Escape Velocity Comparator

Find the speed at which something thrown from Earth never comes back — then use it to explain why Earth still has nitrogen, is slowly losing its hydrogen, and why the Moon has no air at all. Earth: M = 5.972 × 10²⁴ kg, R = 6371 km.

1. “Never comes back” is a statement about energy, not about height. Gravity’s pull weakens with distance but never switches off, so the condition is that kinetic energy is at least as large as the depth of the gravitational well the object is sitting in.

   $$\frac{1}{2}m{v}^2-\frac{GMm}{R}\ge 0$$

2. Set the total to zero — the marginal case, arriving at infinity with nothing left — and solve. The projectile’s own mass cancels, which is the first surprising thing: a bullet and a boulder need exactly the same speed.

   $$v_{\text{esc}}=\sqrt{\frac{2GM}{R}}$$

3. Put Earth’s numbers in. This is the figure the comparator above prints when Earth is selected, and it is a speed, not a thrust: nothing here says how you reach it.

   $$v_{\text{esc}}=\sqrt{\frac{2(6.674\times 10^{-11})(5.972\times 10^{24})}{6.371\times 10^6}}=11186m/s\approx 11.2km/s$$

4. Now use the fact that m cancelled. It applies to a nitrogen molecule as much as to a rocket. Gas molecules do not all move at one speed, but the root-mean-square speed follows from temperature alone.

   $$v_{\text{rms}}=\sqrt{\frac{3kT}{m}}:{\mathrm{H}}_2\text{at}288\mathrm{K}=1.90km/s,{\mathrm{N}}_2=0.51km/s$$

5. A gas is not lost the moment a molecule exceeds escape speed — the fast tail of the distribution leaks away over billions of years. The standard rule of thumb is that a world keeps a gas if escape speed is at least about six times the molecular speed. The Moon’s row is at its own temperature: with no air to spread the heat, its day side reaches about 390 K, which lifts nitrogen to 0.59 km/s.

   $$\frac{v_{\text{esc}}}{v_{\text{rms}}}:\text{Earth},{\mathrm{N}}_2=22\text{Earth},{\mathrm{H}}_2=5.9\text{Moon},{\mathrm{N}}_2=4.0$$

11.2 km/s escapes Earth. It does not escape the Sun: it leaves you in the Sun's grip at Earth's distance, orbiting alongside the planet you just left. Work out what leaving the Solar System actually costs.

1. Start with the figure on the panel, from the energy balance that produced it: kinetic energy exactly cancelling the depth of the gravitational well.

   $$v_{\mathrm{esc}}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.674\times 10^{-11})(5.972\times 10^{24})}{6.371\times 10^6}}=11.19\mathrm{km}{\mathrm{s}}^{-1}$$

2. Now change the centre you are measuring from. Earth itself is falling around the Sun, at a speed fixed by the same law one astronomical unit out.

   $$v_{\oplus }=\sqrt{\frac{G{M}_{\odot }}{a}}=\sqrt{\frac{(6.674\times 10^{-11})(1.989\times 10^{30})}{1.496\times 10^{11}}}=29.79\mathrm{km}{\mathrm{s}}^{-1}$$

3. Escaping the Sun from that distance takes √2 times the circular speed — the identical relation as before, because the arithmetic does not care which body has hold of you.

   $$v_{\mathrm{esc},\odot }(1\mathrm{AU})=\sqrt{2}{v}_{\oplus }=1.4142\times 29.79=42.13\mathrm{km}{\mathrm{s}}^{-1}$$

4. Subtract, and notice what you are being given. You are already travelling at 29.79 and you need 42.13, so only the difference is missing — and you collect it for free by launching along the direction Earth is already moving.

   $$\Delta v=42.13-29.79=12.34\mathrm{km}{\mathrm{s}}^{-1}$$

5. But you still have to climb out of Earth, and it is energies that add, not speeds. That is why the two figures combine as squares: the speed needed at the surface is the hypotenuse, not the sum.

   $$\frac{1}{2}{v}^2-\frac{GM}{R}=\frac{1}{2}\Delta v^2⟹v=\sqrt{v_{\mathrm{esc}}^2+\Delta v^2}=\sqrt{11.19^2+12.34^2}=16.65\mathrm{km}{\mathrm{s}}^{-1}$$

6. Now turn around and launch against Earth's motion. The two speeds stop cancelling and start adding.

   $$v_{\mathrm{retro}}=\sqrt{11.19^2+(42.13+29.79{)}^2}=\sqrt{125.2+5172.5}=72.79\mathrm{km}{\mathrm{s}}^{-1}$$