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.

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.

artwork by John Polgreen

The basic deltaV cost for liftoff and landing is what is needed to achieve orbital (or circular) velocity: Δvo = sqrt[ (G * Pm) / Pr ] where 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 Example: 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.

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

Δ_esc = sqrt[ (2 * G * P_m) / P_r ]

Δ_esc = sqrt[ (1.3346e-10 * P_m) / P_r ]

Example: 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

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

Δ_vd = g_p * t_L

Example: 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.

From ENVIRONMENT OF MANNED SYSTEMS. 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.

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.

TABLE 1.-Gross effects of acceleration forces

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.

You won't use this equation either but tL = Δvo / A where A = spacecraft's acceleration (m/s2) The spacecraft's acceleration will be discussed on the page about blast-off. Example: 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.

Artwork by Ed Emshwiller, 1961

The equation you will use is this:

A_pg = A / g_p

Δ_vd = Δ_esc / A_pg

Example: 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).

Atmospheric drag only occurs on planets with atmospheres. 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. 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. So the total deltaV to lift off from Terra 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.

Dynamic Science Fiction, October.

from We Claim These Stars by Poul Anderson, 1959

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.

Hop Davis has computed the cosmic train schedule for Hohmann railroad towns.

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.

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 are good basic tutorials on orbital mechanics and trajectory here, here and here.

There is a simple listing of the appropriate equations at http://scienceworld.wolfram.com/physics/HohmannTransferOrbit.html and at http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

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).

Here's the deltaV budget for our Polaris mission:

So Solar Guard cruiser Polaris needs a deltaV of at least 47,056 m/s in order to perform the mission specified.

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:

Δ_vTotal = sqrt( Δ_vLiftoff² + Δ_vHohmann²)

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.

Updated deltaV budget for our Polaris mission:

Therefore 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 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.

Chart is from Rockets and Space Transportation. Delta Vs are in kilometers per second. "AB" means "aerobraking", that is, the planet's atmosphere may be used to change delta V instead of expending thrust.

Chart by Wolfkeeper diagraming delta V requirements in cis-Lunar and Martian space.

Oberth Effect 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) where: Vesc = escape velocity at periapsis (m/s) G = Gravitational Constant = 6.67428e-11 (m3 kg-1 s-2) M = mass of planet or sun (kg) r = periapsis (m)

Example: 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.

Example: 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.

To actually calculate the bonus delta V you will get from the Oberth Maneuver:

V_f = sqrt((Δ_v + sqrt(V_h² + V_esc²))² - V_esc²)

Δ_v = sqrt(V_f² + V_esc²) - sqrt(V_h² + V_esc²)

Example: 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.

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!

Isaac Kuo:

Erik Max Francis:

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'

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.