(a) With what velocity does a body falling from outer space…

(a) With what velocity does a body falling from outer space reach the Earth, neglecting air resistance? (b) What is its velocity at a height $H = R$ above the Earth's surface, where $R$ is the Earth's radius?

Step 1: Idea

\"From outer space\" means from a very great distance, where the Earth's potential energy is practically zero, and starting (nearly) from rest. So the total energy is zero, and at any distance $r$ the kinetic energy equals the depth of the potential well there. This is the escape problem run backwards.

\[ \tfrac12mv^2 - \frac{\gamma Mm}{r} = 0, \]

\[ v(r) = \sqrt{\frac{2\gamma M}{r}}. \]

Step 2: (a) At the surface

With $r = R$ and $\gamma M = gR^2$: $v_2 = \sqrt{2gR}$.

\[ v_2 = \sqrt{2 \cdot 9.81 \cdot 6.37 \cdot 10^6}. \]

\[v_2 \approx 11.2\ \mathrm{km/s}\]

This is exactly the escape velocity (second cosmic velocity): a body falling from infinity arrives with the same speed that it would need to leave for infinity.

Step 3: (b) At height $H = R$

Now $r = R + H = 2R$, and comparing with part (a): $v = \sqrt{\frac{2\gamma M}{R + H}} = v_2\sqrt{\frac{R}{R + H}}$.

\[ v = \frac{v_2}{\sqrt2}. \]

\[v \approx 7.9\ \mathrm{km/s}\]

Speed grows as $1/\sqrt r$: halving the distance from the centre increases the speed by a factor of $\sqrt2$.

Check

At $r = 2R$: $ v $ $= \sqrt{2\gamma M/2R} $ $= \sqrt{gR} $ $= v_1 $, numerically the first cosmic velocity $\checkmark$