1. Rockets don't got a "range" like a car running out of gasoline. Instead they have "maximum velocity" called "Delta V".
2. A rocket's propellant tanks are like a "wallet" full of delta-V "money"
3. Each rocket maneuver has a delta-V "cost" which has to be paid with delta-V money. Don't have enough in your wallet? Then you can't afford to do that manuever.
4. A space mission is composed of several manuevers, each with a delta-V cost. The total is the delta-V mission cost.
5. Refilling the propellant tank is like putting more money in the wallet. This is why rocket engineers are trying to figure out how to put gasoline stations in space. Because it sucks trying to do a long trip in your car on one tank.
6. Most rockets are poor, with very little money in their wallet. So they can only afford the cheapest maneuvers
7. One of the cheapest maneuvers is called a "Hohmann transfer." Using one to go from Terra to Mars takes about 5,700 meters per second of delta-V money and 8.6 months of travel time. And about the same to return home.
8. Also you can't just take off on a Mars Hohmann any time you want. Terra and Mars have to be in the proper spots, which happens every 26 months. This is called a "Hohmann Launch Window." And once you reach Mars you have to wait 15.3 months before the return launch window happens.

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.

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.

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.

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.

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.

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).

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").

The planet Mars has a mass of 6.4171×10²³ kg and a mean radius of 3,390,000 m. What is the orbital period of a station at an altitude of 20,073,000 meters?

OrbitalRadius = OrbitalAltitude + PlanetRadius
OrbitalRadius = 20,073,000 + 3,390,000
OrbitalRadius = 23,463,000 m

OrbitalPeriod = (2 * π) * sqrt[ OrbitalRadius³ / (G * PlanetMass) ]
OrbitalPeriod = (2 * 3.14159) * sqrt[ 23,463,000³ / (6.673×10^-11 * 6.4171×10²³) ]
OrbitalPeriod = 6.28318 * sqrt[ 12,916,671,713,847,000,000,000 / 42,821,308,300,000 ]
OrbitalPeriod = 6.28318 * sqrt[ 301,641,220.84628 ]
OrbitalPeriod = 6.28318 * 17,367.82142
OrbitalPeriod = 109,125 seconds = 1,818.8 minutes = 30.3 hours

For this problem I used the altitude of the Martian moon Phobos, which has an orbital period of 30.312 hours. Our result of 30.3 is close enough for government work.

The planet Terra has a mass of 5.98×10²⁴ kg and a mean radius of 6.37×10⁶ m. What is the orbital altitude of a station with an orbital period of 5,559 seconds?

OrbitalRadius = cubeRoot[ (G * PlanetMass * OrbitalPeriod²) / (4 * π²) ]
OrbitalRadius = cubeRoot[ (6.673×10^-11 * 5.98×10²⁴ * 5,559²) / (4 * π²) ]
OrbitalRadius = cubeRoot[ (6.673×10^-11 * 5.98×10²⁴ * 30,902,481) / (4 * 9.86961) ]
OrbitalRadius = cubeRoot[ 12,331,492,891,637,400,000,000 / 39.47844 ]
OrbitalRadius = cubeRoot[ 312,360,186,766,179,210,728.69140725925 ]
OrbitalRadius = 6,785,031 m

OrbitalAltitude = OrbitalRadius - PlanetRadius
OrbitalAltitude = 6,785,031 - 6.37×10⁶
OrbitalAltitude = 415,031 m = 415 km

For this problem I used the orbital period of the International Space Station. It has a mean orbital altitude of 404.55 km, which is close to our calculated 415 km. I presume the discrepancy is due to the fact that the ISS has an inclination of 51.64°, while Deimos has an inclination of about 1°.

A few times a year I find myself confronting a table full of numbers describing the orbits of things in the solar system, and cursing at myself because I've forgotten, again, what all these numbers mean and how to manipulate them to get the particular numbers I want. In particular, despite the fact that determining perihelion and aphelion distances from semimajor axis length and eccentricity could hardly be easier, I still always draw a blank. So I'm sitting down now to write a blog entry that will tell me what these numbers mean and how to use them to get the numbers I want! I'm posting it because I figure it'll be useful for some of you, too. In the following post, I'll show you why I was interested in getting these numbers today.

Here we go. The shape of an elliptical orbit is described by two parameters:

• semi-major axis, a: one half of the ellipse's long axis
  
  
• eccentricity, e: 0 for circular orbits; between 0 and 1 for ellipses

• 

To compute other numbers describing the shape of the orbit, here's what you do:

• Periapsis distance = a(1-e)
  
  
• Apoapsis distance = a(1+e)
  
  
• Orbital period = 2π√(a³/GM)
  
  
• Orbital period (solar orbit, in years, with a in AU) = a^1.5
  (and recall that 1 AU = 149.60×10⁶ km)

To figure out where an object currently is in space requires a few more pieces of information, including inclination, longitude of ascending node, et cetera.

• 

But because of the particular way in which orbital parameters are usually reported, it's actually quite difficult to use these elements to determine a body's current position in space in a way that makes intuitive geometrical sense to me. (I'm a geologist, dangit, not an orbital mechanic!) It's much faster and easier to make JPL's HORIZONS system crunch the numbers. (It makes me feel slightly better that when I asked this question on Twitter, all three Kuiper belt astronomers who follow me said they use HORIZONS too.) You can use HORIZONS via a telnet or email interface but if you don't have too many things to calculate it's easiest just to use the web interface. So:

• Go to http://ssd.jpl.nasa.gov/horizons.cgi
  
  
• Change the Target Body to the one you're interested in (click "change" and search on the name or number or provisional designation)
  
  
• Change Observer Location to "@sun"
  
  
• Go to Table Settings and check "Helio eclip. lon & lat" (or set the list to "18,20" to get heliocentric lat/lon and range)
  
  
• Click Generate Ephemeris and look for "hEcl-Lon" and "hEcl-Lat," which are in degrees, and "delta," which is the range in AU.

Note: It defaults to a one-month time span with a time step of one day, so the output table will have about thirty entries in it, starting today. The time span permits you to choose a wide range of dates (for most Kuiper belt objects, I find it to cover 1600 to 2200), and the time step chooser permits you to pick calendar years and calendar months as well as increments of days, hours, minutes, or seconds.

HORIZONS can also be used to find the distance between Earth and any of these objects, too, obviously; and you can have it spit the results directly to a text file, which is very handy!

• 

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 5.97×10²⁴ kg, the mass of Sol (M) is 1.99×10³⁰ 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( 5.97×10²⁴ / (3 * 1.99×10³⁰) )
r ≈ 149,600,000 * cubrt( 5.97×10²⁴ / 5.97×10³⁰ )
r ≈ 149,600,000 * cubrt( 0.000001 )
r ≈ 149,600,000 * 0.01
r ≈ 1,496,000 kilometers

So 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).

• 

"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.

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

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

• Mass of Sol = 1.9885×10³⁰ kg = M_primary
• Terra's orbital radius = 1.000 AU = 1.496×10¹¹ meters = OrbitRadius_s
• Mars orbital radius = 1.524 AU = 2.280×10¹¹ meters = OrbitRadius_d
• Mass of Terra = 5.9720×10²⁴ kg = M_s
• Mass of Mars = 6.4171×10²³ kg = M_d
• Mean Radius of Terra = 6.3710×10⁶ m = PlanetRadius_s
• Mean Radius of Mars = 3.3895×10⁶ 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 = (1.496×10¹¹ + 2.280×10¹¹) / 2
• SemimajorAxis = 1.888×10¹¹ meters
  
  
• μ_primary = 6.674×10^-11 * M_primary
• μ_primary = 6.674×10^-11 * 1.9885×10³⁰ kg
• μ_primary = 1.32715×10²⁰ m³/s²
  
  
• OrbitVelocity_s = Sqrt[ μ_primary / OrbitRadius_s]
• OrbitVelocity_s = Sqrt[1.32715×10²⁰ / 1.496×10¹¹]
• OrbitVelocity_s = 29,785 meters/sec
  
  
• Velocity_s = sqrt[ μ_primary * ((2 / OrbitRadius_s) - (1 / SemimajorAxis))]
• Velocity_s = sqrt[1.32715×10²⁰ * ((2 / 1.496×10¹¹) - (1 / 1.888×10¹¹))]
• 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 * 5.9720×10²⁴
• μ_s = 3.9857×10¹⁴
  
  
• ParkingOrbitRadius_s = PlanetRadius_s + ParkingOrbitAltitude_s
• ParkingOrbitRadius_s = 6.3710×10⁶ + 300,000
• ParkingOrbitRadius_s = 6.671×10⁶ m
  
  
• ParkingOrbitCircularVel_s = sqrt[ μ_s / ParkingOrbitRadius_s ]
• ParkingOrbitCircularVel_s = sqrt[ 3.9857×10¹⁴ / 6.671×10⁶ ]
• ParkingOrbitCircularVel_s = 7,730 m/s
  
  
• Velocity_escS = sqrt[ (2 * μ_s) / ParkingOrbitRadius_s ]
• Velocity_escS = sqrt[ (2 * 3.9857×10¹⁴) / 6.671×10⁶ ]
• Velocity_escS =10,931 m/s
  
  
• Velocity_hyperS = sqrt[ Velocity_s∞² + Velocity_sesc² ]
• Velocity_hyperS = sqrt[ 2,946² + 10,931² ]
• 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[1.32715×10²⁰ / 2.280×10¹¹]
• OrbitVelocity_d = 24,126 meters/sec
  
  
• Velocity_d = sqrt[ μ_primary * ((2 / OrbitRadius_d) - (1 / SemimajorAxis))]
• Velocity_d = sqrt[1.32715×10²⁰ * ((2 / 2.280×10¹¹) - (1 / 1.888×10¹¹))]
• 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 * 6.4171×10²³
• μ_d = 4.2828×10¹³
  
  
• ParkingOrbitRadius_d = PlanetRadius_d + ParkingOrbitAltitude_d
• ParkingOrbitRadius_d = 3.3895×10⁶ + 300,000
• ParkingOrbitRadius_d = 3.6895×10⁶ m
  
  
• ParkingOrbitCircularVel_d = sqrt[ μ_d / ParkingOrbitRadius_d ]
• ParkingOrbitCircularVel_d = sqrt[ 4.2828×10¹³ / 3.6895×10⁶ ]
• ParkingOrbitCircularVel_d = 3,407 m/s
  
  
• Velocity_escD = sqrt[(2 * μ_d) / ParkingOrbitRadius_d]
• Velocity_escD = sqrt[(2 * 4.2828×10¹³) / 3.6895×10⁶]
• Velocity_escD = 4,818 m/s
  
  
• Velocity_hyperD = sqrt[Velocity_∞d² + Velocity_escD²]
• Velocity_hyperD = sqrt[2,650² + 4,818²]
• 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/sec

So the Polaris has 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.

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

• μ_primary = 1.32715×10²⁰
• SemimajorAxis = 1.888×10¹¹ meters

Doing the math:

• T_h = 0.5 * sqrt[(4 * π² * SemimajorAxis³) / μ_primary]
• T_h = 0.5 * sqrt[(4 * 3.14159² * (1.888×10¹¹)³) / 1.32715×10²⁰]
• T_h = 0.5 * sqrt[2.65684×10³⁵ / 1.32715×10²⁰]
• T_h = 0.5 * sqrt[2,001,915,283,599,362]
• T_h = 0.5 * 44,742,768
• T_h = 22,371,384 seconds = 8.6 months

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

• μ_primary = 1.32715×10²⁰
• OrbitRadius_i = 1.496×10¹¹ meters
• OrbitRadius_s = 2.280×10¹¹ meters

Doing the math:

• OrbitPeriod_i = 2 * π * sqrt[OrbitRadius_i³ / μ_primary]
• OrbitPeriod_i = 2 * 3.14159 * sqrt[(1.496×10¹¹)³ / 1.32715×10²⁰]
• OrbitPeriod_i = 2 * 3.14159 * sqrt[3.348071936×10³³ / 1.32715×10²⁰]
• 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_s³ / μ_primary]
• OrbitPeriod_s = 2 * 3.14159 * sqrt[(2.280×10¹¹)³ / 1.32715×10²⁰]
• OrbitPeriod_s = 2 * 3.14159 * sqrt[1.1852352×10³⁴ / 1.32715×10²⁰]
• 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

• 

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.

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:

• t₀: the time of departure from Earth.
• t₁: the time of arrival at Mars.
• t₂: the time of departure from Mars.
• t₃: the time of arrival at Earth.

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 2π radians of passage over a single planetary year (a period of time P_E and P_M for Earth and Mars, respectively). We are only interested in their values at the four points above, though. E.g. θ_M,2 is the angle of Mars at time t₂.

We need to define how these angles relate to each other. Let's call t_H := t₁-t₀ the duration of the Hohmann transfer; notice that it is the same going out as coming back. For the time period t₀ to t₁, the spacecraft is on a Hohmann transfer from Earth to Mars. The angles relate as:

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 t_W (waiting time). Again, the planets' angular positions advance:

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 π around the center:

The 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 2π of the angles to get them to agree:

We 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 k₁ to just k.)

We 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 t_W, which works out to be:

Because k is just some integer, we can remove that -1. Generalizing the notation a little, so that the home planet has period P₀ and the visited planet has period P₁, we get:

For k, any value can be chosen so long as the resulting t_W is nonnegative.

---

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 k from the integers is obnoxious. It would be nice to choose it from the natural numbers (i.e., ℕ := {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 P₀ > P₁, then we must have k ≥ (-2t_H)/(P₀). Similarly, if P₀ < P₁, then k ≤ (-2t_H)/(P₀). And, of course, if P₀ = P₁, 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).

We now want to generate a sequence of k values from natural-valued n values that produces the correct result either way:

Some 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 P₀ < P₁ case I tested.

---

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 P_i whereas Gregorian calendar used to convert from months):

P₀ = 365.256363004
P₁ = 686.980
t_H =“8.5 months” ≈ 259 days
t_W =“14.9 months” ≈ 454 days

According to our first formula, we get:

k	tW
-3	≈ 1235 days
-2	≈ 455 days
-1	≈ -325 days
0	≈ -1105 days
1	≈ -1885 days
2	≈ -2665 days
3	≈ -3445 days

Therefore, k must be -2 or 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 within 1.5% or so, which is pretty good given the imprecision of t_H and the expected t_W, as well as the astronomical fact that the orbits are imperfect.

For our second formula, we have:

n	tW
1	≈ 455 days
2	≈ 1235 days
3	≈ 2015 days

This confirms that the formula works when P₀ < P₁, but the case for P₀ > P₁ remains untested.

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.

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CONJUNCTION CLASS

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).

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OPPOSITION CLASS

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.

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LOW THRUST

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.

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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.)

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.

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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 which outbound 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 L₂ 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 L₁ 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 L₁ through L₃. For instance, the Earth–Moon L₁ 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 L₄ and L₅. 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 L₁ 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 L₁ 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, L₄ and L₅, are stable, but the halo orbits for L₁ through L₃ 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 Hiten lunar mission in 1991. Another example of the use of the ITN was NASA's 2001–2003 Genesis mission, which orbited the Sun–Earth L₁ point for over two years collecting material, before being redirected to the L₂ Lagrange point, and finally redirected from there back to Earth. The 2003–2006 SMART-1 of 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 L₂ point, then on to fly by the asteroid 4179 Toutatis.

      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.

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     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:

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     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:

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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: SML2

     Since 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:

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     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:

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     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:

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     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.

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     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:

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     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:

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

     Mars is even worse:

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     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:

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     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 = 0.069

     So 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 = 1.0134

     We'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.

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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.

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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.

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.

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 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.)

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 × 10⁸ 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 Mars orbit as 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 how long it 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 × 10²⁷ 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 10²² cm² within the Belt. That means the asteroids lie on an average of 10¹¹ cm apart, which happens to be 10⁶ km or one million kilometers, about three times the distance from Earth to the Moon.

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.

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_esc²)
• 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

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_f² + V_esc²) - sqrt(V_h² + V_esc²)
• Δ_v = sqrt(50,000² + 200,000²) - sqrt(3200² + 200,000²)
• 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/s

So 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 V²/2 per 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 (610²-600²)/2 = 6,050 units. 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 6,050/50 = 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.

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 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

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.

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.

• 

In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by is 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

• 

• 

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 v relative 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 of v, 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 of v + 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.

• 

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

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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 Δv, 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 lift that additional propellant. Thus the liftoff mass requirement increases exponentially with an increase in the required delta-v of 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

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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 a is the comet's semi-major axis, e its eccentricity, i its inclination, and a_J is 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.

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 a_J.) This imposes a constraint on how a gravity assist may be used to alter a spacecraft's orbit.

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.