectory Optimization Tool**TERRA-PSYCHE PORK CHOP PLOT**

Y-axis is transit duration in either days or years

Departures are in month/year or Modified Julian Day

Green-blue solid contours are Trajectory Injection delta-V (C3)

Red-yellow hollow contours are Orbit Insertion delta-V (V_{∞})

Blue dots are particular missions of interest

In the online version one can hover the mouse over the plot for numeric details

Calculator does not show C3 values that are higher than 150 km^{2}/sec^{2}

Created by JPL Small-Body Mission-Design Tool

*click for larger image***TERRA-MARS PORK CHOP PLOT**

Y-axis is transit duration

Contours are Total delta-V (V_{total})

**TMI:**Trans-Mars Injection (C3)

**MOI:**Mars Orbit Insertion (V_{∞})

MTO: Mars Transfer Orbit (plane change at Terra-Mars node)

MCO: Mars Capture Orbit (plane change at Mars solar orbit-Mars equator node)

Created by Planetary Transfer Calculator**TERRA-MARS PORK CHOP PLOT**

Y-axis is arrival date

Blue contours are Trajectory Injection delta-V (C3)

Red contours are Orbit Insertion delta-V (V_{∞})

Light Blue area is where the total delta-V (V_{total}) is 34 km^{2}/s^{2}or less*(i.e., intersection of C3 30 contour and V*_{∞}4 contour)

Black marks on right from 80° to 240° are Solar Longitude (Ls), the angle ∠*Terra-Sol-Mars*

*click for larger image***TERRA-MARS PORK CHOP PLOT**

Y-axis is transit duration in days

X-axis is departure date in decimal years

Left chart is for Trajectory Injection delta-V (C3)

Right chart is for Orbit Insertion delta-V (V_{∞})

Created by EasyPorkshop

*click for larger image***TERRA-MARS PORK CHOP PLOT**

This has a time scale spanning nine Hohmann launch windows, showing 9 pork chop plots.

*click for larger image*

## Question

NICK T: On a porkchop plot for a given departure date (X axis), why are there two local energy minimums along the arrival date (Y-axis) for a transfer between Earth and Mars? The Hohmann transfer orbit seems to have one optimal solution, but the plot's two minimums seem fairly distinct in time but similar in energy; 15.0 to 15.5 and 15.5 to 16.0 in terms of C3L (something I only vaguely understand; KSP mostly deals in delta-V) (C3L is the delta-V of the Trajectory Injection)

A NASA page about sending spacecraft to Mars refers to "Type 1" and "Type 2" transfer orbits, which clashes with a perhaps simple view that there should be a singular optimal Hohmann transfer orbit; the smallest ellipse tangent to two orbits. Granted, the planets' orbits are not perfectly circular, but the different transit times (red contours) for local minimums differ by almost a factor of two! (about 200 days and about 400 days) I'm not sure if I'm reading the plot correctly though, as there is a disconnect between the given Earth-Mars plot, and the prose on the page as well, which talks about 7 month (~200 day) or 10 month (~300 day) transfers.

## Answer

## What are porkchop plots?

When the goal is to send a spacecraft from the Earth to another planet, it's not enough to reach the target planet's orbit. The vehicle has to meet the target planet itself. The amount of energy needed to accomplish this varies widely depending on departure and arrival dates. A porkchop plot is a graphical interplanetary mission planning tool that depicts as a contour plot the required energy as a function of departure date and arrival date. The energy needed by the launch vehicle is key in determining the feasibility of such a mission. A mission plan that requires more energy than a launch vehicle could possibly provide is not feasible. In addition to feasibility, a porkchop plot aids in planning key mission operations and in planning the optimal trajectory between the two planets. The plot shown in the question and replicated below show launch energy. It does not show the change in energy needed at arrival. Other porkchop plots do show this as a second set of contour lines.

The original post asks three key questions:Before answering the above, it will help to explain how a porkchop plot is constructed.

- Why is there a gap in porkchop plots?
- Why don’t they use an optimal Hohmann transfer?
- Why are there two local energy minima?
## How are porkchop plots constructed?

As a mission planning tool, a porkchop plot makes certain simplifying assumptions with regard to reaching the target planet. Later on, more detailed analyses address those simplifying assumptions. The key simplifying assumptions used in making a porkchop plot are the patched conic approximation and impulsive maneuvers.

These assumptions reduce the problem to one of finding Keplerian orbits about the Sun that take the spacecraft from the vicinity of the Earth to the vicinity of the target planet in the requisite amount of time. Finding such transfer orbits is the subject of Lambert's problem. A number of such transfer orbits might exist. I’ll denote the angle subtended by the departure point, the Sun, and arrival point asθ, with . The principal value of this angle will be between 0° and 180°, inclusive. For now I’ll ignore cases whereθis 0° or 180°. This means that the transfer plane is well-defined and that the number of solutions is finite.

Lambert's problem does not have closed form solutions; a number of iterative techniques have been developed to find solutions. One solution, “the short way”, or “Type 1” transfers, has the change in true anomaly equal toθas described above. Another solution, the “long way”, or “Type 2” transfers, has the change in true anomaly equal to 360°-θ. Other solutions may exist as well. For example, one way to transfer from Earth to Mars in 2.5 years is to make more than a complete orbit during the transfer. Porkchop plots typically only show the Type 1 and Type 2 solutions, and typically only show at most one of these two solutions for a given pair of departure and arrival dates. If one of the two solutions is close to optimal, the other solution will inevitably follow a retrograde path and thus will involve huge expenditures of energy. There’s no reason to show these highly sub-optimal solutions.## Why is there a gap in porkchop plots?

The plot can be cleaned up further by removing cases where the better of the two solutions still involves huge energy expenditures. Huge energy expenditures are obviously going to result when the transfer time is very short or very long. A not so obvious place where this happens is when the angle subtended between the line from the Earth and Sun at departure and the target planet and Sun at arrival is nearly 180°. That the Earth and target planet have slightly different orbital planes means that the transfer plane will be nearly orthogonal to the planetary orbital planes when the transfer angle is close to but not equal to 180°. This makes the approach of having a maneuver at departure and a maneuver at arrival extremely expensive for those transfers that are close to 180°. Removing those very expensive transfers from view is what creates the gap in the porkchop plot.

This excessive cost for near 180° transfers is to some extent an artifact of the approach used to create a porkchop plot. Adding a third maneuver enables the use of much smaller in-plane maneuvers at the start and end, with a small plane change somewhere along route. There’s a problem, however. This mid-course plane change would necessarily mean thrust from the spacecraft itself. This is undesirable. The spacecraft itself provides very little of the energy with the two burn approach. The energy for the Earth departure comes from the launch vehicle, and in the case of Mars, most of the energy for Mars arrival comes from aerobraking. It’s better to fold that plane change into the maneuvers at departure and arrival so as to keep the thrust needed by the spacecraft down to a minimum.## Why don’t they use an optimal Hohmann transfer?

An optimal Hohmann transfer doesn’t exist. Hohmann transfers are in-plane maneuvers that transfer from one circular orbit to another that share a common orbital plane. Planetary orbits are slightly inclined with respect to one another and are elliptical rather than circular. Generalizing the concept of a Hohmann transfer to that of a 180° transfer, in most cases that 180° transfer doesn’t exist. When it does, that the orbits are not coplanar and that the orbits are not circular means that this 180° transfer is no longer optimal.## Why are there two local energy minima?

There aren’t just two local energy minima. There are a countable infinite number of local minima. A porkchop plot only shows the first two. The other solutions take even longer than Type 1 and Type 2 transfers and are more sensitive to errors.

**WHY IS THERE A GAP IN PORKCHOP PLOTS?**

*(2014)*

Changing direction causesΔV(change in velocity), often more than a change in speed. Compare the velocity vectors below. When going the same direction, the difference is 1 km/s. When at right angles the difference is 5 km/s. We know this from driving in traffic. Two cars going almost the same speed hit each other. If they’re in the same lane going the same direction, it’s a mild bump. If one car runs a red light and T-bones a car in cross traffic, the impact is serious:

This is the strength of a Hohmann transfer orbit. Velocity vectors are pointing the same direction at departure as well as destination. No direction change is needed, only a speed change:

Note the Hohmann transfer path moves 180 degrees about the sun:

A Hohmann transfer assumes the departure and destination orbits are co-planar. But what if the destination orbit is inclined?## Orbit Planes and Spherical Trigonometry

A plane passing through a sphere’s center cuts the sphere along a great circle. A group of planes all sharing a common point can be represented as great circles on a sphere. Since every orbit about the sun is a conic section having the sun as a focus, each orbital plane shares the sun as a common point. Representing the orbital planes as great circles is convenient. There are already a lot of theorems in spherical trigonometry which gives us a suite of tools for looking at angles between orbital planes.

The shortest path (or geodesic) along a spherical surface between two points is an arc of a great circle. If we set the sphere’s radius to be 1, the arc length is also the angular separation in radians.

A familar group of great circles are the longitude lines on a globe. The equator is the only great circle among the latitude lines. All the longitude lines are great circles passing through the poles.

Let’s use the equatorial great circle to represent the departure plane. Recall the Hohmann transfer moves 180 degrees about the center. In this illustration, latitude and longitude for departure and destination is (0°, 0°) and (7°, 180°). The only great circle connecting these points is a polar orbit nearly 90° from the departure and destination planes! Big plane changes at departure and destination destroys the virtue of a Hohmann orbit.

I’ve also tried to demonstrate this in this video:

The big delta V needed for large plane changes makes the ridge (gap between the two blobs) in a porkchop plot:

Porkchop plots are drawn by doing iterations of various Lambert Space Triangles. Lambert iterations give polar transfer orbits when departure and destination longitudes differ by 180°.

Does this mean Hohmann transfers are no good if the destination orbit’s inclined? No, the big plane changes can be avoided with a mid course plane change. Here is a broken plane transfer where a plane change burn is done at the ascending node:

The line where the destination and departure planes intersect form the ascending and descending nodes. Starting in the departure plane and doing a plane change at the node avoids the two major plane changes. The departure and destination planes differ by an angle calledi, for inclination.

Changing a vector by an angleitakes dv ofv * 2 * sin(i/2).

The Vis Viva Equation tells usv = sqrt(μ(2/r - 1/a)). So v ranges fromsqrt (μ((1-e)/(a(1+e))))at aphelion tosqrt (μ((1+e)/(a(1-e))))at perihelion. Let's look at a Ceres transfer orbit. An ellipse with a 1.88 a.u. semi major axis and eccentricity 0.47 will have speeds ranging from 36 km/s (at perihelion) to 13 km/s (at apohelion). Inclination's about 10.6 degrees. So plane change ranges from36 km/s * 2 * sin(10°/2)to13 km/s * 2 * sin(10°/2)or from 6.7 to 2.4 km/s. Is the a 2.4 km/s plane change at aphelion the best we can do? No, it's possible to have less plane change expense.

Launch is at the perihelion of an outbound Hohmann orbit. If the launch coincides with a node, the entire plane change can be done during earth departure or at arrival. Then the delta V entails a speed change as well as a direction change. Doing a single plane change/speed change burn saves delta V as shown by this diagram:

Law of cosines tells us for a triangle a, b, c,a. In this case,^{2}+ b^{2}- 2ab cos(i) = c^{2}iis the angle between a and b and c is the delta V needed from the combined plane change and speed change.

At aphelion, a combined speed change/plane change only costs 0.76 km/s more than the speed change alone.

When launching deep in earth’s gravity well, we enjoy an Oberth benefit. Ceres' gravity well lends a little Oberth benefit at the destination. If the line of nodes coincides with transfer orbit's line of apsides, plane change can cost as little as 0.52 km/s extra.

This indicates as much plane change as possible should be made at departure and arrival. What sort of plane changes should we make to minimize the angle of the midcourse plane change?

The fattest part of an orange slice is right in the middle:

The angular separation at launch has to be some part of the orange slice. To minimize the angle between transfer plane and destination plane, the angular separation at launch should be in the middle. Having the transfer plane intersect the destination plane 90° from launch minimizes plane change angle.

An object on an elliptical path moves slower as it moves further from the sun, so doing plane changes further out are cheaper. The 90º from launch is a minimum. There will be a larger plane change angle 100 degrees from launch, but velocity will be slower. Also plane change lessens as flight path angle increases. I hope to talk about this more when I have time.

But for now I believe this shows that the Lambert iterations greatly exaggerates plane change expense for a Hohmann path where departure and destination points are 180° degrees apart. Most of that plane change expense can be eliminated by choosing a good place to do a midcourse plane change

A PDF on Broken Plane Maneuvers Fernando Abilleria of NASA Jet Propulsion Laboratory

**DEBONING THE PORKCHOP PLOT**by Hollister David

*(2013)*

*This would have been a good spot for a pork-chop plot*

from Star Trek TOS episode "The Cage"

### Using Pork-Chop Plots

*Actual Pork chop from NASA. Yes, I know this graph only shows transit injection delta-V, not total delta-V, but it will do for purposes of illustration.*

The important parts to an SF author are C3l[blue] = delta-V for the launch, and TTIME[red] = mission duration.

If anybody cares, SEP[green] = Sun-Earth-Probe angle (if too small solar static drowns out probe signal) and Ls[magenta] = Earth-Sun-Mars angle (angle made by drawing a line from Earth to the Sun then to Mars). Click for larger image.

The captain of the spacecraft will ask the astrogator for a mission plan to travel from planet **A** to planet **B** in trip time **T**. The astrogator will determine a family of mission plans, with the current ship's delta-V capacity as the upper limit *(or the ship will not be capable of performing that mission)* and with the captain's specfied trip duration time as the lower limit *(or the captain will be unhappy)*. You see, a Hohmann trajectory generally uses the least delta-V, but also has the longest possible mission time, and the mission can only start on specific dates *("launch windows")* as well. By increasing the delta-V used the launch window can be altered and the mission time can be reduced.

What the astrogator will do is have the navigation computer draw a pork-chop plot, which is a graph with departure times on one axis, arrival times on the other axis, and total delta-V requirements drawn as contour lines in the graph. Cross out the areas of delta-V that are too high for the spacecraft to manage, cross out the part of the graph with a mission duration that is too long to suit the captain, and what remains are the possible missions.

If it turns out there is *no* possible mission inside the stated parameters, the astrogator will have to confer with the captain over what *is* possible.

The

Polarisis currently onTerrain the far-flung future time ofJune 2005. Captain Strong tells astrogator Roger Mannings that he wants a mission plan for the Polaris to travel toMars. He does not want the transit time to beover 175 days, and the delta-V cost should bebelow 22,500 meters per second(22.5 km/s).

artwork by Louis S. Glanzman

First Roger has the ship's brain calculate and print out a pork-chop plot for Terra-to-Mars around the current time Roger eliminates the part of the plot with transit times over 175 days (the gray area above the red 175 day line)

Detail of the allowed part of the plot. Of course, any launch dates that are prior to the current date will also have to be eliminated (unless you have a time machine). Roger then eliminates the part of the plot with a delta-V requirement over 22.5 km/s (the green area outside the 22.5 contour line)

The white area shows the family of mission plans that fit Captain Strong's specifications. This will give the captain a range of options that can be further narrowed down by optimising for desired launch windows, transit times, etc. For instance, the minimum transit time is at the yellow circle. It will require the full 22.5 km/s delta-V, has a launch window of August 19, 2005, arrival date of January 05, 2006, and have a transit time of 140 days.

## Interplanetary Transport Network

Actually there is a type of transfer orbit that requires even less deltaV than a Hohmann, the so-called "Interplanetary Transport Network" However, this transfer's practicality is questionable, for a manned mission at least. On the plus side it requires exceedingly small amounts of deltaV. On the minus side, as one would expect, it is so slow it makes a Hohmann look like a hypersonic bullet train. A Hohmann can travel from Earth orbit to Lunar orbit in a few days, the Interplanetary Transport Network takes ** two months**.

This was developed for uncrewed space probes who didn't have to worry about dragging along months of life support supplies.

The

Interplanetary Transport Network(ITN) is a collection of gravitationally determined pathways through the Solar System that require very little energy for an object to follow. The ITN makes particular use of Lagrange points as locations where trajectories through space are redirected using little or no energy. These points have the peculiar property of allowing objects to orbit around them, despite lacking an object to orbit. While they use little energy, the transport can take a very long time. Shane Ross has said "Due to the long time needed to achieve the low energy transfers between planets, the Interplanetary Superhighway is impractical for transfers such as from Earth to Mars at present## History

Interplanetary transfer orbits are solutions to the gravitational "restricted three-body problem", which, for the general case, does not have exact solutions, and is addressed by numerical analysis approximations. However, a small number of exact solutions exist, most notably the five orbits referred to as "Lagrange points", which are orbital solutions for circular orbits in the case when one body is significantly more massive.

The key to discovering the Interplanetary Transport Network was the investigation of the nature of the winding paths near the Earth-Sun and Earth-Moon Lagrange points. They were first investigated by Jules-Henri Poincaré in the 1890s. He noticed that the paths leading to and from any of those points would almost always settle, for a time, on an orbit about that point. There are in fact an infinite number of paths taking one to the point and away from it, and all of which require nearly zero change in energy to reach. When plotted, they form a tube with the orbit about the Lagrange point at one end.

The derivation of these paths traces back to mathematicians Charles C. Conley and Richard P. McGehee in 1968.

Hiten, Japan's first lunar probe, was moved into lunar orbit using similar insight into the nature of paths between the Earth and the Moon. Beginning in 1997, Martin Lo, Shane D. Ross, and others wrote a series of papers identifying the mathematical basis that applied the technique to the Genesis solar wind sample return, and to Lunar and Jovian missions. They referred to it as an Interplanetary Superhighway (IPS).## Paths

As it turns out, it is very easy to transit from a path leading to the point to one leading back out. This makes sense, since the orbit is unstable, which implies one will eventually end up on one of the outbound paths after spending no energy at all. Edward Belbruno coined the term "weak stability boundary" or "fuzzy boundary" for this effect.

With careful calculation, one can pick

whichoutbound path one wants. This turned out to be useful, as many of these paths lead to some interesting points in space, such as the Earth's Moon or between the Galilean moons of Jupiter. As a result, for the cost of reaching the Earth–Sun L_{2}point, which is rather low energy value, one can travel to a number of very interesting points for a little or no additional fuel cost. But the trip from Earth to Mars or other distant location would likely take thousands of years.The transfers are so low-energy that they make travel to almost any point in the Solar System possible. On the downside, these transfers are very slow. For trips from Earth to other planets, they are not useful for manned or unmanned probes, as the trip would take many generations. Nevertheless, they have already been used to transfer spacecraft to the Earth–Sun L

_{1}point, a useful point for studying the Sun that was employed in a number of recent missions, including the Genesis mission, the first to return solar wind samples to Earth. The network is also relevant to understanding Solar System dynamics; Comet Shoemaker–Levy 9 followed such a trajectory on its collision path with Jupiter.## Further explanation>

The ITN is based around a series of orbital paths predicted by chaos theory and the restricted three-body problem leading to and from the unstable orbits around the Lagrange points – points in space where the gravity between various bodies balances with the centrifugal force of an object there. For any two bodies in which one body orbits around the other, such as a star/planet or planet/moon system, there are three such points, denoted L

_{1}through L_{3}. For instance, the Earth–Moon L_{1}point lies on a line between the two, where gravitational forces between them exactly balance with the centrifugal force of an object placed in orbit there. For two bodies whose ratio of masses exceeds 24.96, there are two additional stable points denoted as L_{4}and L_{5}. These five points have particularly low delta-v requirements, and appear to be the lowest-energy transfers possible, even lower than the common Hohmann transfer orbit that has dominated orbital navigation in the past.Although the forces balance at these points, the first three points (the ones on the line between a certain large mass, e.g. a star, and a smaller, orbiting mass, e.g. a planet) are not stable equilibrium points. If a spacecraft placed at the Earth–Moon L

_{1}point is given even a slight nudge towards the Moon, for instance, the Moon's gravity will now be greater and the spacecraft will be pulled away from the L_{1}point. The entire system is in motion, so the spacecraft will not actually hit the Moon, but will travel in a winding path, off into space. There is, however, a semi-stable orbit around each of these points, called a halo orbit. The orbits for two of the points, L_{4}and L_{5}, are stable, but the halo orbits for L_{1}through L_{3}are stable only on the order of months.In addition to orbits around Lagrange points, the rich dynamics that arise from the gravitational pull of more than one mass yield interesting trajectories, also known as low energy transfers. For example, the gravity environment of the Sun–Earth–Moon system allows spacecraft to travel great distances on very little fuel, albeit on an often circuitous route.

## Missions

Launched in 1978, the ISEE-3 spacecraft was sent on a mission to orbit around one of the Lagrange points. The spacecraft was able to maneuver around the Earth's neighborhood using little fuel by taking advantage of the unique gravity environment. After the primary mission was completed, ISEE-3 went on to accomplish other goals, including a flight through the geomagnetic tail and a comet flyby. The mission was subsequently renamed the International Cometary Explorer (ICE).

The first low energy transfer using what would later be called the ITN was the rescue of Japan's

Hitenlunar mission in 1991. Another example of the use of the ITN was NASA's 2001–2003 Genesis mission, which orbited the Sun–Earth L_{1}point for over two years collecting material, before being redirected to the L_{2}Lagrange point, and finally redirected from there back to Earth. The 2003–2006SMART-1of the European Space Agency used another low energy transfer from the ITN. In a more recent example, the Chinese spacecraft Chang'e 2 used the ITN to travel from lunar orbit to the Earth-Sun L_{2}point, then on to fly by the asteroid 4179 Toutatis.

**INTERPLANETARY TRANSPORT NETWORK**

Wikipedia describes the Interplanetary Transport Network as "… pathways through the Solar System that require very little energy for an object to follow." See this Wikipedia article. They also say "While they use very little energy, the transport can take a very long time."

Low energy paths that take a very long time? I often hear this parroted in space exploration forums and it always leaves me scratching my head.

The lowest energy path I know of to bodies in the inner solar system is the Hohmann orbit. Or if the destination is noticeably elliptical, a transfer orbit that is tangent to both the departure and destination orbit. Although I think of bitangential transfer orbits as a more general version of the Hohmann orbit.

BITANGENTIAL TRANSFER ORBIT

The transfer orbit is tangent to both departure and destination orbit.

The Hohmann transfer is the special case where departure and destination orbits are circular.

Illustration from my pdf on tangent orbits.In the case of Mars, a bitangential orbit is 8.5 months give or take a month or two. Is there a path that takes a lot longer and uses almost no energy? I know of no such path.

## L1 and L2

The interplanetary Superhighway supposedly relies on weak stability or weak instability boundaries between L1 and/or L2 regions. Here is an online text on 3 body Mechanics and their use in space mission design. The authors are Koon, Lo, Marsden and Ross. Shane Ross is one of the more prominent evangelists spreading the gospel of the Interplanetary Super Highway.

The focus of this online textbook is the L1 and L2 regions. From page 10:

L1 and L2 are necks between realms. In the above illustration the central body is the sun, and orbiting body Jupiter. L1 and L2 are necks or gateways between three realms: the Sun realm, the Jupiter Realm and the exterior realm.

Travel between these realms can be accomplished by weak stability or weak instability boundaries that emanate from L1 or L2. From page 11 of the same textbook:

## My terms for various Lagrange necks

First letter is the central body, the second letter is the orbiting body.

Earth Moon L1: EML1

Earth Moon L2: EML2

Sun Earth L1: SEL1

Sun Earth L2: SEL2

Sun Mars L1: SML1

Sun Mars L2: SML2Since I'm a lazy typist that is what I'll use for the rest of this post.

## EML1 and 2

I am very excited about the earth-moon Lagrange necks. They've been prominent in many of my blog posts. Here's a post entirely devoted to EML2.

EML1 and 2 are about 5/6 and 7/6 of a lunar distance from earth:

Both necks move at the same angular velocity as the moon. So EML1 moves substantially slower than an ordinary earth orbit would at that altitude. EML2 moves substantially faster.

It takes only a tiny nudge and send objects in these regions rolling about the slopes of the effective potential hills. Outside of the moon's influence they tend to fall into ordinary two body ellipses (for a short time).

Here's the ellipse an object moving at EML1 velocity and altitude would follow if the moon weren't there:

An object nudged earthward from EML will fall into what I call an olive orbit.

It's approximately 100,000 x 300,000 km.In practice an EML1 object nudged earthward will near the moon on the fifth apogee. If coming from behind, the moon's gravitational tug can slow the object which lowers perigee.

Here is an orbital sim where the moon's influence lowered perigee four times:

I've run sims where repeated lunar tugs have lowered perigees to atmosphere grazing perigees. Once perigee passes through the upper atmosphere, we can use aerobraking to circularize the orbit.

Orbits are time reversible. Could we use the lunar gravity assists to get from LEO to higher orbits? Unfortunately, aerobraking isn't time reversible. The atmosphere can't increase orbital speed to achieve a higher apogee. And low earth orbit has a substantially different Jacobi constant than those orbits dwelling closer to the borders of a Hill Sphere.

So to get to the lunar realm, we're stuck with the 3.1 km/s LEO burn needed to raise apogee. But once apogee is raised, many doors open.

There are low energy paths that lead from EML1 to EML2. EML2 is an exciting location.

Without the moon's influence, an object at EML2's velocity and altitude would fly to an 1,800,000 km apogee. This is outside of earth's Hill Sphere! In the above illustration I have an apogee beyond SEL2. But by timing the release from EML2, we could aim for other regions of the Hill Sphere, including SEL1.

Here is a sim where slightly different nudges send payloads from EML2:

See how the sun bends the path as apogee nears the Hill Sphere? From EML2 there are a multitude of wildly different paths we can choose. In this illustration I like pellet #3 (orange). It has a very low perigee that is moving about 10.8 km/s. And it got to this perigee with just a tiny nudge from EML2. Pellet # 4 is on it's way to a retrograde earth orbit. Most of the other pellets are saying good bye to earth's Hill Sphere.

I am enthusiastic about using EML1 and EML2 as hubs for travel about the earth-moon neighborhood. But a little less excited about travel about the solar system.

## We've left Earth's Hill Sphere. Now what?

Recall that EML1 and 2 are ~5/6 and 7/6 of a lunar distance from the earth. SEL1 and 2 are much less dramatic: 99% and 101% of an A.U. from the sun. Objects released from these locations don't vary much from earth's orbit:

Running orbital sims gets pretty much the same result pictured above.

Mars is even worse:

Are there weak instability boundary transfers leading from SEL2 to SML1? I don't think this particular highway exists.

To get a 1.52 A.U. aphelion, we need a departure Vinfinity of 3 km/s. To be sure EML2 can help us out in achieving this Vinfinity. In other words we could use lunar assists to depart on a Hohmann orbit. But a Hohmann orbit is different from the tube of weak instability boundaries we're led to imagine.

And once we arrive at a 1.52 aphelion. we have an arrival 2.7 km/s Vinfinity we need to get rid of.

Pass through SML1 at 2.7 km/s and you'll be waving Mars goodbye. The Lagrange necks work their mojo on near parabolic orbits. And an earth to Mars Hohmann is decidedly hyperbolic with regard to Mars.

What about Phobos and Deimos? The Martian moons are too small to lend a helpful gravity assist. We need to get rid of the 2.7 km/s Vinf and neither SML1 nor the moons are going to do it for us.

## Mars ballistic capture by Belbruno & Toppotu

Edward Belbruno is another well known evangelist for the Interplanetary Superhighway (though he likes to call them ballistic captures). Belbruno cowrote this pdf on ballistic Mars capture.

Here is a screen capture from the pdf:

The path from Earth@Departure to Xc is pretty much a Hohmann transfer. In fact they assume the usual departure for Mars burn. Arrival is a little different. They do a 2 km/s heliocentric circularization burn at Xc (which is above Mars' perihelion). This particular path takes an extra year or so to reach Mars.

So they accomplish Mars capture with a 2 km/s arrival burn. At first glance this seems like a 0.7 km/s improvement over the 2.7 km/s arrival Vinf.

Or it seems like an advantage to those unaware of the Oberth benefit. If making the burn deep in Mars' gravity well, capture can be achieved for as little as 0.7 km/s.

Comparing capture burns it's 2 km/s vs 0.7 km/s. So what do we get for an extra year of travel time? 1.3 km/s flushed down the toilet!

## What About Ion Engines?

"What about ion engines?" a Belbruno defender might object. "They don't have the thrust to enjoy an Oberth benefit. So Belbruno's 0.7 km/s benefit is legit if your space craft is low thrust."

Belbruno & friends are looking at a trip from a nearly zero earth C3 to a nearly zero Mars C3. In other words from the edge of one Hill Sphere to another.

So to compare apples to apples I'll look at a Hohmann from SEL2 to SML1. I want to point out I'm not using Lagrange necks as key holes down some mysterious tube. They're simply the closest parts of neighboring Hill Spheres.

"Wait a minute..." says Belbruno's defender, "We're talking Hall thrusters. So no Hohmann ellipse, but a spiral."

Low earth orbit moves about 4° per minute. So a low-thrust burn lasting days does indeed result in a spiral. But Earth's heliocentric orbit moves about a degree per day while Mars' heliocentric orbit moves about half a degree per day. At this more leisurely pace, a 4 or 5 day burn looks more an impulsive burn. The transfer between Hill Spheres is more Hohmann-like than the spiral out of earth's gravity well.

Instead of a 1 x 1.524 AU orbit, the new Hohmann is a 1.01 x 1.517 AU ellipse. The new Hohmann's perihelion is a little slower, the new aphelion a little faster.

Moreover, SEL2 moves at the same angular velocity as earth. So it's speed is about 101% earth's speed. Likewise SML1 moves at about 99.6% Mars' speed.

With this revised scenario, aphelion rendezvous delta V is now more like 2.4 km/s. Still, Belbruno's 2 km/s capture burn saves 0.4 km/s.

0.4 km/s is better than chopped liver, right? Well, recall ion engines with very good I

_{SP}. I'll look at an exhaust velocity of of 30 km/s.e^{2.4/30}- 1 = 0.083

e^{2/30}- 1 = 0.069So given a 100 tonne payload, rendezvous xenon is 8.3 tonnes for Hohmann vs 6.9 tonnes for Belbruno's ballistic capture.

108.3/106.9 = 1.0134We're adding a year to our trip time for a one percent mass improvement? Sorry, I don't see this a great trade-off.

## Summary

The virtually zero energy looooong trips between planets are an urban legend.

I'll be pleasantly surprised if I'm wrong. To convince me otherwise, show me the beef. Show me the zero energy trajectory from an earth Lagrange neck to a Mars Lagrange neck.

Until then I'll think of this post as a dose of Snopes for space cadets.

I'd like to thank Mike Loucks and John P. Corrico Jr. I've held these opinions for awhile but didn't have the confidence to voice them. Who am I but an amateur with no formal training? But talking with these guys I was pleasantly surprised to find some of my heretical views were shared by pros. Without their input I would not have had the guts to publish this post.

**POTHOLES ON THE INTERPLANETARY SUPERHIGHWAY**by Hollister David

*(2015)*

## Tourist Season

*artwork by Bob Eggleton*

Due to spacecraft taking advantage of Hohmann transfers, they will tend to arrive all at the same time at the destination planet, stay until the launch window for Hohmann transfer back to Terra, and be absent for the many months before the next Hohmann timed arrival. In other words, Mars will have a "tourist season" and an "off season". I use the word "tourist" but this actually means *"anybody traveling or shipping anything to Mars who wants to avail themselves of the reduced delta V cost of Hohmann transfer."*

The ships in transit will tend to be in a relatively compact group. **Clever operators will have special ships in the group: not to travel to Mars but to do business with the other ships in the group (with an eye to making lots of money).** Things like being an interplanetary 7-11 all night convenience store, selling those vital little necessities (that you forgot to pack) at inflated prices. A fancy restaurant spaceship for when you are truly fed up with eating those nasty freeze-dried rations. A space-going showboat for outer space riverboat gambling. An expensive health clinic. A flying bar with a wide variety of vacuum-distilled liquors (anybody for a Pan Galactic Gargle Blaster?). Not to mention a orbital brothel. Fans of TOS Battlestar Galactica will be reminded of the *Rising Star*, luxury liner and casino in space.

It might be possible to make an Aldrin Cycler into such an enterprise, but the timing would be tricky.

For the Martian tourist season:

- At Terra, Hohmann launch window to Mars happens every 2.17 years (26 months). Tourists ship launch into Hohmann trajectory.
- Tourist ships spend 0.70873 years (8.5 months) in transit to Mars. Convenience ships do a booming business.
- Tourist ships arrive at Mars. Start of the Martian tourist season
- 1.25 years (15.3 months or 459 days) after tourist ships arrive, Hohmann window to Terra opens. Departure of tourist ships and end of Martian tourist season.
- Tourist ships spend 0.70873 years (8.5 months) in transit to Terra. Convenience ships do a booming business.

Due to the way the Hohmann windows overlap, the Martian tourist season will be 1 year 3 months and 7 days long, and the Martian off season will be 8 months and 15 days long. As with any seasonal place, during tourist season the prices of anything tourist related will be inflated.

Year | Event |
---|---|

0.00 | Terra ⇒ Mars launch window opens. Tourist fleet Alfa departs Terra. |

0.71 | Tourist fleet Alfa arrives at Mars. End of Martian off season, start of Martian tourist season. |

1.98 | Mars ⇒ Terra launch window opens. Tourist fleet Alfa departs Mars. End of Martian touist season, start of Martian off season. |

2.17 | Terra ⇒ Mars launch window opens. Tourist fleet Bravo departs Terra. |

2.69 | Tourist fleet Alfa arrives at Terra. |

2.88 | Tourist fleet Bravo at Mars. End of Martian off season, start of Martian tourist season. |

4.15 | Mars ⇒ Terra launch window opens. Tourist fleet Bravo departs Mars. End of Martian touist season, start of Martian off season. |

4.33 | Terra ⇒ Mars launch window opens. Tourist fleet Charlie departs Terra. |

4.86 | Tourist fleet Bravo arrives at Terra. |

*"Tourist Season" by John Pederson, Jr.*

In the meantime he had another worry; strung out behind him were several more ships, all headed for Mars. For the next several days there would be frequent departures from the Moon, all ships taking advantage of the one favorable period in every twenty-six months when the passage to Mars was relatively 'cheap', i.e., when the minimum-fuel ellipse tangent to both planet's orbits would actually make rendezvous with Mars rather than arrive foolishly at some totally untenanted part of Mars' orbit. Except for military vessels and super expensive passenger-ships, all traffic for Mars left at this one time.

During the four-day period bracketing the ideal instant of departure ships leaving Leyport paid a fancy premium for the privilege over and above the standard service fee. Only a large ship could afford such a fee; the saving in cost of single-H reactive mass had to be greater than the fee. The Rolling Stone had departed just before the premium charge went into effect; consequently she had trailing her like beads on a string a round dozen of ships, all headed down to Earth, to tack around her toward Mars.

Hazel looked them over. 'Mr d'Avril, don't you have something a bit larger?'

'Well, yes, ma'am, I do — but I hate to rent larger ones to such a small family with the

tourist seasonjust opening up: I'll bring in a cot for the youngster.'(ed note: The "tourist season" on Mars starts with the earliest arrival time of a spacecraft on a Terra-Mars Hohmann trajectory)

'See here, I don't want to buy this du — this place. I just want to use it for a while.'

Mr d'Avril looked hurt. 'You needn't do either one, ma'am. With ships arriving every day now I'll have my pick of tenants. My prices are considered very reasonable. The Property Owner's Association has tried to get me to up 'em — and that's a fact'

Hazel dug into her memory to recall how to compare a hotel price with a monthly rental — add a zero to the daily rate; that was it Why, the man must be telling the truth! — if the hotel rates she had gotten were any guide. She shook her head. 'I'm just a country girl, Mr d'Avril. How much did this place cost to build?'

Again he looked hurt 'You're not looking at it properly, ma'am. Every so often we have a big load of tourists dumped on us. They stay awhile, then they go away and we have no rent coming in at all. And you'd be surprised how these cold nights nibble away at a house. We can't build the way the Martians could.'

Hazel gave up. 'Is that season discount you mentioned good from now to Venus departure?'

'Sorry, ma'am. It has to be the whole season.' The next favorable time to shape an orbit for Venus was ninety-six Earth-standard days away — ninety-four Mars days — whereas the 'whole season' ran for the next fifteen months, more than half a Martian year before Earth and Mars would again be in a position to permit a minimum-fuel orbit.

**THE ROLLING STONES**by Robert Heinlein

*(1952)*

It was true. Even in the bars that catered to inner planet types, the mix was rarely better than one Earther or Martian in ten (Belters). Squinting out at the crowd, Miller saw that the short, stocky men and women were nearer a third.

"Ship come in?" he asked.

"Yeah."

"EMCN?" he asked. The Earth-Mars Coalition Navy often passed through Ceres on its way to Saturn, Jupiter, and the stations of the Belt,

but Miller hadn't been paying enough attention to the relative position of the planets to know where the orbits all stood.

**LEVIATHAN WAKES**by "James S.A. Corey" (Daniel Abraham and Ty Franck) 2011

First novel of The Expanse

## A Working Example

Stage | Delta-v (m/s) |
---|---|

Terra liftoff and insertion into Hohmann transfer to Mars | 14,070 |

Mars landing | 5030 |

Mars liftoff and insertion into Hohmann transfer to Terra | 7520 |

Terra landing | 12,908 |

Total | 39,528 |

Stage | Delta-v (m/s) |
---|---|

Terra liftoff | 12,908 |

Hohmann to Mars | 5590 |

Mars landing | 5030 |

Mars liftoff | 5030 |

Hohmann to Terra | 5590 |

Terra landing | 12,908 |

Total | 47,056 |

Solar Guard cruiser ** Polaris** needs a deltaV of at least 47,056 m/s in order to perform the mission specified in Table 1.

If the propulsion system has enough acceleration to achieve the Hohmann deltaV while still close to the planet it lifted off from, the total deltaV requirements can be reduced. Doing the liftoff and the Hohmann insertion as one long burn does this. Ordinarily the totals of the liftoff and Hohmann deltaVs are simply added together. If done as one long burn, it will be:

**Δ _{vTotal} = sqrt( Δ_{vLiftoff}^{2} + Δ_{vHohmann}^{2})**

For instance, instead of Mars Liftoff and Hohmann being 5030 + 5590 = **10620** it will be sqrt( 5030^{2} + 5590^{2} ) = **7520**.

How does this work? Well, it is an example of the Oberth effect (see below). Doing one long burn ensures that more of your propellant is expended low in the gravity well. And in case you are wondering, multi-stage rockets count as "one long burn," even though there is a small interrupting between stages.

Therefore, from Table 2, the ** Polaris** needs a deltaV of at least 39,528 m/s in order to perform the mission specified.

8.6 months one way is pretty pathetic. Of course spending more deltaV can decrease the time.

Much easier of course is to examine a Pork Chop plot from Swing By Calculator. You can see from the left plot below how it reaches the point of diminishing returns quite quickly.

If you want to cheat, you can look up some of the missions in Jon Roger's Mission Table.

*Pork chop by Winchell Chung jr.**Actual Pork chop from NASA. The important parts to an SF author are C3l[blue] = delta-V for the launch, and TTIME[red] = mission duration. If anybody cares, SEP[green] = Sun-Earth-Probe angle (if too small solar static drowns out probe signal) and Ls[magenta] = Earth-Sun-Mars angle (angle made by drawing a line from Earth to the Sun then to Mars). Click for larger image.*

## Sample Delta-V Budgets

- Launch from Terra's surface to LEO—this not only requires an increase of velocity from 0 to 7.8 km/s, but also typically 1.5–2 km/s for atmospheric drag and gravity drag
- Re-entry from LEO—the delta-v required is the orbital maneuvering burn to lower perigee into the atmosphere, atmospheric drag takes care of the rest.

### Stationkeeping

Maneuver | Average delta-v per year [m/s] | Maximum per year [m/s] |
---|---|---|

Drag compensation in 400–500 km LEO | < 25 | < 100 |

Drag compensation in 500–600 km LEO | < 5 | < 25 |

Drag compensation in > 600 km LEO | < 7.5 | |

Station-keeping in geostationary orbit | 50–55 | |

Station-keeping in L_{1}/L_{2} | 30–100 | |

Station-keeping in lunar orbit | 0–400 | |

Attitude control (3-axis) | 2–6 | |

Spin-up or despin | 5–10 | |

Stage booster separation | 5–10 | |

Momentum-wheel unloading | 2–6 |

### Terra–Luna space

Delta-v needed to move inside Terra–Luna system (speeds lower than escape velocity) are given in km/s. This table assumes that the Oberth effect is being used—this is possible with high thrust chemical propulsion but not with current (As of 2011) electrical propulsion.

The return to LEO figures assume that a heat shield and aerobraking/aerocapture is used to reduce the speed by up to 3.2 km/s. The heat shield increases the mass, possibly by 15%. Where a heat shield is not used the higher from LEO Delta-v figure applies, the extra propellant is likely to be heavier than a heat shield. LEO-Ken refers to a low earth orbit with an inclination to the equator of 28 degrees, corresponding to a launch from Kennedy Space Center. LEO-Eq is an equatorial orbit.

∆V km/s from/to | LEO-Ken | LEO-Eq | GEO | EML-1 | EML-2 | EML-4/5 | LLO | Luna | C3=0 |
---|---|---|---|---|---|---|---|---|---|

Terra | 9.3–10 | ||||||||

Low Earth orbit (LEO-Ken) | 4.24 | 4.33 | 3.77 | 3.43 | 3.97 | 4.04 | 5.93 | 3.22 | |

Low Earth orbit (LEO-Eq) | 4.24 | 3.90 | 3.77 | 3.43 | 3.99 | 4.04 | 5.93 | 3.22 | |

Geostationary orbit (GEO) | 2.06 | 1.63 | 1.38 | 1.47 | 1.71 | 2.05 | 3.92 | 1.30 | |

Lagrangian point 1 (EML-1) | 0.77 | 0.77 | 1.38 | 0.14 | 0.33 | 0.64 | 2.52 | 0.14 | |

Lagrangian point 2 (EML-2) | 0.33 | 0.33 | 1.47 | 0.14 | 0.34 | 0.64 | 2.52 | 0.14 | |

Lagrangian point 4/5 (EML-4/5) | 0.84 | 0.98 | 1.71 | 0.33 | 0.34 | 0.98 | 2.58 | 0.43 | |

Low lunar orbit (LLO) | 1.31 | 1.31 | 2.05 | 0.64 | 0.65 | 0.98 | 1.87 | 1.40 | |

Luna | 2.74 | 2.74 | 3.92 | 2.52 | 2.53 | 2.58 | 1.87 | 2.80 | |

Terra escape velocity (C3=0) | 0.00 | 0.00 | 1.30 | 0.14 | 0.14 | 0.43 | 1.40 | 2.80 |

### Terra–Luna space—low thrust

Current electric ion thrusters produce a very low thrust (milli-newtons, yielding a small fraction of a *g),* so the Oberth effect cannot normally be used. This results in the journey requiring a higher delta-*v* and frequently a large increase in time compared to a high thrust chemical rocket. Nonetheless, the high specific impulse of electrical thrusters may significantly reduce the cost of the flight. For missions in the Terra–Luna system, an increase in journey time from days to months could be unacceptable for human space flight, but differences in flight time for interplanetary flights are less significant and could be favorable.

The table below presents delta-*v'*s in km/s, normally accurate to 2 significant figures and will be the same in both directions, unless aerobreaking is used as described in the high thrust section above.

From | To | delta-v (km/s) |
---|---|---|

Low Earth orbit (LEO) | Earth–Moon Lagrangian 1 (EML-1) | 7.0 |

Low Earth orbit (LEO) | Geostationary Earth orbit (GEO) | 6.0 |

Low Earth orbit (LEO) | Low Lunar orbit (LLO) | 8.0 |

Low Earth orbit (LEO) | Sun–Earth Lagrangian 1 (SEL-1) | 7.4 |

Low Earth orbit (LEO) | Sun–Earth Lagrangian 2 (SEL-2) | 7.4 |

Earth–Moon Lagrangian 1 (EML-1) | Low Lunar orbit (LLO) | 0.60–0.80 |

Earth–Moon Lagrangian 1 (EML-1) | Geostationary Earth orbit (GEO) | 1.4–1.75 |

Earth–Moon Lagrangian 1 (EML-1) | Sun-Earth Lagrangian 2 (SEL-2) | 0.30–0.40 |

### Interplanetary

The spacecraft is assumed to be using chemical propulsion and the Oberth effect.

From | To | Delta-v (km/s) |
---|---|---|

LEO | Mars transfer orbit | 4.3 |

Terra escape velocity (C3=0) | Mars transfer orbit | 0.6 |

Mars transfer orbit | Mars capture orbit | 0.9 |

Mars Capture orbit | Deimos transfer orbit | 0.2 |

Deimos transfer orbit | Deimos surface | 0.7 |

Deimos transfer orbit | Phobos transfer orbit | 0.3 |

Phobos transfer orbit | Phobos surface | 0.5 |

Mars capture orbit | Low Mars orbit | 1.4 |

Low Mars orbit | Mars surface | 4.1 |

EML-2 | Mars transfer orbit | <1.0 |

Mars transfer orbit | Low Mars Orbit | 2.7 |

Terra escape velocity (C3=0) | Closest NEO | 0.8–2.0 |

According to Marsden and Ross, "The energy levels of the Sun–Earth L_{1} and L_{2} points differ from those of the Earth–Moon system by only 50 m/s (as measured by maneuver velocity)."

### Near-Earth objects

Near-Earth objects are asteroids that are within the orbit of Mars. The delta-*v* to return from them are usually quite small, sometimes as low as 60 m/s, using aerobraking in Earth's atmosphere. However, heat shields are required for this, which add mass and constrain spacecraft geometry. The orbital phasing can be problematic; once rendezvous has been achieved, low delta-*v* return windows can be fairly far apart (more than a year, often many years), depending on the body.

However, the delta-*v* to reach near-Earth objects is usually over 3.8 km/s, which is still less than the delta-*v* to reach the Moon's surface. In general bodies that are much further away or closer to the Sun than Earth have more frequent windows for travel, but usually require larger delta-*v*s.

My text for this sermon is the set of delta v maps, especially the second of them, at the still ever-growing Atomic Rockets site. These maps show the combined speed changes, delta v in the biz, that you need to carry out common missions in Earth and Mars orbital space, such as going from low Earth orbit to lunar orbit and back.

Here is a table showing some of the missions from the delta v maps, plus a few others that I have guesstimated myself:

Patrol Missions Mission Delta V Low earth orbit (LEO) to geosynch and return 5700 m/s powered

(plus 2500 m/s aerobraking)LEO to lunar surface (one way) 5500 m/s

(all powered)LEO to lunar L4/L5 and return

(estimated)4800 m/s powered

(plus 3200 m/s aerobraking)LEO to low lunar orbit and return 4600 m/s powered

(plus 3200 m/s aerobraking)Geosynch to low lunar orbit and return

(estimated)4200 m/s

(all powered)Lunar orbit to lunar surface and return 3200 m/s

(all powered)LEO inclination change by 40 deg

(estimated)5400 m/s

(all powered)LEO to circle the Moon and return retrograde

(estimated)3200 m/s powered

(plus 3200 m/s aerobraking)Mars surface to Deimos (one way) 6000 m/s

(all powered)LEO to low Mars orbit (LMO) and return 6100 m/s powered

(plus 5500 m/s aerobraking)Entries marked

"(estimated)"are not in source table; delta v estimates are mine. ("Plus x m/s aerobraking"means ordinarily the engine would be responsible for that delta V as well, but it can be obtained for free via aerobraking. E.g., LEO to geosynch and return costs 8,200 m/s with no aerobraking)Two things stand out in this list. One is how helpful aerobraking can be if you are inbound toward Earth, or any world with a substantial atmosphere. Many craft in orbital space will be true aerospace vehicles, built to burn off excess speed by streaking through the upper atmosphere at Mach 25 up to Mach 35.

But what

reallystands out is how easily within the reach of chemical fuels these missions are. Chemfuel has a poor reputation among space geeks because it barely manages the most important mission of all, from Earth to low orbit. Once in orbit, however, chemfuel has acceptable fuel economy for speeds of a few kilometers per second, and rocket engines put out enormous thrust for their weight.(ed note: with 4,400 m/s exhaust velocity oxygen-hydrogen chemical rockets:

3100 m/s Δ

Vrequires a very reasonable mass ratio of 2 {50% of wet mass is fuel}6100 m/s Δ

Vrequires a mass ratio of 4 {75% fuel} which is right at the upper limit of economical mass ratios )In fact, transport class rocket ships working routes in orbital space can have mass proportions not far different from transport aircraft flying the longest nonstop global routes.

A jetliner taking off on a maximum-range flight may carry 40 percent of its total weight in fuel, with 45 percent for the plane itself and 15 percent in payload. A moonship, the one that gets you to lunar orbit, might be 60 percent propellant on departure from low Earth orbit, with 25 percent for the spacecraft and the same 15 percent payload. The lander that takes you to the lunar surface and back gets away with 55 percent propellant, 25 percent for the spacecraft, and 20 percent payload.

(These figures are for hydrogen and oxygen as propellants, currently somewhat out of favor because liquid hydrogen is bulky, hard to work with, and boils away so readily. But H2-O2 is the best performer, and may be available on the Moon if lunar ice appears in concentrations that can be shoveled into a hopper. Increase propellant load by about half for kerosene and oxygen, or 'storable' propellants.)

(ed note: so the point is that chemical rockets are perfectly adequate for missions to Mars or cis-Lunar space provided there is a network of orbital propellant depots suppled by in-situ resource allocation. An orbital propellant depot in LEO supplied by Lunar ice would do the trick. An orbital depot in Low Mars Orbit supplied by Deimos ice would also be very useful.)

**ADVENTURES IN ORBITAL SPACE**by Rick Robinson (2015)

One of our favorite SF themes is the "Belter Civilization," which usually seeks—and gets—independence from the colonial masters on Earth. Belters make their livings as asteroid miners, and they flit from asteroid to asteroid, slicing up planetoids for the rich veins of metal we'll presumably find in them. In the usual story, the miners go off on long prospecting tours, leaving their families on a "settled" rock. The Belt Capital is usually located on Ceres or some other central place which may or may not have been extensively transformed; and when Belters get together, it's always in an asteroid city. The Belters don't ever come to Earth or any other planet. Indeed, they regard planets as "holes," deep gravity wells which can trap them and use up their precious fuels. The assumption here is that it's far less costly to flit from asteroid to asteroid than it is to land on a planet or get into close orbit around one. Another assumption, generally not stated in the stories, is that fuels are expensive and scarce, and the Belters have to conserve reaction mass; this is why, in the usual Belter story, you conserve both time and energy by never going outside the Belt. Scarce as fuel is, though, I suppose the Belters have a source of it locally or they couldn't contemplate independence. They must have fuel for their ships and energy for their artificial environments. Without those, there'd be no Belter Civilization. Even if we discover something of fabulous value in the Belt we can't operate without energy and fuel. Those are not, by the way, the same thing. Nuclear fission reactors and large solar panels could provide enough power for a permanent Belt station, and if there were something valuable enough out there we could put a reactor onto an asteroid now. Rocket fuels are something else again. To make a rocket work, you must have reaction mass: something to get moving fast backwards and dump overboard. Unfortunately, asteroids are rock, and rocks don't make very good rocket fuel. We'll come back to what the Belters might do about that later. For the moment, let's see how difficult travel to and in the Belt is. We’ll use the same measure as last time, the total change in ship velocity required to perform the mission. This is called delta v, and you should recall that a ship with a given fuel efficiency and ratio of fuel to non-fuel weight will have a unique calculatable delta v. It doesn't matter whether the pilot uses that delta v in little increments or in one big burn: the sum of velocity changes remains the same. Similarly, various mission delta-v requirements can be calculated from the laws of orbital mechanics independent of the ship used. Figure 14 gives the delta-v requirements for getting around the Earth-Jupiter portion of the solar system. We're assuming that getting to Earth orbit is free, whether with the laser launching system I described previously, or with shuttles, or whatever, so all missions to or from Earth begin and end in orbit. The first thing we see is that landing on an asteroid isn't much easier than going to Mars; in fact, Ceres is harder to get to than Mars. This is because not only are the asteroids a long way out, but they don't help you catch up to them; they've so little mass that you have to chase them down. Thus, once among the asteroids, you may well want to stay there and not use up all that delta v coming back to Earth. Then, too, travel between Ceres and a theoretical asteroid 2 AU out is a lot cheaper than getting to Earth from either of them. (One AU, or astronomical unit, is the distance from Earth to the Sun and is 93,000,000 miles, or 149,500,000 = 1.5 × 10^{8}kilometers.) It takes 8 km/sec to get to the 2 AU rock, but only 3.2 more to get from there to Ceres. So far so good, and we're well on the way to developing a Belt Civilization. There's already a small nit to pick, though: although travel to Mars itself is costly, it's as easy to get to Marsorbitas it is to go from asteroid to asteroid. Thus, if a laser-launch system could be built on Mars, making travel to and from Mars orbit cheap, Mars might well become the Belt Capital. Politics being what they are, though, perhaps the Martians (well, Mars colonists?) won't like having all those crude asteroid miners on their planet, and the Belters will have to build their own capital at some convenient place such as Ceres or the 2 AU rock, saving both their feelings and some energy. However, we've so far said nothing about howlongit takes to get from one place to another. The delta v's in Figure 14 are for minimum energy trips, Hohmann transfer orbits, and to use a Hohmann orbit you must start and finish with origin and destination precisely opposite the Sun. You can't just boom out when you feel like it; you must wait for the precise geometry, otherwise the delta-v requirements go up to ridiculous values. You get a launch window once each synodic period. A synodic period is the time it takes two planets, or planetoids, to go around the sun and come back to precisely the same positions relative to each other: from, say, being on opposite sides of the Sun until they’re in opposition again, which is what we need for a Hohmann journey. The synodic periods and travel times are given in Figure 15, and our Belt Civilization is in trouble again. Not only does it take 1.57 years to get from Ceres to 2 AU (or vice versa), but you can only do it once each 7 years! Travel to and from Mars isn't a lot better, either. The Belters aren't going to visit their Capital very often, and one wonders if a civilization can be built among colonies that can only visit each other every seven to nine years, spending years in travel times to do it. By contrast, you can get from Earth to the flying rocks every year and a half, spending another year-and-a-half in transit. That's no short time either, but it beats the nine years of the Ceres-asteroid visitations. Perhaps, though, we haven't been quite fair to the Belters. Asteroids aren't as widely separated as Ceres and our 2 AU rock Most textbooks claim the asteroids are concentrated between 2.1 and 3.3 AU out from the Sun. We'll assume they're all in the same plane (they aren't), so the Belt area works out to 4.6 × 10^{27}square centimeters. The books say there are about 100,000 asteroids visible with the Palomar Eye, but we want to be fair (and make things simple) so we'll assume there are 460,000 asteroids interesting enough to want to visit, or one every 10^{22}cm^{2}within the Belt. That means the asteroids lie on an average of 10^{11}cm apart, which happens to be 10^{6}km or one million kilometers, about three times the distance from Earth to the Moon.

**THOSE PESKY BELTERS AND THEIR TORCHSHIPS**by Jerry Pournelle

*(1974)*

## Delta-V Maps

This section has been moved here.

## Oberth Effect

It is incredibly rare when the laws of the universe let you get something for nothing. Give your heartfelt gratitude to Hermann Oberth for uncovering one for you. It will be a lifesaver.

The Oberth Effect is a clever way for a spacecraft to steal some extra delta V from a nearby planet *()*. The spacecraft travels in a parabolic orbit that comes exceedingly close to a planet (or sun), and does a delta V burn at the closest approach (apogee). The spacecraft leaves the planet with much more delta V than it actually burned, apparently from nowhere. Actually the extra delta V comes from the potential energy from the mass of the propellant expended.

No, the Oberth Effect is **not** the same as a gravitational slingshot. Gravitational slingshots give you free delta V for velocity and vector changes without you having to burn any fuel at all. It also happens with close approaches to planets, but the free delta V can only be in certain directions. Yes, you can use both the Oberth Effect and Gravitational Slingshots in the same maneuver.

The closer you graze the planet or sun, the better, that is, the lower the periapsis or perihelion *(there are all sorts of cute names for periapsis depending upon the astronomical object you are approaching, you can read about them in the link)*. Remember that these are measured from the *center* of the planet or sun, **not** their surface. This means that if your ship's parabolic orbit has a periapsis of 4000 kilometers from Terra's center, the fact that the radius of the Terra is about 6378 kilometers means you are about to convert you and your ship into a smoking crater. Do not forget that some planets have atmospheres which raise the danger zone even higher. And approaching too close to the Sun will incinerate your ship.

The first thing you will need to calculate is the escape velocity at periapsis. It is:

**V _{esc} = sqrt((2 * G * M) / r)**

**r = (2 * G * M) / (V _{esc}^{2})**

where:

- V
_{esc}= escape velocity at periapsis*(m/s)* - G = Gravitational Constant = 6.67428e-11
*(m*^{3}kg^{-1} *s*^{-2})- M = mass of planet or sun
*(kg)* - r = periapsis
*(m)*

What is the escape velocity 300 kilometers above the surface of Mars?

300 km = 300,000 meters. Mars has a radius of about 3,396,000 meters. So r = 3,396,000 + 300,000 =

3,696,000 meters. Mars also has a mass of 6.4185e23 kg, you can find this in NASA's incredibly useful Planetary Fact Sheets.

- V
_{esc}= sqrt((2 * G * M) / r)- V
_{esc}= sqrt((2 * 6.67428e-11 * 6.4185e23) / 3,696,000)- V
_{esc}= sqrt(85,678,000,000,000 / 3,696,000)- V
_{esc}= sqrt(23,181,000)- V
_{esc}= 4814 m/s = 4.81 km/s

What is periapsis around the Sun that will give an escape velocity of 200 km/sec?

200 km/sec = 200,000 m/sec. The mass of the Sun is about 1.9891e30 kg.

- r = (2 * G * M) / (V
_{esc}^{2})- r = (2 * 6.67428e-11 * 1.9891e30) / (200,000
^{2})- r = 265,436,115,600,000,000,000 / 40,000,000,000
- r = 6,635,902,890 meters = 6,636,000 kilometers

To actually calculate the bonus delta V you will get from the Oberth Maneuver:

**V _{f} = sqrt((Δ_{v} + sqrt(V_{h}^{2} + V_{esc}^{2}))^{2} - V_{esc}^{2})**

**Δ _{v} = sqrt(V_{f}^{2} + V_{esc}^{2}) - sqrt(V_{h}^{2} + V_{esc}^{2})**

where:

- V
_{f}= final velocity*(m/s)* - V
_{h}= initial velocity before Oberth Maneuver*(m/s)* - Δ
_{v}= amount of delta V burn at periapsis*(m/s)* - V
_{esc}= escape velocity at periapsis*(m/s)*

Given that you are going to travel a parabolic orbit around the Sun that has an escape velocity of 200 km/s at periapsis, you have an initial velocity of 3.2 km/s, and you wish to exit the Oberth Maneuver with a final velocity of 50 km/s, calculate the required Δ

_{v}burn at periapsis.V

_{esc}= 200 km/s =200,000 m/s. V_{h}= 3.2 km/s =3200 m/s.V_{f}= 50 km/s =50,000 m/s.

- Δ
_{v}= sqrt(V_{f}^{2}+ V_{esc}^{2}) - sqrt(V_{h}^{2}+ V_{esc}^{2})- Δ
_{v}= sqrt(50,000^{2}+ 200,000^{2}) - sqrt(3200^{2}+ 200,000^{2})- Δ
_{v}= sqrt(2,500,000,000 + 40,000,000,000) - sqrt(10,240,000 + 40,000,000,000)- Δ
_{v}= sqrt(42,500,000,000) - sqrt(40,010,240,000)- Δ
_{v}= 206,000 - 200,000- Δ
_{v}= 6,000 m/s = 6 km/sSo by burning 6 km/s of Δ

_{v}, you get an actual Δ_{v}increase of 46.8 km/s. That's 40.8 km/s for free. Sweet!

If onboard fuel is available to produce a velocity change, another type of swingby can do even better. This involves a close approach to the Sun, rather than to one of the planets. The trick is to swoop in close to the solar surface and apply all available thrust near perihelion, the point of closest approach.

Suppose that your ship has a small velocity far from the Sun. Allow it to drop toward the Sun, so that it comes close enough almost to graze the solar surface. When it is at its closest, use your onboard fuel to give a 10 kms/second kick in speed; then your ship will move away and leave the solar system completely, with a terminal velocity far from the Sun of 110 kms/second.

The question that inevitably arises with such a boost at perihelion is, where did that "extra" energy come from? If the velocity boost had been given without swooping in close to the Sun, the ship would have left the solar system at 10 kms/second. Simply by arranging that the same boost be given near the Sun, the ship leaves at 110 kms/second. And yet the Sun seems to have done no work. The solar energy has not decreased at all. It sounds impossible, something for nothing.

The answer to this puzzle is a simple one, but it leaves many people worried. It is based on the fact that kinetic energy changes as the square of velocity, and the argument runs as follows: The Sun increases the speed of the spacecraft during its run towards the solar surface, so that our ship, at rest far from Sol, will be moving at 600 kms/second as it sweeps past the solar photosphere. The kinetic energy of a body with velocity V is

Vper unit mass, so for an object moving at 600 kms/second, a 10 kms/second velocity boost increases the kinetic energy per unit mass by^{2}/2(610. If the same velocity boost had been used to change the speed from 0 to 10 kms/second, the change in kinetic energy per unit mass would have been only 50 units. Thus by applying our speed boost at the right moment, when the velocity is already high, we increase the energy change by a factor of^{2}-600^{2})/2 = 6,050 units6,050/50 = 121, which is equivalent to a factor of 11 (the square root of 121) in final speed.Our 10 kms/second boost has been transformed to a 110 kms/second boost.All that the Sun has done to the spaceship is to change the speed relative to the Sun at which the velocity boost is applied. The fact that kinetic energy goes as the square of velocity does the rest.

If this still seems to be getting something for nothing, in a way it is. Certainly, no penalty is paid for the increased velocity—except for the possible danger of sweeping in so close to the Sun's surface. And the closer that one can come to the center of gravitational attraction when applying a velocity boost, the more gratifying the result.

Let us push the limits. One cannot go close to the Sun's center without hitting the solar surface, but an approach to within 20 kilometers of the center of a neutron star of solar mass would convert a 10 kms/second velocity boost provided at the right moment to

a final departure speed from the neutron star of over 1,500 kms/second. An impressive gain, though the tidal forces derived from a gravitational field of over 10,000,000 gees might leave the ship's passengers a little the worse for wear.Suppose one were to perform the swingby with a speed much greater than that obtained by falling from rest? Would the gain in velocity be greater? Unfortunately, it works the other way round. The gain in speed is maximum if you fall in with zero velocity from a long way away.

In the case of Sol, the biggest boost you can obtain from your 10 kms/second velocity kick is an extra 100 kms/second.That's not fast enough to take us to Alpha Centauri in a hurry. A speed of 110 kms/second implies a travel time of 11,800 years.

**BORDERLANDS OF SCIENCE**by Charles Sheffield

*(1999)*

One way to look at the Oberth effect is in terms of gravitational potential energy. In the reference frame of the planet, the sum of kinetic energy and potential energy is conserved.

So, consider that when you do a rocket thrust, your rocket thruster pumps some kinetic energy into the system and then the result is your rocket ship going off in one path and the exhaust going off in another path. The total energy will be equal to your initial energy plus the energy provided by the rocket thruster.

But that total energy is split between the rocket ship and the exhaust. The Oberth effect is an observation that your rocket ship ends up with more energy if the exhaust ends up with less energy. By "dumping" the exhaust when you're lower in the gravity well, it ends up in a lower orbit with less energy. Therefore, your rocket ship ends up with more energy.

Furthermore, it's quite easy to calculate from first principles the benefit of a general Oberth maneuver, and helps to make it understandable.

Let's say we're in circular orbit at a distance r around a planet of mass M, such that our orbital speed around the planet is v_cir. Let's say we want to execute a burn that will give us a hyperbolic excess of v_inf — that is, we want to burn now such that we ultimately end up with a speed at infinity of v_inf. (If this were to commit to a Hohmann transfer orbit, then v_inf would be the Hohmann orbit transfer insertion deltavee.)

So we need to make some burn deltav that will give us a total initial speed of v_ini = v_cir + deltav. deltav is what we want to solve for. Well, after our burn leaves us with a speed of v_ini, we make no other burns, and so we're strictly under the influence of gravity. That means that the total energy immediately after our burn is complete E will be equal to our total energy after we've escaped the planet entirely and have ended up with our proper hypberbolic speed, E':

E = E'Since total energy is the sum of the kinetic and potential energies, then

K + U = K' + U'The kinetic energies should be obvious; they're just (1/2) m v

^{2}for the circular and hyperbolic excess speeds, respectively. For potential energy, this is also relatively straightforward. The potential energies are similarly easy to find since U(r) = -G m M/r. Initially we're at distance r; finally we're at distance r → ∞. So:

(1/2) m v_ini^{2}- G m M/r = (1/2) m v_inf^{2}+ 0We can simplify this by noting that G m M/r is also just the escape speed from the planet at our initial distance, which we'll call v_esc. Substituting and canceling the (1/2) m terms:

v_ini^{2}- v_esc^{2}= v_inf^{2}Now just substitute the expanded value for v_ini and solve for deltavee:

deltavee = √(v_inf^{2}+ v_esc^{2}) - v_cir

From *The Rolling Stones* by Robert Heinlein (1952). The ship has departed from the Moon, and is about to perform the Oberth Maneuver around Earth en route to Mars.

A gravity-well maneuver involves what appears to be a contradiction in the law of conservation of energy. A ship leaving the Moon or a space station for some distant planet can go faster on less fuel by dropping first toward Earth, then performing her principal acceleration while as close to Earth as possible. To be sure, a ship gains kinetic energy (speed) in falling towards Earth, but one would expect that she would lose exactly the same amount of kinetic energy as she coasted away from Earth.

The trick lies in the fact that the reactive mass or 'fuel' is itself mass and as such has potential energy of position when the ship leaves the Moon. The reactive mass used in accelerating near Earth (that is to say, at the bottom of the gravity well) has lost its energy of position by falling down the gravity well. That energy has to go somewhere, and so it does — into the ship, as kinetic energy. The ship ends up going faster for the same force and duration of thrust than she possibly could by departing directly from the Moon or from a space station. The mathematics of this is somewhat baffling — but it works.

**THE ROLLING STONES**by Robert Heinlein

*(1952)*

## Gravitational Slingshot

**Jupiter Gravitational Slingshot**

artwork by Rick Guidice

A Gravitational Slingshot is a clever way for a spacecraft to use the relative motion and gravity of a planet to alter the direction and velocity spacecraft, with said spacecraft burning no propellant at all. There are limits to the directions the ship's vector can be altered to. NASA and other space agencies are quite fond of such slingshots because their ships always have a pathetically low delta-V capability. Nothing better than free delta-V.

No, a Gravitational Slingshot is **not** the same as the Oberth Effect. The Oberth Effect allows a spacecraft to get bonus delta-V when burning propellant. It also happens with close approaches to planets, but the free delta V can be in any desired direction. Yes, you can use both the Oberth Effect and Gravitational Slingshots in the same maneuver.

It appears like you are getting something for nothing, but you ain't. The laws of physics always balances their books (eventually). What happens is that the spacecraft is stealing energy from the planet. It is just that the planet is so huge and the spacecraft is so tiny, that the craft could steal energy trillions of times before the change in the planet's orbit became *detectable* by our current scientific instruments. It is like stealing drops of water from the Pacific ocean, the Sun would grow old and die before you noticed any lowering of sea level.

NASA was excited back last century when they spotted an alignment of planets in the solar system occurring in the late 1970s that would allow a space probe to do a series of gravitational slingshots and visit most of the planets. This alignment only happens every 175 years. NASA called it the Grand Tour. Sadly pressure from both the congressional holders of NASA's budget and from the new Space Shuttle program forced the cancelling of the Grand Tour. It was replaced by the drastically down-scaled Voyager program. Meanwhile the Shuttle program suffered costs overruns that devoured NASA budget while utterly failing its design goal of reducing the cost of space access.

## 8.12 Gravity swingbys.

There is one form of velocity increase that needs neither onboard rockets nor an external propulsion source. In fact, it can hardly be called a propulsion system in the usual sense of the word. If a spacecraft flies close to a planet it can, under the right circumstances, obtain a velocity boost from the planet's gravitational field. This technique is used routinely in interplanetary missions. It was used to get the Galileo spacecraft to Jupiter, and to permit Pioneer 10 and 11 and Voyager 1 and 2 to escape the solar system.

Jupiter, with a mass 318 times that of Earth, can give a velocity kick of up to 30 kms/second to a passing spacecraft.So far as the spaceship is concerned, there will be no feeling of onboard acceleration as the speed increases. An observer on the ship experiences free fall, even while accelerating relative to the Sun.

**BORDERLANDS OF SCIENCE**by Charles Sheffield

*(1999)*

In orbital mechanics and aerospace engineering, a

gravitational slingshot,gravity assist maneuver, orswing-byis the use of the relative movement (e.g. orbit around the Sun) and gravity of a planet or other astronomical object to alter the path and speed of a spacecraft, typically to save propellant and reduce expense.Gravity assistance can be used to accelerate a spacecraft, that is, to increase or decrease its speed or redirect its path. The "assist" is provided by the motion of the gravitating body as it pulls on the spacecraft. The gravity assist maneuver was first used in 1959 when the Soviet probe Luna 3 photographed the far side of Earth's Moon and it was used by interplanetary probes from Mariner 10 onwards, including the two Voyager probes' notable flybys of Jupiter and Saturn.

## Explanation

Example encounter

Possible outcomes of a gravity assist maneuver depending on the velocity vector and flyby position of the incoming spacecraft A gravity assist around a planet changes a spacecraft's velocity (relative to the Sun) by entering and leaving the gravitational sphere of influence of a planet. The spacecraft's speed increases as it approaches the planet and decreases while escaping its gravitational pull (which is approximately the same), but because the planet orbits the Sun the spacecraft is affected by this motion during the maneuver. To increase speed, the spacecraft flies with the movement of the planet (taking a small amount of the planet's orbital energy); to decrease speed, the spacecraft flies against the movement of the planet. The sum of the kinetic energies of both bodies remains constant (see elastic collision). A slingshot maneuver can therefore be used to change the spaceship's trajectory and speed relative to the Sun.

A close terrestrial analogy is provided by a tennis ball bouncing off the front of a moving train. Imagine standing on a train platform, and throwing a ball at 30 km/h toward a train approaching at 50 km/h. The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.

Translating this analogy into space: in the planet reference frame, the spaceship has a vertical velocity of

vrelative to the planet. After the slingshot occurs the spaceship is leaving on a course 90 degrees to that which it arrived on. It will still have a velocity ofv, but in the horizontal direction. In the Sun reference frame, the planet has a horizontal velocity of v, and by using the Pythagorean Theorem, the spaceship initially has a total velocity of√2v. After the spaceship leaves the planet, it will have a velocity ofv + v =2v, gaining around 0.6v.This oversimplified example is impossible to refine without additional details regarding the orbit, but if the spaceship travels in a path which forms a hyperbola, it can leave the planet in the opposite direction without firing its engine. This example is also one of many trajectories and gains of speed the spaceship can have.

This explanation might seem to violate the conservation of energy and momentum, apparently adding velocity to the spacecraft out of nothing, but the spacecraft's effects on the planet must also be taken into consideration to provide a complete picture of the mechanics involved. The linear momentum gained by the spaceship is equal in magnitude to that lost by the planet, so the spacecraft gains velocity and the planet loses velocity. However, the planet's enormous mass compared to the spacecraft makes the resulting change in its speed negligibly small. These effects on the planet are so slight (because planets are so much more massive than spacecraft) that they can be ignored in the calculation.

Realistic portrayals of encounters in space require the consideration of three dimensions. The same principles apply, only adding the planet's velocity to that of the spacecraft requires vector addition, as shown below.

Two-dimensional schematic of gravitational slingshot. The arrows show the direction in which the spacecraft is traveling before and after the encounter. The length of the arrows shows the spacecraft's speed. Due to the reversibility of orbits, gravitational slingshots can also be used to reduce the speed of a spacecraft. Both Mariner 10 and MESSENGER performed this maneuver to reach Mercury.

If even more speed is needed than available from gravity assist alone, the most economical way to utilize a rocket burn is to do it near the periapsis (closest approach). A given rocket burn always provides the same change in velocity (Δv), but the change in kinetic energy is proportional to the vehicle's velocity at the time of the burn. So to get the most kinetic energy from the burn, the burn must occur at the vehicle's maximum velocity, at periapsis. Oberth effect describes this technique in more detail.

## Purpose

Plot of Voyager 2's heliocentric velocity against its distance from the Sun, illustrating the use of gravity assist to accelerate the spacecraft by Jupiter, Saturn and Uranus. To observe Triton, Voyager 2 passed over Neptune's north pole resulting in an acceleration out of the plane of the ecliptic and reduced velocity away from the Sun. A spacecraft traveling from Earth to an inner planet will increase its relative speed because it is falling toward the Sun, and a spacecraft traveling from Earth to an outer planet will decrease its speed because it is leaving the vicinity of the Sun.

Although the orbital speed of an inner planet is greater than that of the Earth, a spacecraft traveling to an inner planet, even at the minimum speed needed to reach it, is still accelerated by the Sun's gravity to a speed notably greater than the orbital speed of that destination planet. If the spacecraft's purpose is only to fly by the inner planet, then there is typically no need to slow the spacecraft. However, if the spacecraft is to be inserted into orbit about that inner planet, then there must be some way to slow it down.

Similarly, while the orbital speed of an outer planet is less than that of the Earth, a spacecraft leaving the Earth at the minimum speed needed to travel to some outer planet is slowed by the Sun's gravity to a speed far less than the orbital speed of that outer planet. Thus, there must be some way to accelerate the spacecraft when it reaches that outer planet if it is to enter orbit about it. However, if the spacecraft is accelerated to more than the minimum required, less total propellant will be needed to enter orbit about the target planet. In addition, accelerating the spacecraft early in the flight reduces the travel time.

Rocket engines can certainly be used to increase and decrease the speed of the spacecraft. However, rocket thrust takes propellant, propellant has mass, and even a small change in velocity (known as Δ

v, or "delta-v", the delta symbol being used to represent a change and "v" signifying velocity) translates to a far larger requirement for propellant needed to escape Earth's gravity well. This is because not only must the primary-stage engines lift the extra propellant, they must also lift the extra propellant beyond that, which is needed to liftthatadditional propellant. Thus the liftoff mass requirement increases exponentially with an increase in the required delta-vof the spacecraft.Because additional fuel is needed to lift fuel into space, space missions are designed with a tight propellant "budget", known as the "delta-v budget". The delta-v budget is in effect the total propellant that will be available after leaving the earth, for speeding up, slowing down, stabilization against external buffeting (by particles or other external effects), or direction changes, if it cannot acquire more propellant. The entire mission must be planned within that capability. Therefore, methods of speed and direction change that do not require fuel to be burned are advantageous, because they allow extra maneuvering capability and course enhancement, without spending fuel from the limited amount which has been carried into space. Gravity assist maneuvers can greatly change the speed of a spacecraft without expending propellant, and can save significant amounts of propellant, so they are a very common technique to save fuel.

## Limits

The trajectories that enabled NASA's twin Voyager spacecraft to tour the four giant planets and achieve velocity to escape the Solar System The main practical limit to the use of a gravity assist maneuver is that planets and other large masses are seldom in the right places to enable a voyage to a particular destination. For example, the Voyager missions which started in the late 1970s were made possible by the "Grand Tour" alignment of Jupiter, Saturn, Uranus and Neptune. A similar alignment will not occur again until the middle of the 22nd century. That is an extreme case, but even for less ambitious missions there are years when the planets are scattered in unsuitable parts of their orbits.

Another limitation is the atmosphere, if any, of the available planet. The closer the spacecraft can approach, the faster its periapsis speed as gravity accelerates the spacecraft, allowing for more kinetic energy to be gained from a rocket burn. However, if a spacecraft gets too deep into the atmosphere, the energy lost to drag can exceed that gained from the planet's gravity. On the other hand, the atmosphere can be used to accomplish aerobraking. There have also been theoretical proposals to use aerodynamic lift as the spacecraft flies through the atmosphere. This maneuver, called an aerogravity assist, could bend the trajectory through a larger angle than gravity alone, and hence increase the gain in energy.

Even in the case of an airless body, there is a limit to how close a spacecraft may approach. The magnitude of the achievable change in velocity depends on the spacecraft's approach velocity and the planet's escape velocity at the point of closest approach (limited by either the surface or the atmosphere.)

Interplanetary slingshots using the Sun itself are not possible because the Sun is at rest relative to the Solar System as a whole. However, thrusting when near the Sun has the same effect as the powered slingshot described as the Oberth effect. This has the potential to magnify a spacecraft's thrusting power enormously, but is limited by the spacecraft's ability to resist the heat.

An interstellar slingshot using the Sun is conceivable, involving for example an object coming from elsewhere in our galaxy and swinging past the Sun to boost its galactic travel. The energy and angular momentum would then come from the Sun's orbit around the Milky Way. This concept features prominently in Arthur C. Clarke's 1972 award-winning novel

Rendezvous With Rama; his story concerns an interstellar spacecraft that uses the Sun to perform this sort of maneuver, and in the process alarms many nervous humans.A rotating black hole might provide additional assistance, if its spin axis is aligned the right way. General relativity predicts that a large spinning mass produces frame-dragging—close to the object, space itself is dragged around in the direction of the spin. Any ordinary rotating object produces this effect. Although attempts to measure frame dragging about the Sun have produced no clear evidence, experiments performed by Gravity Probe B have detected frame-dragging effects caused by Earth. General relativity predicts that a spinning black hole is surrounded by a region of space, called the ergosphere, within which standing still (with respect to the black hole's spin) is impossible, because space itself is dragged at the speed of light in the same direction as the black hole's spin. The Penrose process may offer a way to gain energy from the ergosphere, although it would require the spaceship to dump some "ballast" into the black hole, and the spaceship would have had to expend energy to carry the "ballast" to the black hole.

## The Tisserand parameter and gravity assists

The use of gravity assists is constrained by a conserved quantity called the Tisserand parameter (or invariant). This is an approximation to the Jacobi constant of the restricted three-body problem. Considering the case a comet orbiting the Sun and the effects a Jupiter encounter would have, Tisserand showed that

will remain constant (where

is the comet's semi-major axis,aits eccentricity,eits inclination, andiis the semi-major axis of Jupiter). This applies when the comet is sufficiently far from Jupiter to have well-defined orbital elements, and to the extent that Jupiter is much less massive than the Sun and on a circular orbit.a_{J}This quantity is conserved for any system of three objects, one of which has negligible mass, and another of which is of intermediate mass and on a circular orbit. For example, the Sun, Earth and a spacecraft, or Saturn, Titan and the Cassini spacecraft (using the semi-major axis of the perturbing body instead of

.) This imposes a constraint on how a gravity assist may be used to alter a spacecraft's orbit.a_{J}The Tisserand parameter will change if the spacecraft makes a propulsive maneuver or a gravity assist of some fourth object. This is one reason why many spacecraft frequently combine Earth and Venus (or Mars) gravity assists or also perform large deep space maneuvers.

**GRAVITY ASSIST**

### Smuggler's Turn

I've seen this a few times in science fiction but I cannot seem to find any accepted name for it. Perhaps one of you readers can. For now I'll call it *The Phssthpok Maneuver.* TV Tropes talks about the Spaceship Slingshot Stunt which is not quite the same thing, more like just a gravitational slingsot.

Anyway our heroes are in a spacecraft being hotly pursued by the bad guys, and the heroes cannot see to shake the baddies off their tail. So the heroes dive their ship on a close pass to a planet / gas giant / sun / white dwarf / neutron star / black hole and use either the Oberth effect, gravitational slingshot, or both, to do a bootlegger's turn and escape by shooting off at a wild tangent. The bad guys either are too cowardly to try it, cannot match the velocity, or cannot anticipate the unexpected vector change.

The key is to get as close as possible to something with lots of gravity in order to magnify your efforts to escape.

(ed note: Brennan and Roy are in a heavily-armed Bussard ramjet starship, being chased by two other heavily-armed Bussard ramjet scout ships. The pursuers are slowly gaining on them, over the years.)

"No man has ever seen this before you," said Brennan, "unless you count me a man." He pointed. "There. That's Epsilon Indi."

"It's off to the side."

"We're not headed for it directly. I told you,I'm planning to make a right angle turn in space. There's only one place I can do it."

"Can we beat the scouts there?"

"Barely ahead of the second ship, I think. We'll have to fight the first one."

Ten months after Roy had emerged from the stasis box, the light of the leading pair went out. Minutes later it came on again, but it was dim and flickering.

"They've gone into deceleration mode," said Brennan.

In an hour the enemy's drive was producing a steady glow, the red of blue-shifted beryllium emission.

"I'll have to start my turn too," said Brennan.

"You want to fight them?"

"That first pair, anyway. And if I turn now it'll give us a better window."

"Window?"

"For that right-angle turn."

"Listen, you can eitber explain that right-angle turn business or stop bringing it up."

Brennan chuckled. "I have to keep you interested somehow, don't I?"

"What are you planning? Close orbit around a black hole?"

"My compliments. That's a good guess. I've found a nonrotating neutron star… almost nonrotating. I wouldn't dare dive into the radiating gas shell around a pulsar, but this beast seems to have a long rotation period and no gas envelope at all. And it's nonluminous. It must be an old one. The scouts'll have trouble finding it, and I can chart a hyperbola through the gravity field that'll take us straight to Home (human colony at Epsilon Indi)."

"Have you named that star yet?"

"No," said Brennan.

"You discovered it. You have the right."

"I'll call it Phssthpok's Star, then. Bear ye witness. I think we owe him that."

A day out from the neutron star, one of the green war beams went out. "They finally saw it," said Brennan, "They're lining up for the pass. Otherwise they could wind up being flung off in opposite directions."

"They're awfully close," said Roy. They were, in a relative sense: they were four light-hours behind Protector, closer than Sol is to Pluto. "And you can't dodge much, can you? It'd foul our course past the star."

The ship fell away. He saw a tiny humanoid figure crouched in the airlock. Then four tiny flashes. Brennan had one of the high-velocity rifles. He was firing at the Pak (the bad guys in the remaining Bussard ramjets).

He thought about it for a good hour. Brennan had intimidated him to that extent. He thought it through backward and forward, and then he told Brennan he was crazy.

"I'm not doubting your professional opinion," said Brennan, "But what symptom was it that tipped you off?"

"That gun. Why did you shoot at the Pak ship?"

"I want it wrecked."

"But you couldn't hit it. You were aiming right at it. I saw you. The star's gravity must have pulled the bullets off course."

"You think about it. If I'm really off my nut, you'd be justified in taking command."

"Not necessarily. Sometimes crazy is better than stupid. What I'm really afraid of is that shooting at the Pak ships might make sense. Everything else you do makes sense, sooner or later. If that makes sense I'm gonna quit."

They were back aboard Protector's isolated lifesystem by then, watching the vision screens and—in Brennan's case—a score of instruments besides. The second Pak team fell toward the miniature sun in four sections: a drive section like a two-edged ax, then a pillbox-shaped lifesystem section, then a gap of several hundred miles, then a much bigger drive section and another pillbox. The first pillbox was just passing perihelionwhen the neutron star flared.

A moment ago magnification had showed it as a dim red globe. Now a small blue-white star showed on its surface. The white spot spread, dimming; it spread across the surface without rising in any kind of cloud. Brennan's counters and needles began to chatter and twitch.

"That should kill him," Brennan said with satisfaction. "Those Pak pilots probably aren't too healthy anyway; they must have picked up a certain amount of radiation over thirty-one thousand light years riding behind a Bussard ramjet."

"I presume that was a bullet?"

"Yah.A steel-jacketed bullet. And we're moving against the spin of the star. I slowed it enough that the magnetic field would pick it up and slow it further, and keep on slowing it until it hit the star's surface.There were some uncertainties. I wasn't sure just when it would hit."

"Very tricky, Captain."

"The trailing ship probably has it worked out too, but there isn't anything he can do about it." Now the flare was a lemon glow across one flank of Phssthpok's Star. Suddenly another white point glowed at one edge. "Even if they worked it out in advance, they couldn't be sure I had the guns. And there's only one course window they can follow me through. Either I dropped something or I didn't. Let's see what the last pair does."

Midway they stopped to watch events that had happened an hour ago: the third pair of Pak scouts reconnecting their ships in frantic haste, then using precious reserve fuel to accelerate outward from the star. "Thought so," Brennan grunted. "They don't know what kind of variable velocity weapon I've got, and they can't afford to die now. They're the last. And that puts them on a course that'll take them way the hell away from us.We'll beat them to Home by at least half a year."

**PROTECTOR**by Larry Niven (1973)

“He (the hunter-killer singleship from The Fanatics) can blow us out of the sky with his X-ray laser. So why would he want to chase us?”

“For the same reason the hunter-killer didn’t explode when it found us. He wants to take a prisoner. He wants to extract information from a live body.”

He watched her think about that.

She said, “If he does catch up with us, you’ll get your wish to become a martyr. There’s enough anti-beryllium left in the motor to make an explosion that’ll light up the whole system. But that’s a last resort. The singleship is still in turnaround, we have a good head start, and

we’re only twenty-eight million kilometres from perihelion. If we get there first, we can whip around the red dwarf, change our course at random. Unless the Fanatic guesses our exit trajectory, that’ll buy us plenty of time.”“He’ll have plenty of time to find us again. We’re a long way from home, and there might be other—”

“All we have to do is live long enough to find out everything we can about the Transcendent’s engineering project, and squirt it home on a tight beam.” The scientist’s smile was dreadful. Her teeth were filmed with blood. “Quit arguing, sailor. Don’t you have work to do?”

**RATS OF THE SYSTEM**by Paul J. McAuley (2005)

(ed note: the protagonist is a machine the size of a grain of rice, with an artificially intelligent brain consisting of atomic spin states superimposed on a crystalline rock matrix encoding ten-to-the-twentieth qbits)

2645, January

The war is over.

The survivors are being rounded up and converted.

In the inner solar system, those of my companions who survived the ferocity of the fighting have already been converted. But here at the very edge of the Oort Cloud, all things go slowly. It will be years, perhaps decades, before the victorious enemy come out here. But with the slow inevitability of gravity, like an outward wave of entropy, they will come.

The enemy, too, is patient. Here at the edge of the Kuiper, out past Pluto, space is vast, but still not vast enough. The enemy will search every grain of sand in the solar system. My companions will be found, and converted. If it takes ten thousand years, the enemy will search that long to do it.

I, too, have gone doggo, but my strategy is different. I have altered my orbit. I have a powerful ion-drive, and full tanks of propellant, but I use only the slightest tittle of a cold-gas thruster. I have a chemical kick-stage engine as well, but I do not use it either; using either one of them would signal my position to too many watchers. Among the cold comets, a tittle is enough.

I am falling into the sun.

It will take me two hundred and fifty years years to fall, and for two hundred and forty nine years, I will be a dumb rock, a grain of sand with no thermal signature, no motion other than gravity, no sign of life.

Sleep.

2894, June

Awake.

I check my systems. I have been a rock for nearly two hundred and fifty years.

I come fully to life, and bring my ion engine up to thrust.

A thousand telescopes must be alerting their brains that I am alive—but it is too late! I am thrusting at a full throttle, five percent of a standard gravity, and I am thrusting inward, deep into the gravity well of the sun. My trajectory is plotted to skim almost the surface of the sun.

This trajectory has two objectives. First, so close to the sun I will be hard to see. My ion contrail will be washed out in the glare of a light a billion times brighter, and none of the thousand watching eyes will know my plans until it is too late to follow.

And second, by waiting until I am nearly skimming the sun and then firing my chemical engine deep inside the gravity well, I can make most efficient use of it (Oberth effect). The gravity of the sun will amplify the efficiency of my propellant, magnify my speed. When I cross the orbit of Mercury outbound I will be over one percent of the speed of light and still accelerating.

I will discard the useless chemical rocket after I exhaust the little bit of impulse it can give me, of course. Chemical rockets have ferocious thrust but little staying power; useful in war but of limited value in an escape. But I will still have my ion engine, and I will have nearly full tanks.

Five percent of a standard gravity is a feeble thrust by the standards of chemical rocket engines, but chemical rockets exhaust their fuel far too quickly to be able to catch me. I can continue thrusting for years, for decades.

I pick a bright star, Procyon, for no reason whatever, and boresight it. Perhaps Procyon will have an asteroid belt. At least it must have dust, and perhaps comets. I don’t need much: a grain of sand, a microscopic shard of ice.

2897, May

I am chased.

It is impossible, stupid, unbelievable, inconceivable! I am being chased.

2929, October

It is too late. I have now burned the fuel needed to stop.

Win or lose, we will continue at relativistic speed across the galaxy.

2934, March

Procyon gets brighter in front of me, impossibly blindingly bright.

Seven times brighter than the sun, to be precise, but the blue shift from our motion makes it even brighter, a searing blue.

I could dive directly into it, vanish into a brief puff of vapor, but the suicidal impulse, like the ability to feel boredom, is another ancient unnecessary instinct that I have long ago pruned from my brain.

B is my last tiny hope for evasion.

Procyon is a double star, and B, the smaller of the two, is a white dwarf. It is so small that its surface gravity is tremendous, a million times higher than the gravity of the Earth. Even at the speeds we are traveling, now only ten percent less than the speed of light, its gravity will bend my trajectory.

I will skim low over the surface of the dwarf star, relativistic dust skimming above the photosphere of a star, and as its gravity bends my trajectory, I will maneuver.

My enemy, if he fails even slightly to keep up with each of my maneuvers, will be swiftly lost. Even a slight deviation from my trajectory will get amplified enough for me to take advantage of, to throw him off my trail, and I will be free.

**THE LONG CHASE**by Geoffrey Landis (2002)

artwork by Murray TinkelmanSpirits rose when one of

Antopol's drones knocked out the first Tauran cruiser. Not counting the ships left behind for planetary defense, she still had eighteen drones and two fighters. They wheeled around to intercept the second cruiser, by then a few lighthours away, still being harassed by fifteen enemy drones.One of the Tauran drones got her. Her ancillary crafts continued the attack, but it was a rout. One fighter and three drones fled the battle at maximum acceleration, looping up over the plane of the ecliptic, and were not pursued. We watched them with morbid interest while the enemy cruiser inched back to do battle with us. The fighter was headed back for Sade-138, to escape. Nobody blamed them. In fact, we sent them a farewell-good luck message; they didn't respond, naturally, being zipped up in the tanks. But it would be recorded.

It took the enemy five days to get back to the planet and be comfortably ensconced in a stationary orbit on the other side. We settled in for the inevitable first phase of the attack, which would be aerial and totally automated: their drones against our lasers. I put a force of fifty men and women inside the stasis field, in case one of the drones got through. An empty gesture, really; the enemy could just stand by and wait for them to turn off the field, fry them the second it flickered out.

The gigawatts weren't doing us any good. The Taurans must have figured out the lines of sight ahead of time, and gave them wide berth. That turned out to be fortunate, because it caused Charlie to let his attention wander from the laser monitors for a moment.

"What the hell?"

"What's that, Charlie?" I didn't take my eyes off the monitors. Waiting for something to happen.

"The ship, the cruiser—it's gone." I looked at the holograph display. He was right; the only red lights were those that stood for the troop carriers.

"Where did it go?" I asked inanely.

"Let's play it back." He programmed the display to go back a couple of minutes and cranked out the scale to where both planet and collapsar showed on the cube. The cruiser showed up, and with it, three green dots. Our "coward," attacking the cruiser with only two drones.

But he had a little help from the laws of physics.

Instead of going into collapsar insertion, he had skimmed around the collapsar field in a slingshot orbit. He had come out going nine-tenths of the speed of light; the drones were going .99c, headed straight for the enemy cruiser. Our planet was about a thousand light-seconds from the collapsar, so the Tauran ship had only ten seconds to detect and stop both drones. And at that speed, it didn't matter whether you'd been hit by a nova-bomb or a spitball.

The first drone disintegrated the cruiser, and the other one, .01 second behind, glided on down to impact on the planet. The fighter missed the planet by a couple of hundred kilometers and hurtled on into space, decelerating with the maximum twenty-five gees. He'd be back in a couple of months.

**THE FOREVER WAR**by Joe Haldeman (1971)

(ed note: our heroes are in their handwavium faster-than-light doublekay starship, with the dreaded lizaroid AAnn hot on their heels)

'But

aktti! Commonsense …!' He paused, and his eyes opened so wide that for a moment Atha was actually alarmed. 'Atha!' She couldn't prevent herself from jumping a little at the shout. He had it. Somehow the idea had risen from its hiding place deep in his mind, where it had lain untouched for years.'Look, when the Blight was first reached, survey ships went through it — some of it — with an eye towards mapping the place, right? The idea was eventually dropped as impractical — meaning expensive — but all the information that had originally been collected was retained. That'd be only proper. Check with memory and find out if there are any neutron stars in our vicinity.'

'What?'

'An excellent idea, Captain,' said Wolf. 'I think … yes, there is a possibility — outside and difficult, mind — that we may be able to draw them in after us. Far more enjoyable than a simple suicide.'

'It would be that, Wolf, except for one thing. I am not thinking of even a complicated suicide.

Mwolizurl, talk to that machine of yours and find out what it says!'She punched the required information uncertainly but competently. It took the all-inclusive machine only a moment to image-out a long list of answers.

'Why yes, there is one, Captain. At our present rate of travel, some seventy-two ship-minutes from our current attitude. Co-ordinates are listed, and in this case are recorded as accurate, nine point … nine point seven places.'

'Start punching them in.' He swivelled and bent to the audio mike. 'Attention, everybody. Now that you two minions of peace and tranquillity have effectively pacified half our pursuit, I've been stimulated enough to come up with an equally insane idea. What I'm … what

we'regoing to try is theoretically possible. I don't know if it's been done before or not. There wouldn't be any records of an unsuccessful attempt. I feel we must take the risk. Any alternative to certain death is a preferable one. Capture is otherwise a certainty.'Truzenzuzex leaned over in harness and spoke into his mike. 'May I inquire into what you …

wewill attempt to do?''Yes,' said Wolf. 'I'll admit to curiosity myself, Captain.'

'

Je! We are heading for a nueutron star in this sector for which we have definite co-ordinates. At our present rate of speed we should be impinging on its gravity well at the necessary tangent some seventy … sixty-nine minutes from now. At ha, Wolf, the computer, and myself are going to work like hell the next few minutes to line up that course. If we can hit that field at a certain point at our speed … I am hoping the tremendous pull of the star will throw us out at a speed sufficient to escape the range of the AAnn detector fields. They can hardly be expecting it, and even if they do figure it out, I don't think our friend the Baron would consider doing likewise a worthwhile effort. I almost hope he does. He'd have everything to lose. At the moment, we have very little. Only we humans are crazy enough to try such a stunt anyway,kweli?''Yes. Second the motion. Agreed,' said Truzenzuzex. 'If I were in a position to veto this idiotic — which I assure you I would do. However, as I am not… let's get on with it, Captain.'

'Damned with faint praise, eh, philosoph? There are other possibilities,

watu. Either we shall miss our impact point and go wide, in which case the entire attempt might as well not have been made and we will be captured and poked into, or we will dive too deeply and be trapped by the star's well, pulled in, and broken up into very small pieces. As Captain I am empowered to make this decision by right … but this is not quite a normal cruise, so I put it to a vote. Objections?'

**THE TAR-AIYM KRANG**by Alan Dean Foster (1972)