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Delta-v budgets

Gravity and orbits · level 8, lesson 1 of 6

A question to start

A mission planner does not ask "how far is Mars?" but "how much delta-v does it take?" Why is that the right question, and what is the answer, step by step, from the launch pad on Earth to the ground on Mars?

The idea

Delta-v (\(\Delta v\)) is the total change of speed a ship must make with its engines. It is the right currency because the rocket equation (Rockets, level 6) turns it straight into fuel: \(m_0/m_1 = e^{\Delta v / v_e}\). Distance is almost free in space; changes of speed are not.

A budget adds the burns of a mission in order. For Earth to Mars, with figures typical of mission studies:

Step \(\Delta v\) (km/s) How it is worked out
Earth's surface to low Earth orbit 9.3 to 9.5 7.8 for orbital speed, plus about 1.5 lost to gravity and drag on the way up
Low Earth orbit to the transfer to Mars 3.6 \(\sqrt{v_\text{esc}^2 + v_\infty^2} - v_\text{circ}\), with \(v_\infty = 2.94\)
Capture into low Mars orbit 2.1 by engine, or almost 0 using the air \(\sqrt{v_\text{esc}^2 + v_\infty^2} - v_\text{circ}\) at Mars, with \(v_\infty = 2.65\)
Low Mars orbit to the surface about 0.7 to 1 by engine most of the 3.4 to 5.6 km/s removed by the air; engines for the last part
Mars's surface to low Mars orbit 4.1 3.4 for orbital speed, plus losses (about 0.4 for gravity, small for drag, less for spin)
Low Mars orbit to the transfer home 2.1 the same sum, with \(v_\infty = 2.65\)
Earth arrival almost 0 the air of Earth brakes the capsule

Two things stand out. First, getting off Earth is the biggest step; once in low Earth orbit, the surface of Mars is a little over 4 km/s away with the air of Mars doing the braking. Second, leaving Mars costs 4.1 km/s, far less than leaving Earth, which is why people plan to make return fuel on Mars.

The Oberth effect. Why is leaving Earth from low orbit only 3.6 km/s when the excess speed needed is 2.94 km/s and escape from low orbit alone needs another 3.2 km/s? Because a burn deep in a gravity well, where the ship is already fast, adds more kinetic energy per km/s: \(\Delta(\tfrac12 v^2) = v\,\Delta v\). Energy needed: \(\tfrac12(v_\text{esc}^2 + v_\infty^2) - \tfrac12 v_\text{circ}^2\); at 7.7 km/s each km/s of \(\Delta v\) buys a lot of it. Burning at the lowest, fastest point is always best.

Worked examples

1. The departure burn, without Oberth. If a ship first escaped (3.20 km/s, from 7.73 to 10.93 km/s) and then, far away, added 2.94 km/s, the total would be 6.14 km/s. Doing it in one burn at the bottom: 3.59 km/s. The difference, 2.5 km/s, is the Oberth effect.

2. Capture at Mars. At 300 km, \(v_\text{esc} = 4.82\), \(v_\text{circ} = 3.41\) km/s. \(\sqrt{4.82^2 + 2.65^2} = 5.50\), so capture takes \(5.50 - 3.41 = 2.09\) km/s. Capturing into a long ellipse instead (periapsis 300 km, apoapsis near the Hill sphere) takes only \(5.50 - 4.82 \approx 0.7\) km/s; the orbit can then be lowered with aerobraking.

3. The Moon against Mars. From low Earth orbit to the Moon's surface: about 3.1 (to the Moon) + 0.8 (into lunar orbit) + 1.7 (landing) \(\approx\) 5.6 km/s, all by engine, since the Moon has no air. To Mars's surface: about 3.6 + 0.7 to 1 \(\approx\) 4.5 km/s by engine. In delta-v, Mars's surface is closer than the Moon's.

Try it

  1. Add up the budget for a one-way trip from low Earth orbit to Mars's surface, with the air doing the capture and most of the braking.
  2. Find the capture burn into a 500 km circular orbit round Mars (\(v_\infty = 2.65\) km/s).
  3. A ship with \(v_e = 3.7\) km/s must make the Mars ascent of 4.1 km/s. What mass ratio does it need?
  4. Explain in one sentence why burning at periapsis saves fuel.
  5. Why is a delta-v budget more useful than a distance for planning a trip?
Answers
  1. About \(3.6 + 0 + (0.7 \text{ to } 1) \approx 4.3\) to 4.6 km/s.
  2. \(r = 3889.5\) km: \(v_\text{circ} = 3.32\), \(v_\text{esc} = 4.69\) km/s. \(\sqrt{4.69^2 + 2.65^2} = 5.39\), so 2.07 km/s.
  3. \(e^{4.1/3.7} = e^{1.108} = 3.03\).
  4. The gain in kinetic energy is \(v\,\Delta v\), so the same \(\Delta v\) gives the most energy where \(v\) is largest.
  5. Fuel depends on changes of speed, not on distance, and coasting is free.

Common mistake

Adding \(v_\text{esc}\) and \(v_\infty\). Speeds combine through energy: \(v^2 = v_\text{esc}^2 + v_\infty^2\), not \(v = v_\text{esc} + v_\infty\). Adding them overestimates the burn by kilometres per second.

Where it leads