Escape Velocity

Energy decides escape

Launch a body of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\), with speed \(v\). Ignore the air. After launch only gravity acts, and gravity is conservative. So the total energy \(E = K + U\) stays constant along the whole flight.

At the surface \(E = \tfrac12 mv^2 - GMm/R\). Far away \(U\) approaches zero, so there \(E\) is all kinetic energy, which cannot be negative. So the body can reach infinity only if \(E \ge 0\). The least launch speed that does it makes \(E = 0\) exactly: the body arrives at infinity with no speed to spare. This least speed is the escape speed \(v_e\), usually called the escape velocity.

For the Earth, \(v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = 1.12 \times 10^4\,\text{m/s}\), that is \(11.2\,\text{km/s}\). That is about \(40\,000\,\text{km/h}\), some forty times the speed of a passenger jet. For the Moon, with a mass of \(7.35 \times 10^{22}\,\text{kg}\) and a radius of \(1.74 \times 10^6\,\text{m}\), the same formula gives only \(2.4\,\text{km/s}\).

The well, drawn

In the chapter on Potential Energy and Conservation of Energy you read energy diagrams. A curve shows \(U\), a level line shows \(E\), and the gap between them is \(K\). Draw one for a body and the Earth. The curve \(U = -GMm/r\) starts at \(-GMm/R\) at the surface and rises towards zero far away, but never reaches it. It is a well with a sloping wall and no rim, the gravitational well.

Set \(E = -0.5\,GMm/R\) and press play. The body leaves the surface with the gap \(K\) between the line and the curve. As it rises, the gap closes. At \(r = 2R\) the line meets the curve, so \(K = 0\), and the body starts to fall. A negative total energy always meets the curve somewhere, at \(r = GMm/|E|\). So the body is bound: it can never go beyond that distance.

Now raise \(E\) to zero or above. The curve approaches zero only from below, so the line never meets it. The gap \(K\) never closes, and the body climbs out for ever. It is unbound. The dividing line is \(E = 0\), and on the Earth that needs exactly \(11.2\,\text{km/s}\) at launch.

For a body launched straight up with \(v < v_e\), energy gives the greatest height \(h\) at once. Put \(\tfrac12 mv^2 - GMm/R = -GMm/(R + h)\) and use \(v_e^2 = 2GM/R\). The result is \(h = Rv^2/(v_e^2 - v^2)\). At \(v = 7.9\,\text{km/s}\), where \(v^2 = v_e^2/2\), the body rises to \(h = R\), a height of \(6400\,\text{km}\).

Mass and direction do not matter

In the derivation, the mass \(m\) of the body cancelled. Both its kinetic energy and its potential energy are proportional to \(m\). So a pebble and a loaded rocket need the same \(11.2\,\text{km/s}\). The rocket needs far more energy, of course, but the same speed.

Energy is a scalar, so the direction of launch never entered either. A body thrown upward, sideways or at a slant with the same speed has the same total energy. As long as its path does not strike the ground, it escapes. The paths differ in shape, but every one of them climbs away for ever.

A launch faster than \(v_e\) leaves some kinetic energy over at infinity. There \(U = 0\), so \(\tfrac12 mv_\infty^2 = \tfrac12 mv^2 - \tfrac12 mv_e^2\). This gives \(v_\infty = \sqrt{v^2 - v_e^2}\). A launch at \(15\,\text{km/s}\) keeps only \(\sqrt{225 - 125} = 10\,\text{km/s}\) far from the Earth. The speed on the way out is lost much faster than a simple subtraction of \(11.2\) suggests.

Why the Moon has no air

The molecules of a gas move in all directions at many speeds. Oxygen molecules at \(120\,°\text{C}\), the temperature of the Moon's surface at noon, have a typical speed of about \(0.55\,\text{km/s}\). A very small fraction of them move three or four times faster than that. The escape speed does not depend on what is moving. So a molecule at the top of an atmosphere that exceeds \(v_e\) and heads upward is lost for ever.

The fast molecules that leave are replaced, as collisions give a new few a high speed. So an atmosphere leaks slowly, and the leak is fast when \(v_e\) is only a few times the typical molecular speed. A useful rule of thumb follows. A planet keeps a gas for billions of years only if its \(v_e\) exceeds about six times the gas's typical speed.

The hook asked for the speed at which a thrown body never returns. From the Earth's surface it is \(11.2\,\text{km/s}\), whatever the body and whatever the direction. The Moon, with \(v_e = 2.4\,\text{km/s}\), cannot hold even the slow-moving molecules of oxygen for long, and so its sky is black by day.