First slide

Motion of planets and satellites

Question

Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Moderate

A

$v = \frac{1}{2 R} \sqrt{(\frac{1}{Gm})}$
B

$v = \sqrt{(\frac{Gm}{2 R})}$
C

$v = \frac{1}{2} \sqrt{(\frac{Gm}{R})}$
D

$v = \sqrt{(\frac{4 Gm}{R})}$

Solution

$\begin{array}{l} F_{G} = \frac{{Gm}^{2}}{4 R^{2}} \Rightarrow \frac{m v^{2}}{R} = \frac{{Gm}^{2}}{4 R^{2}} \\ ∴ v = \frac{1}{2} \sqrt{\frac{Gm}{R}} \end{array}$

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