Suppose the gravitational force varies inversely as the  nth power of distance. Then the time period of a planet in circular orbit of radius  R  around the sun will be proportional to :

Answers (1)

For motion of a planet in circular orbit,

Centripetal force = Gravitational force

\therefore \; \; mR\omega ^{2}=\frac{GMm}{R^{n}}\; \; or\; \; \omega =\sqrt{\frac{GM}{R^{n+1}}}

\therefore \; \; \; T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R^{n+1}}{GM}}=\frac{2\pi }{\sqrt{GM}}R^{\left ( \frac{n+1}{2} \right )}

\therefore \; \; T\; is\; proportional\; to\; R^{\left ( \frac{n+1}{2} \right )}

 

 

Newton's Law of Gravitation -

F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}

F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}

F\rightarrow Force    

G\rightarrow Gravitalional constant

m_1,m_2\rightarrow  Masses

r\rightarrow  Distance between masses

- wherein

Force is along the line joining the two masses

 

m\omega ^2R = \frac{Gm_{m}}{R^{n}}

W = \sqrt{\frac{Gm}{R^{n+1}}}

\therefore T= \frac{2\pi }{w} = 2\pi \sqrt{\frac{R^{n+1}}{Gm}}

= \frac{2\pi }{\sqrt{Gm}}R

T\ is\ Proportional\ to\ R \left ( \frac{n+1}{2} \right )

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