Worked exercise: two asteroids falling into each other

Problem

Two asteroids of mass $m_1 = 6\cdot 10^{24}\ \text{kg}$ (like the Earth) and $m_2 = 10^{10}\ \text{kg}$ are initially at rest at a distance $d_A = 10^9\ \text{m}$. Find their speed when they meet ($d_B = R_\text{Terra} = 6{,}4\cdot 10^6\ \text{m}$), assuming both bodies are spherical and that $m_1$ is essentially stationary.

Solution

Conservation of mechanical energy. Since $m_1 \gg m_2$, the centre of mass remains approximately at $m_1$, which we can treat as fixed; all the kinetic energy goes to $m_2$. Between the initial state A (at rest, distance $d_A$) and the final state B (contact, distance $d_B$): $\underbrace{0}_{E_\text{cin,A}} - \frac{Gm_1 m_2}{d_A} = \tfrac{1}{2}m_2 v_B^2 - \frac{Gm_1 m_2}{d_B}$

Cancelling $m_2$ and solving for $v_B$: $v_B = \sqrt{2Gm_1\left(\frac{1}{d_B} - \frac{1}{d_A}\right)} = \sqrt{\frac{2\cdot 6{,}67\cdot 10^{-11}\cdot 6\cdot 10^{24}}{6{,}4\cdot 10^6}\cdot\left(1 - \frac{6{,}4\cdot 10^6}{10^9}\right)}$

The term $6{,}4\cdot 10^6/10^9 \approx 0{,}006$ is negligible: starting from such a distant $d_A$ is almost equivalent to starting from infinity. $v_B \approx \sqrt{1{,}25\cdot 10^8}$

$\ev{v_B \approx 11{,}2\ \text{km/s}}$

This is the same order of magnitude as the escape velocity from the Earth, as expected: starting far away and falling to the surface is the reverse process of escaping.

Links

Topics: Gravitation Concepts: Conservation of mechanical energy · Universal gravitational potential energy · Escape velocity Skills: Conservation of energy