2008: Problem 18

Topic: Gravity
Concepts: Conservation of energy, Gravitational potential energy


Solution:

Recall the gravitational potential energy between two particles of masses mm and MM separated by distance rr is

U=GMmrU=-\frac{GMm}{r}

In our case, the particle starts very far away so there is no potential energy in the beginning,

Ui=0U_i=0

The particle maximizes its speed when it minimizes its potential energy. This occurs when the particle is closest to the ring i.e. located at the center. At this point, the distance between the particle and all parts of the ring is the radius RR so the potential energy is

Uf=GMmRU_f=-\frac{GMm}{R}

where MM is the mass of the ring. By conservation of energy,

Ui=K+UfU_i=K+U_f
0=12mv2GMmR0=\frac{1}{2}mv^2-\frac{GMm}{R}

so we have

v=2GMR=4πGλv=\sqrt{\frac{2GM}{R}}=\sqrt{4\pi G\lambda}

using M=2πRλM=2\pi R\lambda with λ\lambda being the linear mass density. Since the maximum speed is independent of radius, the answer is C.

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