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Escape speed

Gravity and orbits · level 7, lesson 2 of 6

A question to start

In level 4 you read that escape speed is 11.2 km/s for Earth and 5.0 km/s for Mars. Where do those numbers come from, and how much more must a ship already in orbit round Mars add to leave it?

The idea

A ship leaving with speed \(v\) from distance \(r\) has total energy \(\tfrac12 mv^2 - GMm/r\). Coasting, it slows as it climbs, but its energy stays the same. To reach "infinity" with nothing to spare, it must arrive with zero kinetic energy where \(U\) is also zero: total energy zero. So

\[\tfrac12 m v_\text{esc}^2 = \frac{GMm}{r} \qquad\Rightarrow\qquad v_\text{esc} = \sqrt{\frac{2GM}{r}}.\]

Comparing with \(v_\text{circ} = \sqrt{GM/r}\):

\[v_\text{esc} = \sqrt2\, v_\text{circ}.\]

Escaping takes only 41% more speed than orbiting at the same distance.

Mars Earth Moon Phobos
\(GM\) (km³/s²) 42,828 398,600 4,905 0.00071
Radius (km) 3,389.5 6,371 1,737 11.1
Escape speed from the surface 5.03 km/s 11.19 km/s 2.38 km/s 11 m/s
Low orbit speed 3.55 km/s 7.91 km/s 1.68 km/s 8 m/s

(The low-orbit speeds are for an orbit skimming the surface, \(r = R\).)

From orbit. A ship in a 300 km circle round Mars is already going 3.41 km/s. Escape speed there is \(\sqrt2 \times 3.41 = 4.82\) km/s. So it needs to add only 1.41 km/s, fired along its direction of motion.

Leaving with speed to spare. If a ship leaves faster than escape speed, it keeps some speed when it is far away, called the hyperbolic excess speed \(v_\infty\). Energy gives

\[v^2 = v_\text{esc}^2 + v_\infty^2.\]

Lesson 5 uses this for leaving the Earth on the way to Mars.

The direction does not matter for escaping (ignoring air and the ground in the way): energy has no direction. But for going somewhere particular afterwards, it matters a lot.

Worked examples

1. Mars from the surface. \(\sqrt{2 \times 42828 / 3389.5} = \sqrt{25.27} = 5.03\) km/s.

2. Leaving Earth for Mars. A ship needs \(v_\infty = 2.94\) km/s. From a 300 km orbit (\(v_\text{esc} = 10.93\) km/s, \(v_\text{circ} = 7.73\) km/s) it must reach \(\sqrt{10.93^2 + 2.94^2} = 11.32\) km/s: a burn of \(11.32 - 7.73 = 3.59\) km/s.

3. A rock knocked off Mars. Meteorites from Mars have been found on Earth. They were thrown off by big impacts at more than 5 km/s. Escaping Earth's gravity would take more than twice that, which is one reason we have Mars meteorites but Mars probably has few from Earth.

Try it

  1. Find the escape speed from the surface of Venus (\(GM = 324{,}860\) km³/s², \(R = 6{,}052\) km).
  2. A ship is in a circle 1,000 km above Mars. How much must it add to escape?
  3. Show that \(v_\text{esc} = \sqrt2\,v_\text{circ}\).
  4. A probe leaves Mars at 6 km/s from the surface. What is its speed when far away?
  5. Phobos's escape speed is 11 m/s. Could an astronaut jump off it?
Answers
  1. \(\sqrt{2 \times 324860 / 6052} = \sqrt{107.4} = 10.4\) km/s.
  2. \(v_\text{circ} = 3.12\) km/s; \(v_\text{esc} = 4.42\) km/s; add 1.29 km/s.
  3. \(\sqrt{2GM/r} = \sqrt2\sqrt{GM/r}\).
  4. \(v_\infty = \sqrt{36 - 25.3} = 3.27\) km/s.
  5. A standing jump on Earth leaves at about 2.5 to 3 m/s, so not from a standing jump; a running jump at 11 m/s would be beyond an athlete. But a hard throw of a ball (25 m/s) would leave Phobos.

Common mistake

Thinking escape needs the speed all the way. Escape speed is the speed at the start, for a ship coasting with no more push. A rocket that keeps firing can leave slowly; it just uses more fuel.

Where it leads