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").

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 = ParkingOrbitRadius_s
• Mars Parking Orbit Altitude = 300 km = 300,000 m = ParkingOrbitRadius_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.

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 = 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]
• 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]
• 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 = 2.41 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.

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.

• 

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.

• 

• 

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.

• 

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.

• 

     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:

• 

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:

• 

     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:

• 

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

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

• 

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

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

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

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

• 

     In the above illustration I have an apogee beyond SEL2. But by timing the release from EML2, we could aim for other regions of the Hill Sphere, including SEL1.

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

• 

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

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

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

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

• 

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

     Mars is even worse:

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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.

• 

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

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

---

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

'Well, yes, ma'am, I do — but I hate to rent larger ones to such a small family with the tourist season just opening up: I'll bring in a cot for the youngster.'

(ed note: The "tourist season" on Mars starts with the earliest arrival time of a spacecraft on a Terra-Mars Hohmann trajectory)

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

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

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

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

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

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

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.)

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!

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.