What's The Mission?
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Going The Distance
The main way to get a handle on your ship definition is to decide what kinds of missions it will be capable of. Let's decide that the Solar Guard cruiser Polaris will be capable of taking off from Terra, travelling to Mars, landing on Mars, taking off from Mars, travelling to Terra, and landing on Terra. All without re-fuelling.
Keep in mind that this is an incredibly silly sort of ship to design. Any real spacecraft designer would design two craft: one surface to orbit shuttle, and one orbit to orbit vehicle.
All those cute starship spec sheets you see with moronic entries like "range" or "maximum distance" betray a dire lack of spaceflight knowledge. Spacecraft ain't automobiles, if they run out of gas they don't drift to a halt. DeltaV is the key.
The main number of interest is deltaV. This means "change of velocity" and is usually measured in meters per second (m/s) or kilometers per second (km/s). A spacecraft's maximum deltaV can be though of as how fast it will wind up traveling at if it keeps thrusting until the propellant tanks are dry.
If that means nothing to you, don't worry. The important thing is that a "mission" can be rated according to how much deltaV is required. For instance: lift off from Terra, Hohmann orbit to Mars, and Mars landing, is a mission which would take a deltaV of about 18,290 m/s. If the spacecraft has equal or more deltaV capacity than the mission, it is capable of performing that mission. The sum of all the deltaV requirements in a mission is called the deltaV budget.
This is why it makes sense to describe a ship's performance in terms of its total deltaV capacity, instead of its "range" or some other factor equally silly and meaningless. In Michael McCollum's classic Antares Dawn, when the captain asks the helmsman how much propellant they have, the helmsman replies that they have only 2200 kps (kilometers per second) left in the tanks.
The basic deltaV cost for liftoff and landing is what is needed to achieve orbital (or circular) velocity:
Δvo = sqrt[ (G * Pm) / Pr ]
- Δvo = deltaV to lift off into orbit or land on a planet from orbit (m/s)
- G = 0.00000000006673 or 6.673e-11 (gravitational constant, don't ask)
- Pm = planet's mass (kg)
- Pr = planet's radius (m)
- sqrt[x] = square root of x
Mercury's mass is 3.302e23 kg and radius is 2.439e6 m. sqrt[ (6.673e-11 * 3.302e23) / 2.439e6] = 3006 m/s liftoff deltaV.
Δvo is what you will use for missions like the Space Shuttle, where you just climb into orbit, deliver or pick up something, then land from orbit. However, if the mission involved travelling to other planets, you will have to use Δesc instead. This is "escape velocity", and is also the delta V required to land from deep space instead of landing from orbit.
Δesc = sqrt[ (2 * G * Pm) / Pr ]
Δesc = sqrt[ (1.3346e-10 * Pm) / Pr ]
- Δesc = deltaV for escape velocity from a planet (m/s)
- G = 0.00000000006673 or 6.673e-11 (gravitational constant)
- 1.3346e-10 = 2 * G
- Pm = planet's mass (kg)
- Pr = planet's radius (m)
Mercury's mass is 3.302e23 kg and radius is 2.439e6 m. sqrt[ (1.3346e-10 * 3.302e23) / 2.439e6] = 4251 m/s escape velocity deltaV.
So for our Polaris mission, basic deltaV for Terra escape or capture: 11,180 m/s, basic deltaV for Mars escape or capture: 5030 m/s
Please note that Δesc already includes the deltaV for Δvo. In other words, when figuring the total deltaV for a given mission, you will add in either Δesc or Δvo, but not both. Use Δvo for surface-to-orbit missions and use Δesc for planet-to-planet missions
The above equation does not take into account gravitational drag or atmospheric drag. Both are very difficult to estimate.
Gravitational drag depends on the planet's gravity, the angle of the flight path, and the acceleration of the spacecraft. For Terra, the first approximation is 762 m/s (acceleration of ten gees). You won't use this equation, but the actual first approximation is
Δvd = gp * tL
- Δvd = deltaV to counteract gravitational drag (m/s)
- gp = acceleration due to gravity on planet's surface (m/s2)(this assumes that the majority of the burn is close to the ground)
- tL = duration of liftoff or duration of liftoff burn (seconds)
Mercury's surface gravity is 3.70 m/s. Say that the duration of liftoff burn is 30 seconds. Then the gravitational drag would be 3.70 * 30 = 110 m/s Gravitational Drag.
Arthur Harrill has made a nifty Excel Spreadsheet that calculates the liftoff deltaV for any given planet.
Gravitational drag grows worse with each second of burn, so one wants to reduce the burn time. Unfortunately reducing the burn time is the same as increasing the acceleration, and there is a limit to what the human frame can stand. Thorarinn Gunnarsson noted that the eyes are very vulnerable to high-gravity acceleration, second only to bad hearts and full bladders.
You won't use this equation either but
tL = Δvo / A
- A = spacecraft's acceleration (m/s2)
The spacecraft's acceleration will be discussed on the page about blast-off.
Mercury's basic deltaV cost for liftoff is 3006 m/s. If the spacecraft has an acceleration of 10 g, or 98.1 m/s, then 3006 / 98.1 = 30 seconds Liftoff Burn Duration.
The equation you will use is this:
Apg = A / gp
Δvd = Δesc / Apg
- Apg = acceleration of spacecraft in terms of planetary gravities
Say the spacecraft can accelerate at 10 g (Terra gravities), or 98.1 m/s. Mercury's surface gravity is 3.70 m/s. So the spacecraft can accelerate at 98.1 / 3.70 = 26.5 Mercury Gravities. Since the deltaV for escape velocity is 4251 m/s, then 4251 / 26.5 = 160 m/s Gravitational Drag (which is close enough for government work to 110 m/s).
For our Polaris mission, with an acceleration of 10 g, gravitational drag during Terra lift off will be 11,180 m/s / 10 = 1,118 m/s.
Atmospheric drag only occurs on planets with atmospheres (Δva). There ain't many planets in the solar system with atmospheres. At least none that you'd care to land on. Landing on Jupiter is a quick way to convert your spacecraft into a tiny ball of crumpled metal. The same holds true for Venus, except that the tiny ball will be acid-etched. So for a planet with no atmosphere, Δva will be zero.
For Terra, the first approximation is 610 m/s. It is not possible to give a general equation for atmospheric drag due to the large number of factors and variables. You can probably get away with proportional scaling, comparing atmospheric density, assuming you can find data on planetary atmospheric density (translation: I don't know how to do it).
The total lift-off or landing deltaV is the basic deltaV plus the extra deltaV due to atmospheric drag (if any) and gravitational drag.
Δtvo = Δvo + Δvd + Δva
Δtesc = Δesc + Δvd + Δva
- Δtvo = total orbital deltaV (m/s2)
- Δtesc = total escape deltaV (m/s2)
- Δvo = basic deltaV cost for liftoff and orbital landing (m/s2)
- Δesc = basic deltaV cost for escape and deep space landing (m/s2)
- Δvd = deltaV to counteract gravitational drag (m/s2)
- Δva = deltaV to counteract atmospheric drag (m/s2)
So the total deltaV to lift off from Terra for our Polaris mission is 11,180 + 1118 + 610 = 12,908 m/s. Maybe 13,058 if you add in about 150 m/s for course corrections and as a safety margin.
Effects of Acceleration Forces
In the following, "positive longitudinal" means sitting in a chair with your feet on the ground and your head above your heart. "Transverse forces prone" means lying face down. "Transverse forces supine" means lying on one's back.
TABLE 1.-Gross effects of acceleration forces Effects: g's Weightlessness 0 Earth normal (32.2 feet/second) 1 Hands and feet heavy;
walking and climbing difficult
2 Walking and climbing impossible;
crawling difficult; soft tissues sag
3 Movement only with great effort;
crawling almost impossible
4 Only slight movements of arms and head possible 5
The relative position or orientation of the subject is of prime importance in determining tolerable levels of gravitational or acceleration force, or "g force.' As the g force is gradually increased, certain effects are observed.
Figure 5 shows the time-tolerance relationships for positive longitudinal forces and for transverse forces (either prone or supine, prone being the position of lying face down and supine being the position of lying on one's back).
For the transverse position, human subjects in Germany during World War II were subjected to 17 g's for as long as 4 minutes reportedly with no harmful effects and no loss of consciousness. The curves indicated for very long periods of time are extrapolations and are speculative, since no data are available on long-term effects. Col. John Stapp, Air Force Missile Development Center, has investigated extreme g loadings, up to 45 g's, sustained for fractions of a second; These are the kind of accelerations or decelerations that would be experienced in crash landings. For these brief high g loadings, the rate of change of g exceeds 500 g's per second.
As a matter of interest, the beaded line on the figure indicates the approximate accelerations that would be experienced by a man in a vehicle designed to reach escape velocity with three stages of chemical burning, each stage having a similar load-factor-time pattern. This curve enters the critical region for positive g's. Most individuals would probably black out and some would become unconscious. However, for individuals in the transverse position, this acceleration could be tolerated and the individual would not lose consciousness.
Positive longitudinal g's, short duration
(blood forced from head toward feet):
Effects: g's Visual symptoms appear 2.5 - 7.0 Blackout 3.5 - 8.0 Confusion, loss of consciousness 4.0 - 8.5 Structural damage, especially to spine > 18 - 23
Transverse g's, short duration
(head and heart at same hydrostatic level):
Effects: g's No visual symptoms or loss of consciousness 0 - 17 Tolerated 28 - 30 Structural damage may occur > 30 - 45
Lift-off Acceleration Profile
So you want to keep the acceleration at a maximum of 4g or so otherwise the astronauts cannot manipulate the controls. But you want to spend as little time as possible getting into orbit in order to minimize gravitational drag. Therefore you want to maintain a steady 4g (throttling back the thrust as the mass of the propellant drops) until you get into orbit, right?
Well, I found that it was not that simple. You see, if you are lifting off from a planet with an atmosphere, you have to have to keep your spacecraft's speed such that the maximum dynamic pressure (or "Max Q") is too low to shred the ship into titanium confetti. The Space Shuttle's acceleration profile keeps Max Q below about 700 pounds per square foot, but a more sturdy spacecraft could probably survive 800 pounds per square foot.
On the NASA Spaceflight forum I asked what the optimal "acceleration profile" would be for an atomic rocket with a thrust-to-weight ratio above 1, an unreasonable specific impulse of 20,000 (a NSWR), single-stage surface to orbit.
A gentleman who goes by the Internet handle of "Strangequark" was kind enough to answer me.
Well, for the very unreasonable case where you have an infinitely throttleable rocket, and a 4 g upper limit, you would do something like what’s in the attached. Basically, you keep your accel as high as you can, while staying under a maximum dynamic pressure limit (0.5 * air density * velocity2). Once you’re through the thick part of the atmosphere, density drops off, and you can punch it again. I chose 800 psf for the attached graph, because it is a reasonable upper limit on maximum dynamic pressure (or "Max Q").
The acceleration profile says that the spacecraft takes off and accelerates at 4g for about five seconds. From second 5 to second 8 it drastically throttles back to an acceleration of about 0.25g. From second 8 to second 50 it gradually increases acceleration until it is back to 4g. It then stays at 4g until second 215, where it achieves orbit and the engine is shut off.
Hohmann Transfer Orbits
In 1925 Walter Hohmann made your life incredibly easier when he discovered Hohmann transfer orbits. Be grateful.
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.
Now we need to figure the deltaV for Terra-Mars transits.
Current spacecraft propulsion systems are so feeble that they cannot manage much more than the lowest deltaV missions. So they tend to use a lot of "Hohmann transfer orbits".
A Hohmann orbit between two planets is guaranteed to take the smallest amount of deltaV possible. For the Terra-Mars Hohmann, deltaV is 5590 m/s.
Notice that the deltaV required to get into orbit is 11180 m/s while the Terra-Mars deltaV is only 5590 m/s. As Robert Heinlein noted, once one gets into Earth orbit, you are "halfway to anywhere."
Unfortunately a Hohmann orbit also takes the maximum amount of transit time. For the Terra-Mars Hohmann mission, transit time is about 8.6 months.
The other drawback is that there are only certain times that one can depart for a given mission, the so-called "Synodic period" or Hohmann launch window. The start and destination planets have to be in the correct positions. For the Terra-Mars Hohmann mission, the Hohmann launch windows occur only every 26 months! If you do not launch at the proper time, when you get to the destination planet's orbit the planet won't be there. And then your life span is the same as your rapidly dwindling oxygen supply.
Actually there is a type of transfer orbit that requires even less deltaV than a Hohmann, the so-called "Interplanetary Transport Network" However, this transfer's practicality is questionable, for a manned mission at least. On the plus side it requires exceedingly small amounts of deltaV. On the minus side, as one would expect, it is so slow it makes a Hohmann look like a hypersonic bullet train. A Hohmann can travel from Earth orbit to Lunar orbit in a few days, the Interplanetary Transport Network takes two years.
Expending more deltaV than a Hohmann requires can also allow a ship to depart more often than the Hohmann's limit of one per synodic period, but this is hideously complicated to calculate (no, I don't know how to do this either). This is displayed in something called a C3 or "pork-chop" plot.
Luckily (for Windows users at least) there is a Windows program called Swing-by Calculator by http://www.jaqar.com which can calculate all the orbits over a series of dates and export a datafile, which can be imported into Excel, which can then draw the pork chop plot. Full instructions on how to do this is included with the software, which currently is free so long as it is not used for commercial purposes. Unfortunately Swing-By calculator seems to have vanished. As a replacement, Windows users can use Trajectory Optimization Tool by Adam Harden.
Alternatively there is an old-school Windows command-line program with no graphic user interface that only does Earth-Mars porkchop plots here. Go to section A Computer Program for Creating Pork Chop Plots of Ballistic Earth-to-Mars Trajectories, and download PDF document, Zipped file of executable program, and JPL DE421 ephemeris binary data file. It is actually written in Fortran.
On the plot, departure times are on one axis, and arrival times are on another. Every intersection corresponds to an orbit. The color of the intersection tells the amount of deltaV required. The spot at the center of the big bulls-eye is the Hohmann orbit. By varying your departure time you can see the deltaV cost of launching at other than the proper synodic period. By varying your arrival time you can see the deltaV cost of shortening the duration of the trip.
On this plot I also drew diagonal white lines for trip duration. For instance, all the intersections between the "9 month" and "10 month" lines have a duration between nine months and ten months.
Yes, I know that this chart shows the Hohmann mission taking about 16,000 m/s of deltaV and ten and a half months. instead of 5590 m/s and 8.6 months. My only conclusion is that the Swing-by Calculator is taking many more variables into account, or I made some rather drastic mistakes while calculating the numbers by hand. But, as Tom Lehrer explained in his song New Math, "But the idea is the important thing."
For step by step instructions on calculating the deltaV of a given Hohmann mission, go here
There is a more in-depth example of calculating both Hohmann and more energetic orbits using Fundamentals of Astrodynamics at the incomparable Voyage to Arcturus. The entries in question are here, here, and here. The discussion is about the superiority of Nuclear-Ion propulsion as compared to Nuclear-Thermal propulsion.
There is a simple listing of the appropriate equations at http://scienceworld.wolfram.com/physics/HohmannTransferOrbit.html and at http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
There is a java applet at http://www.exodusproject.com/NavComp/index.html
Here is an Excel spreadsheet called "Pesky Belter" which will calculate Hohmann deltaV, transit times, and synodic periods.
Erik Max Francis has written a freeware Hohmann orbit calculator in Python, available here. Be warned that the documentation is rudimentary, and operating the calculator requires a beginners knowledge of the Python language.
Information about the mass and semi-major axes of various planets can be found here: http://nssdc.gsfc.nasa.gov/planetary/planetfact.html
The Windows utility program Swing-by calculator can be found at http://www.jaqar.com
There is a freeware Windows program called Orbiter that allows one to fly around the solar system using real physics. A gentleman named Steven Ouellette has created an Orbiter add-on that re-creates the Rolling Stone from the Heinlein novel of the same name, along with the mission it flew (follow the above link).
A Working Example
|Hohmann to Mars||5590|
|Hohmann to Terra||5590|
|Terra liftoff and
insertion into Hohmann transfer to Mars
|Mars liftoff and
insertion into Hohmann transfer to Terra
Solar Guard cruiser Polaris needs a deltaV of at least 47,056 m/s in order to perform the mission specified in Table 1.
If the propulsion system has enough acceleration to achieve the Hohmann deltaV while still close to the planet it lifted off from, the total deltaV requirements can be reduced. Doing the liftoff and the Hohmann insertion as one long burn does this. Ordinarily the totals of the liftoff and Hohmann deltaVs are simply added together. If done as one long burn, it will be:
ΔvTotal = sqrt( ΔvLiftoff2 + ΔvHohmann2)
For instance, instead of Mars Liftoff and Hohmann being 5030 + 5590 = 10620 it will be sqrt( 50302 + 55902 ) = 7520.
How does this work? Well, it is an example of the Oberth effect (see below). Doing one long burn ensures that more of your propellant is expended low in the gravity well. And in case you are wondering, multi-stage rockets count as "one long burn," even though there is a small interrupting between stages.
Therefore, from Table 2, the Polaris needs a deltaV of at least 39,528 m/s in order to perform the mission specified.
8.6 months one way is pretty pathetic. Of course spending more deltaV can decrease the time. The http://www.exodusproject.com/NavComp/index.html applet can calculate that for you. What you do is alter the value of the "semi-latus rectum" hit "calculate" and see the new time and deltaV. (For Terra-Mars Hohmann, the semi-latus rectum is 1.207607)
Much easier of course is to examine a Pork Chop plot from Swing By Calculator. You can see from the left plot below how it reaches the point of diminishing returns quite quickly.
If you want to cheat, you can look up some of the missions in Jon Roger's Mission Table.
These are "maps" of the delta-V cost to move from one "location" to another. Keep in mind that some of the locations are actually orbits. If there is a planet with an atmosphere involved, "aerobraking" may be used (i.e., diving through the planet's atmosphere to use friction to burn off delta-V for free in lieu of expending expensive propellant).
- LEO: Low Earth Orbit. Earth orbit from 160 kilometers to 2,000 kilometers from the Earth's surface (below 200 kilometers Earth's atmosphere will cause the orbit to decay). The International Space Station is in an orbit that varies from 320 km to 400 km.
- GEO: Geosynchronous Earth Orbit. Earth orbit at 42,164 km from the Earth's center (35,786 kilometres from Earth's surface). Where the orbital period is one sidereal day. A satellite in GEO where the orbit is over the Earth's equator is in geostationary orbit. Such a satellite as viewed from Earth is in a fixed location in the sky, which is intensely desirable real-estate for telecommunications satellites. These are called "Clarke orbits" after Sir. Arthur C. Clarke. Competition is fierce for slots in geostationary orbit, slots are allocation by the International Telecommunication Union.
- EML1: Earth-Moon Lagrangian point 1. On the line connecting the centers of the Earth and the Moon, the L1 point is where the gravity of the two bodies cancels out. It allows easy access to both Earth and Lunar orbits, and would be a good place for an orbital propellant depot and/or space station. It has many other uses. It is about 344,000 km from Earth's center.
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 lifesafer.
The Oberth Effect is a clever way for a spacecraft to steal some extra delta V from a nearby planet (No, this is not the same as a gravitational slingshot). The spacecraft travels in a parabolic orbit that comes exceedingly close to a planet (or sun), and does a delta V burn at the closest approach. The spacecraft leaves the planet with much more delta V than it actually burned, apparently from nowhere. Actually the extra delta V comes from the potential energy from the mass of the propellant expended.
The closer you graze the planet or sun, the better, that is, the lower the periapsis or perihelion (there are all sorts of cute names for periapsis depending upon the astronomical object you are approaching, you can read about them in the link). Remember that these are measured from the center of the planet or sun, not their surface. This means that if your ship's parabolic orbit has a periapsis of 4000 kilometers from Terra's center, the fact that the radius of the Terra is about 6378 kilometers means you are about to convert you and your ship into a smoking crater. Do not forget that some planets have atmospheres which raise the danger zone even higher. And approaching too close to the Sun will incinerate your ship.
The first thing you will need to calculate is the escape velocity at periapsis. It is:
Vesc = sqrt((2 * G * M) / r)
r = (2 * G * M) / (Vesc2)
- Vesc = escape velocity at periapsis (m/s)
- G = Gravitational Constant = 6.67428e-11 (m3 kg-1
- M = mass of planet or sun (kg)
- r = periapsis (m)
What is the escape velocity 300 kilometers above the surface of Mars?
300 km = 300,000 meters. Mars has a radius of about 3,396,000 meters. So r = 3,396,000 + 300,000 = 3,696,000 meters. Mars also has a mass of 6.4185e23 kg, you can find this in NASA's incredibly useful Planetary Fact Sheets.
- Vesc = sqrt((2 * G * M) / r)
- Vesc = sqrt((2 * 6.67428e-11 * 6.4185e23) / 3,696,000)
- Vesc = sqrt(85,678,000,000,000 / 3,696,000)
- Vesc = sqrt(23,181,000)
- Vesc = 4814 m/s = 4.81 km/s
What is periapsis around the Sun that will give an escape velocity of 200 km/sec?
200 km/sec = 200,000 m/sec. The mass of the Sun is about 1.9891e31 kg.
- r = (2 * G * M) / (Vesc2)
- r = (2 * 6.67428e-11 * 1.9891e31) / (200,0002)
- r = 2,655,200,000,000,000,000,000 / 40,000,000,000
- r = 66,380,000,000 meters = 66,380,000 kilometers
To actually calculate the bonus delta V you will get from the Oberth Maneuver:
Vf = sqrt((Δv + sqrt(Vh2 + Vesc2))2 - Vesc2)
Δv = sqrt(Vf2 + Vesc2) - sqrt(Vh2 + Vesc2)
- Vf = final velocity (m/s)
- Vh = initial velocity before Oberth Maneuver(m/s)
- Δv = amount of delta V burn at periapsis (m/s)
- Vesc = escape velocity at periapsis (m/s)
Given that you are going to travel a parabolic orbit around the Sun that has an escape velocity of 200 km/s at periapsis, you have an initial velocity of 3.2 km/s, and you wish to exit the Oberth Maneuver with a final velocity of 50 km/s, calculate the required Δv burn at periapsis.
Vesc = 200 km/s = 200,000 m/s. Vh = 3.2 km/s = 3200 m/s.Vf = 50 km/s = 50,000 m/s.
- Δv = sqrt(Vf2 + Vesc2) - sqrt(Vh2 + Vesc2)
- Δv = sqrt(50,0002 + 200,0002) - sqrt(32002 + 200,0002)
- Δv = sqrt(2,500,000,000 + 40,000,000,000) - sqrt(10,240,000 + 40,000,000,000)
- Δv = sqrt(42,500,000,000) - sqrt(40,010,240,000)
- Δv = 206,000 - 200,000
- Δv = 6,000 m/s = 6 km/s
So by burning 6 km/s of Δv, you get an actual Δv increase of 46.8 km/s. That's 40.8 km/s for free. Sweet!
One way to look at the Oberth effect is in terms of gravitational potential energy. In the reference frame of the planet, the sum of kinetic energy and potential energy is conserved.
So, consider that when you do a rocket thrust, your rocket thruster pumps some kinetic energy into the system and then the result is your rocket ship going off in one path and the exhaust going off in another path. The total energy will be equal to your initial energy plus the energy provided by the rocket thruster.
But that total energy is split between the rocket ship and the exhaust. The Oberth effect is an observation that your rocket ship ends up with more energy if the exhaust ends up with less energy. By "dumping" the exhaust when you're lower in the gravity well, it ends up in a lower orbit with less energy. Therefore, your rocket ship ends up with more energy.
Furthermore, it's quite easy to calculate from first principles the benefit of a general Oberth maneuver, and helps to make it understandable.
Let's say we're in circular orbit at a distance r around a planet of mass M, such that our orbital speed around the planet is v_cir. Let's say we want to execute a burn that will give us a hyperbolic excess of v_inf -- that is, we want to burn now such that we ultimately end up with a speed at infinity of v_inf. (If this were to commit to a Hohmann transfer orbit, then v_inf would be the Hohmann orbit transfer insertion deltavee.)
So we need to make some burn deltav that will give us a total initial speed of v_ini = v_cir + deltav. deltav is what we want to solve for. Well, after our burn leaves us with a speed of v_ini, we make no other burns, and so we're strictly under the influence of gravity. That means that the total energy immediately after our burn is complete E will be equal to our total energy after we've escaped the planet entirely and have ended up with our proper hypberbolic speed, E':
E = E'
Since total energy is the sum of the kinetic and potential energies, then
K + U = K' + U'
The kinetic energies should be obvious; they're just (1/2) m v2 for the circular and hyperbolic excess speeds, respectively. For potential energy, this is also relatively straightforward. The potential energies are similarly easy to find since U(r) = -G m M/r. Initially we're at distance r; finally we're at distance r → ∞. So:
(1/2) m v_ini2 - G m M/r = (1/2) m v_inf2 + 0
We can simplify this by noting that G m M/r is also just the escape speed from the planet at our initial distance, which we'll call v_esc. Substituting and canceling the (1/2) m terms:
v_ini2 - v_esc2 = v_inf2
Now just substitute the expanded value for v_ini and solve for deltavee:
deltavee = √(v_inf2 + v_esc2) - v_cir
From The Rolling Stones by Robert Heinlein (1952). The ship has departed from the Moon, and is about to perform the Oberth Maneuver around Earth en route to Mars.
A gravity-well maneuver involves what appears to be a contradiction in the law of conservation of energy. A ship leaving the Moon or a space station for some distant planet can go faster on less fuel by dropping first toward Earth, then performing her principal acceleration while as close to Earth as possible. To be sure, a ship gains kinetic energy (speed) in falling towards Earth, but one would expect that she would lose exactly the same amount of kinetic energy as she coasted away from Earth.
The trick lies in the fact that the reactive mass or 'fuel' is itself mass and as such has potential energy of position when the ship leaves the Moon. The reactive mass used in accelerating near Earth (that is to say, at the bottom of the gravity well) has lost its energy of position by falling down the gravity well. That energy has to go somewhere, and so it does - into the ship, as kinetic energy. The ship ends up going faster for the same force and duration of thrust than she possibly could by departing directly from the Moon or from a space station. The mathematics of this is somewhat baffling - but it works.