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's orbital radius 1.000 AU and Mars orbital radius 1.524 AU:

• μ = 1.32715×10²⁰
• OrbitRadius_s = 1.000 * 1.496×10¹¹ = 1.496×10¹¹ meters
• OrbitRadius_d = 1.524 * 1.496×10¹¹ = 2.280×10¹¹ meters

Doing the math:

• smAxis = (OrbitRadius_s + OrbitRadius_d) / 2
• smAxis = (1.496×10¹¹ + 2.280×10¹¹) / 2
• smAxis = 1.888×10¹¹ meters
  
  
• OrbitVelocity_s = Sqrt[μ / OrbitRadius_s]
• OrbitVelocity_s = Sqrt[1.32715×10²⁰ / 1.496×10¹¹]
• OrbitVelocity_s = 29,785 meters/sec
  
  
• Velocity_i = sqrt[μ * ((2 / OrbitRadius_s) - (1 / smAxis))]
• Velocity_i = sqrt[1.32715×10²⁰ * ((2 / 1.496×10¹¹) - (1 / 1.888×10¹¹))]
• Velocity_i = 32,731 meters/sec
  
  
• DeltaV_i = Velocity_i - OrbitVelocity_s
• DeltaV_i = 32,731 - 29,785
• DeltaV_i = 2,946 meters/sec
  
  
• OrbitVelocity_d = sqrt[μ / OrbitRadius_d]
• OrbitVelocity_d = sqrt[1.32715×10²⁰ / 2.280×10¹¹]
• OrbitVelocity_d = 24,126 meters/sec
  
  
• Velocity_a = sqrt[μ * ((2 / OrbitRadius_d) - (1 / smAxis))]
• Velocity_a = sqrt[1.32715×10²⁰ * ((2 / 2.280×10¹¹) - (1 / 1.888×10¹¹))]
• Velocity_a = 21,476 meters/sec
  
  
• DeltaV_a = Velocity_a - OrbitVelocity_d
• DeltaV_a = 21,476 - 24,126
• DeltaV_a = -2,650 meters/sec (Note that negative just means you are slowing down, not speeding up)
  
  
• DeltaV = abs[2,946] + abs[-2,650]
• DeltaV = 2,946 + 2,650
• DeltaV = 5,596 meters/sec

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

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

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

Doing the math:

• T_h = 0.5 * sqrt[(4 * π² * smAxis³) / μ]
• 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).

---

• 

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.

---

• 

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.

• 

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.