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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?
- sqrt(G*(M + m)/(2*d))
- sqrt(G*(M + m)/d)
- sqrt(4*G*(M + m)/d)
- 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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