First slide

Gravitaional potential energy

Question

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

Easy

A

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

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

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

$(\frac{4 Gm}{R})$

Solution

$\begin{array}{l} \frac{Gmm}{(2 R)^{2}} = \frac{{mv}^{2}}{R} \\ ∴ v^{2} = \frac{Gm}{4 R} or v = \frac{1}{2} \sqrt{(\frac{Gm}{R})} \end{array}$

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