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.


RocketCat sez

All those cute starship spec sheets you see with moronic entries like "range" or "maximum distance" betray a dire lack of spaceflight knowledge. Spacecraft ain't automobiles, if they run out of gas they don't drift to a halt. DeltaV is the key.

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:

Δvo = sqrt[ (G * Pm) / Pr ]


  • Δvo = 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)
  • Pm = planet's mass (kg)
  • Pr = planet's radius (m)
  • sqrt[x] = square root of x

Mercury's mass is 3.302e23 kg and radius is 2.439e6 m. sqrt[ (6.673e-11 * 3.302e23) / 2.439e6] = 3006 m/s liftoff deltaV.

Δvo 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 Δesc instead. This is "escape velocity", and is also the delta V required to land from deep space instead of landing from orbit.

Δesc = sqrt[ (2 * G * Pm) / Pr ]

Δesc = sqrt[ (1.3346e-10 * Pm) / Pr ]


  • Δesc = deltaV for escape velocity from a planet (m/s)
  • G = 0.00000000006673 or 6.673e-11 (gravitational constant)
  • 1.3346e-10 = 2 * G
  • Pm = planet's mass (kg)
  • Pr = planet's radius (m)

Mercury's mass is 3.302e23 kg and radius is 2.439e6 m. sqrt[ (1.3346e-10 * 3.302e23) / 2.439e6] = 4251 m/s escape velocity deltaV.

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 Δesc already includes the deltaV for Δvo. In other words, when figuring the total deltaV for a given mission, you will add in either Δesc or Δvo, but not both. Use Δvo for surface-to-orbit missions and use Δesc for planet-to-planet missions


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

Δvd = gp * tL


  • Δvd = deltaV to counteract gravitational drag (m/s)
  • gp = acceleration due to gravity on planet's surface (m/s2)(this assumes that the majority of the burn is close to the ground)
  • tL = duration of liftoff or duration of liftoff burn (seconds)

Mercury's surface gravity is 3.70 m/s. Say that the duration of liftoff burn is 30 seconds. Then the gravitational drag would be 3.70 * 30 = 110 m/s Gravitational Drag.

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

tL = Δvo / A


  • A = spacecraft's acceleration (m/s2)

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


Mercury's basic deltaV cost for liftoff is 3006 m/s. If the spacecraft has an acceleration of 10 g, or 98.1 m/s, then 3006 / 98.1 = 30 seconds Liftoff Burn Duration.

The equation you will use is this:

Apg = A / gp

Δvd = Δesc / Apg


  • Apg = acceleration of spacecraft in terms of planetary gravities

Say the spacecraft can accelerate at 10 g (Terra gravities), or 98.1 m/s. Mercury's surface gravity is 3.70 m/s. So the spacecraft can accelerate at 98.1 / 3.70 = 26.5 Mercury Gravities. Since the deltaV for escape velocity is 4251 m/s, then 4251 / 26.5 = 160 m/s Gravitational Drag (which is close enough for government work to 110 m/s).

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 (Δva). 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, Δva 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.

Δtvo = Δvo + Δvd + Δva

Δtesc = Δesc + Δvd + Δva


  • Δtvo = total orbital deltaV (m/s)
  • Δtesc = total escape deltaV (m/s)
  • Δvo = basic deltaV cost for liftoff and orbital landing (m/s)
  • Δesc = basic deltaV cost for escape and deep space landing (m/s)
  • Δvd = deltaV to counteract gravitational drag (m/s)
  • Δva = 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.

Well, for the very unreasonable case where you have an infinitely throttleable rocket, and a 4 g upper limit, you would do something like what’s in the attached. Basically, you keep your accel as high as you can, while staying under a maximum dynamic pressure limit (0.5 * air density * velocity2). Once you’re through the thick part of the atmosphere, density drops off, and you can punch it again. I chose 800 psf for the attached graph, because it is a reasonable upper limit on maximum dynamic pressure (or "Max Q").

From Strangequark

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.

Hohmann Transfer Orbits

RocketCat sez

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

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

Hohmann Transfers

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.


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:

  1. Insertion Burn: A large burn to leave circular orbit around starting planet and enter the Hohmann transfer
  2. A long Coasting Phase where the spacecraft travels on an elliptical orbit with engines off
  3. Arrival Burn: A large burn to match the target planet's velocity (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.

Calculating Hohmann Delta V

First you need "μ" ("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 = 1.32715×1020 m3/s2

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

μ = 6.674×10-11 * Mprimary

where Mprimary = 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 1,898.3×1024 kilograms, therefore its μ is 1.2669×1017

The only other values you need to do the calculations are OrbitRadiuss and OrbitRadiusd, the orbital radius of the start and destination planets (in meters). Remember 1 AU = 1.496×1011 meters, since very few astronomical books are silly enough to give orbital radii in meters. So take the radius in AU and multiply it by 1.496×1011 to convert it into meters.

Now for the Hohmann delta V calculation:

smAxis = (OrbitRadiuss + OrbitRadiusd) / 2

OrbitVelocitys = sqrt[μ / OrbitRadiuss]

Velocityi = sqrt[μ * ((2 / OrbitRadiuss) - (1 / smAxis))]

DeltaVi = Velocityi - OrbitVelocitys

OrbitVelocityd = sqrt[μ / OrbitRadiusd]

Velocitya = sqrt[μ * ((2 / OrbitRadiusd) - (1 / smAxis))]

DeltaVa = Velocitya - OrbitVelocityd

DeltaV = abs[DeltaVi] + abs[DeltaVa]


  • sqrt[x] = square root of x
  • abs[x] = absolute value of x, that is, remove any negative sign
  • smAxis = Semi-major axis of Hohmann Transfer orbit (meters)
  • OrbitVelocitys = orbital velocity of the starting planet (m/s), i.e., free delta V
  • Velocityi = velocity needed for Insertion Burn (m/s)
  • DeltaVi = actual deta V needed for Insertion Burn (m/a) after taking advantage of the free delta v
  • OrbitVelocityd = orbital velocity of the destination planet (m/s), i.e., free delta V
  • Velocitya = velocity needed for Arrival Burn (m/s)
  • DeltaVa = actual deta V needed for Arrival Burn (m/a) after taking advantage of the free delta v
  • 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

For a Terra-Mars Hohmann with central body being Sol, Terra's orbital radius 1.000 AU and Mars orbital radius 1.524 AU:

  • μ = 1.32715×1020
  • OrbitRadiuss = 1.000 * 1.496×1011 = 1.496×1011 meters
  • OrbitRadiusd = 1.524 * 1.496×1011 = 2.280×1011 meters

Doing the math:

  • smAxis = (OrbitRadiuss + OrbitRadiusd) / 2
  • smAxis = (1.496×1011 + 2.280×1011) / 2
  • smAxis = 1.888×1011 meters

  • OrbitVelocitys = Sqrt[μ / OrbitRadiuss]
  • OrbitVelocitys = Sqrt[1.32715×1020 / 1.496×1011]
  • OrbitVelocitys = 29,785 meters/sec

  • Velocityi = sqrt[μ * ((2 / OrbitRadiuss) - (1 / smAxis))]
  • Velocityi = sqrt[1.32715×1020 * ((2 / 1.496×1011) - (1 / 1.888×1011))]
  • Velocityi = 32,731 meters/sec

  • DeltaVi = Velocityi - OrbitVelocitys
  • DeltaVi = 32,731 - 29,785
  • DeltaVi = 2,946 meters/sec

  • OrbitVelocityd = sqrt[μ / OrbitRadiusd]
  • OrbitVelocityd = sqrt[1.32715×1020 / 2.280×1011]
  • OrbitVelocityd = 24,126 meters/sec

  • Velocitya = sqrt[μ * ((2 / OrbitRadiusd) - (1 / smAxis))]
  • Velocitya = sqrt[1.32715×1020 * ((2 / 2.280×1011) - (1 / 1.888×1011))]
  • Velocitya = 21,476 meters/sec

  • DeltaVa = Velocitya - OrbitVelocityd
  • DeltaVa = 21,476 - 24,126
  • DeltaVa = -2,650 meters/sec (Note that negative just means you are slowing down, not speeding up)

  • DeltaV = abs[2,946] + abs[-2,650]
  • DeltaV = 2,946 + 2,650
  • DeltaV = 5,596 meters/sec

So the Polaris has to be capable of 5,596 m/s of delta V in order to do the Terra-Mars Hohmann transfer.

Calculating Hohmann Travel Time

Th = 0.5 * sqrt[(4 * π2 * smAxis3) / μ]


  • Th = Hohmann travel time (seconds)
  • smAxis = Semi-major axis of Hohmann Transfer orbit (meters) from above calculation
  • μ = given above, depends on mass of central body
  • sqrt[x] = square root of x
  • x2 = square of x
  • x3 = raise x to the third power
  • π = 3.14159...


  • 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 = 1.32715×1020
  • smAxisAU = 1.888×1011 meters

Doing the math:

  • Th = 0.5 * sqrt[(4 * π2 * smAxis3) / μ]
  • Th = 0.5 * sqrt[(4 * 3.141592 * (1.888×1011)3) / 1.32715×1020]
  • Th = 0.5 * sqrt[2.65684×1035 / 1.32715×1020]
  • Th = 0.5 * sqrt[2,001,915,283,599,362]
  • Th = 0.5 * 44,742,768
  • Th = 22,371,384 seconds = 8.6 months

Calculating Hohmann Launch Windows

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

OrbitPeriodi = 2 * π * sqrt[OrbitRadiusi3 / μ]

OrbitPeriods = 2 * π * sqrt[OrbitRadiuss3 / μ]

SynodicPeriod = 1 / ( (1/OrbitPeriodi) - (1/OrbitPeriods))


  • SynodicPeriod = time delay between Hohmann launch windows (seconds)
  • OrbitRadiusi = orbital radius of planet closer to central object (meters)
  • OrbitRadiuss = orbital radius of planet further away from central object (meters)
  • OrbitPeriodi = one planetary year for the inferior planet (seconds)
  • OrbitPeriods = one planetary year for the superior planet (seconds)
  • μ = given above, depends on mass of central body
  • x3 = raise x to the third power
  • π ≅ 3.14159...


  • 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 = 1.32715×1020
  • OrbitRadiusi = 1.496×1011 meters
  • OrbitRadiuss = 2.280×1011 meters

Doing the math:

  • OrbitPeriodi = 2 * π * sqrt[OrbitRadiusi3 / μ]
  • OrbitPeriodi = 2 * 3.14159 * sqrt[(1.496×1011)3 / 1.32715×1020]
  • OrbitPeriodi = 2 * 3.14159 * sqrt[3.348071936×1033 / 1.32715×1020]
  • OrbitPeriodi = 2 * 3.14159 * sqrt[25,227,532,200,580]
  • OrbitPeriodi = 2 * 3.14159 * 5,022,702
  • OrbitPeriodi = 31,558,565 seconds

  • OrbitPeriods = 2 * π * sqrt[OrbitRadiuss3 / μ]
  • OrbitPeriods = 2 * 3.14159 * sqrt[(2.280×1011)3 / 1.32715×1020]
  • OrbitPeriods = 2 * 3.14159 * sqrt[1.1852352×1034 / 1.32715×1020]
  • OrbitPeriods = 2 * 3.14159 * sqrt[89,306,800,286,328]
  • OrbitPeriods = 2 * 3.14159 * 9,450,228
  • OrbitPeriods = 59,377,531 seconds

  • SynodicPeriod = 1 / ( (1 / OrbitPeriodi) - (1 / OrbitPeriods))
  • SynodicPeriod = 1 / ( (1 / 31,558,565) - (1 / 59,377,531))
  • SynodicPeriod = 67,359,430 seconds = 26 months = 2.41 years

Calculating Launch Timing

This is for calculating two things:

  1. What is the configuration of the two planets indicating it is time to launch?
  2. 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]))
α = π * (1 - ( 0.35355 * sqrt[(r1/r2 + 1)3]))


  • α = Phase Angle, or angle StartPlanet-CenterObject-DestPlanet (radians). If negative, DestPlanet is behind StarPlanet, otherwise it is ahead.
  • x3 = raise x to the third power
  • π ≅ 3.14159...
  • r1 = OrbitRadiuss = orbital radius of start planet (meters)
  • r2 = OrbitRadiusd = 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...


Angle between planets for Terra-Mars Hohmann.

  • r1 = 1.496×1011 meters
  • r2 = 2.280×1011 meters

Doing the math:

  • α = π * (1 - ( (1/(2*sqrt[2])) * sqrt[(r1/r2 + 1)3]))
  • α = 3.14159 * (1 - (0.35355 * sqrt[(1.496×1011/2.280×1011 + 1)3]))
  • α = 3.14159 * (1 - (0.35355 * sqrt[(0.65617 + 1)3]))
  • α = 3.14159 * (1 - (0.35355 * sqrt[1.656173]))
  • 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.

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

Hohmanns In More Detail

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

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 Randolph are 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 Randolph lie 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 Randolph also move eastward nearly 7000 miles per hour -- 1.93 miles per second, to be finicky. The pilot of the Bolivar had to arrive at the Randolph precisely 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 the Randolph and 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.

From Space Cadet by Robert Heinlein (1948)

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 Pegasus was 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 Pegasus and 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 Pegasus were 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.

From Earthlight by Arthur C. Clarke (1955)

Holden leaned back in his chair and listened to the creaks of the Canterbury's final 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.

From Leviathan Wakes by "James S.A. Corey" (Daniel Abraham and Ty Franck) 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


For 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).


To 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 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.

From HOW TO GO TO MARS by George Musser and Mark Alpert (Scientific American March 2000)

A free-return trajectory is 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 term free).

(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.)


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).


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.

From the Wikipedia entry for FREE-RETURN TRAJECTORY

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.

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). This is displayed in something called a C3 or "pork-chop" plot. Basically more deltaV will widen the width of the launch window.

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. As a replacement, Windows users can use Trajectory Optimization Tool by Adam Harden.

Alternatively 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.

On the plot, departure times are on one axis, and arrival times are on another. Every intersection corresponds to an orbit. The color of the intersection tells the amount of deltaV required. The spot at the center of the big bulls-eye is the Hohmann orbit. 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.

On this plot I also drew diagonal white lines for trip duration. For instance, all the intersections between the "9 month" and "10 month" lines have a duration between nine months and ten months.

Yes, I know that this chart shows the Hohmann mission taking about 16,000 m/s of deltaV and ten and a half months. instead of 5590 m/s and 8.6 months. My only conclusion is that the Swing-by Calculator is taking many more variables into account, or I made some rather drastic mistakes while calculating the numbers by hand. But, as Tom Lehrer explained in his song New Math, "But the idea is the important thing."

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!

Tourist Season

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.

0.00Terra ⇒ Mars launch window opens. Tourist fleet Alfa departs Terra.
0.71Tourist fleet Alfa arrives at Mars. End of Martian off season, start of Martian tourist season.
1.98Mars ⇒ Terra launch window opens. Tourist fleet Alfa departs Mars. End of Martian touist season, start of Martian off season.
2.17Terra ⇒ Mars launch window opens. Tourist fleet Bravo departs Terra.
2.69Tourist fleet Alfa arrives at Terra.
2.88Tourist fleet Bravo at Mars. End of Martian off season, start of Martian tourist season.
4.15Mars ⇒ Terra launch window opens. Tourist fleet Bravo departs Mars. End of Martian touist season, start of Martian off season.
4.33Terra ⇒ Mars launch window opens. Tourist fleet Charlie departs Terra.
4.86Tourist fleet Bravo arrives at Terra.

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 season just 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.

From 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.


"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.

From Leviathan Wakes by "James S.A. Corey" (Daniel Abraham and Ty Franck) 2011. First novel of The Expanse

A Working Example

Table 2: DeltaV budget using Hohmann transfers
Terra liftoff and
insertion into
Hohmann transfer
to Mars
Mars landing5030
Mars liftoff and
insertion into
Hohmann transfer
to Terra
Terra landing12,908
Table 1: DeltaV budget for our Polaris mission.
Terra liftoff12,908
Hohmann to Mars5590
Mars landing5030
Mars liftoff5030
Hohmann to Terra5590
Terra landing12,908

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:

ΔvTotal = sqrt( ΔvLiftoff2 + ΔvHohmann2)

For instance, instead of Mars Liftoff and Hohmann being 5030 + 5590 = 10620 it will be sqrt( 50302 + 55902 ) = 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.

Sample Delta-V Budgets

From the Wikipedia article Delta-v Budget.
  • 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.


ManeuverAverage 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 orbit50–55
Station-keeping in L1/L230–100
Station-keeping in lunar orbit0–400
Attitude control (3-axis)2–6
Spin-up or despin5–10
Stage booster separation5–10
Momentum-wheel unloading2–6

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/toLEO-KenLEO-EqGEOEML-1EML-2EML-4/5LLOLunaC3=0
Low Earth orbit (LEO-Ken)4.244.333.773.433.974.045.933.22
Low Earth orbit (LEO-Eq)4.243.903.773.433.994.045.933.22
Geostationary orbit (GEO)2.061.631.381.471.712.053.921.30
Lagrangian point 1 (EML-1)0.770.771.380.140.330.642.520.14
Lagrangian point 2 (EML-2)0.330.331.470.140.340.642.520.14
Lagrangian point 4/5 (EML-4/5)0.840.981.710.330.340.982.580.43
Low lunar orbit (LLO)1.311.312.050.640.650.981.871.40
Terra escape velocity (C3=0)

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.

FromTodelta-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


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

FromToDelta-v (km/s)
LEOMars transfer orbit4.3
Terra escape velocity (C3=0)Mars transfer orbit0.6
Mars transfer orbitMars capture orbit0.9
Mars Capture orbitDeimos transfer orbit0.2
Deimos transfer orbitDeimos surface0.7
Deimos transfer orbitPhobos transfer orbit0.3
Phobos transfer orbitPhobos surface0.5
Mars capture orbitLow Mars orbit1.4
Low Mars orbitMars surface4.1
EML-2Mars transfer orbit<1.0
Mars transfer orbitLow Mars Orbit2.7
Terra escape velocity (C3=0)Closest NEO0.8–2.0

According to Marsden and Ross, "The energy levels of the Sun–Earth L1 and L2 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-vs.


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
MissionDelta V
Low earth orbit (LEO) to geosynch and return5700 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
4800 m/s powered
(plus 3200 m/s aerobraking)
LEO to low lunar orbit and return4600 m/s powered
(plus 3200 m/s aerobraking)
Geosynch to low lunar orbit and return
4200 m/s
(all powered)
Lunar orbit to lunar surface and return3200 m/s
(all powered)
LEO inclination change by 40 deg
5400 m/s
(all powered)
LEO to circle the Moon and return retrograde
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 return6100 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 really stands 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 ΔV requires a very reasonable mass ratio of 2 {50% of wet mass is fuel}

6100 m/s ΔV requires 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.)

From ADVENTURES IN ORBITAL SPACE by Rick Robinson (2015)

Delta-V Maps

This section has been moved here.

Oberth Effect

RocketCat sez

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 (No, this is not the same as a gravitational slingshot). 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.

Gravtational slingshots on the other hand 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:

Vesc = sqrt((2 * G * M) / r)

r = (2 * G * M) / (Vesc2)


  • Vesc = escape velocity at periapsis (m/s)
  • G = Gravitational Constant = 6.67428e-11 (m3 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.

  • Vesc = sqrt((2 * G * M) / r)
  • Vesc = sqrt((2 * 6.67428e-11 * 6.4185e23) / 3,696,000)
  • Vesc = sqrt(85,678,000,000,000 / 3,696,000)
  • Vesc = sqrt(23,181,000)
  • Vesc = 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) / (Vesc2)
  • r = (2 * 6.67428e-11 * 1.9891e30) / (200,0002)
  • 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:

Vf = sqrt((Δv + sqrt(Vh2 + Vesc2))2 - Vesc2)

Δv = sqrt(Vf2 + Vesc2) - sqrt(Vh2 + Vesc2)


  • Vf = final velocity (m/s)
  • Vh = initial velocity before Oberth Maneuver(m/s)
  • Δv = amount of delta V burn at periapsis (m/s)
  • Vesc = 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 Δv burn at periapsis.

Vesc = 200 km/s = 200,000 m/s. Vh = 3.2 km/s = 3200 m/s.Vf = 50 km/s = 50,000 m/s.

  • Δv = sqrt(Vf2 + Vesc2) - sqrt(Vh2 + Vesc2)
  • Δv = sqrt(50,0002 + 200,0002) - sqrt(32002 + 200,0002)
  • Δv = sqrt(2,500,000,000 + 40,000,000,000) - sqrt(10,240,000 + 40,000,000,000)
  • Δv = sqrt(42,500,000,000) - sqrt(40,010,240,000)
  • Δv = 206,000 - 200,000
  • Δv = 6,000 m/s = 6 km/s

So by burning 6 km/s of Δv, you get an actual Δv increase of 46.8 km/s. That's 40.8 km/s for free. Sweet!

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.

Isaac Kuo

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 v2 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_ini2 - G m M/r = (1/2) m v_inf2 + 0

We 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_ini2 - v_esc2 = v_inf2

Now just substitute the expanded value for v_ini and solve for deltavee:

deltavee = √(v_inf2 + v_esc2) - 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.

From The Rolling Stones by Robert Heinlein (1952)

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