First slide

Universal law of gravitation

Question

Three point masses each of mass m are at the corners of an equilateral triangle of side L. The system rotates about the centre of the triangle with the separation of masses not changing during rotation. If T be the time period of rotation, then

Difficult

A

$T \propto L^{3 / 2}$
B

$T \propto \sqrt{L}$
C

$T \propto M$
D

$T \propto \sqrt{M}$

Solution

The force of attraction between any two point masses is responsible for providing the necessary centripetal force to a mass to rotate in a circle of radius R From figure,
$\cos 30^{\circ} = \frac{(L / 2)}{R} = \frac{L}{2 R}$
or $R = \frac{L}{2 \cos 30^{\circ}} = \frac{L}{\sqrt{3}}$ …………..(1)
Now, ${MRω}^{2} = GMM / L^{2}$ ………….(2)
Substituting the value of R from eq. (1) in eq. (2), we get
$M (\frac{L}{\sqrt{3}}) ω^{2} = \frac{{GM}^{2}}{L^{2}}$
or $ω = {[\frac{\sqrt{3} GM}{L^{3}}]}^{1 / 2}$
Therefore, time period $T = \frac{2 π}{ω} = 2 π {[\frac{L^{3}}{\sqrt{3} GM}]}^{1 / 2}$
$∴ T \propto L^{3 / 2}$

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