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Two particles of masses m and M start from rest, separated by a very large distance, and are drawn together by their mutual gravitational attraction. When their separation has reduced to d, what is the speed of one particle relative to the other?

1. sqrt(G*(M + m)/(2*d))
2. sqrt(G*(M + m)/d)
3. sqrt(4*G*(M + m)/d)
4. sqrt(2*G*(M + m)/d)

Correct answer: sqrt(2*G*(M + m)/d)

Solution

Energy conservation: the PE released going from infinity to separation d is G*M*m/d, which equals total KE. With momentum conservation, m*v1 = M*v2, and relative speed v_rel = v1 + v2. The KE = (1/2)*mu*v_rel² where mu = mM/(M+m) is the reduced mass. So G*M*m/d = (1/2)*(mM/(M+m))*v_rel². Solving: v_rel² = 2*G*(M+m)/d, so v_rel = sqrt(2*G*(M+m)/d).

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