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# How to solve this problem- 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 :

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 :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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As we learnt in

For motion of a planet in circular orbit,

Centripetal force = Gravitational force

$\dpi{100} \therefore \; \; mR\omega ^{2}=\frac{GMm}{R^{n}}\; \; or\; \; \omega =\sqrt{\frac{GM}{R^{n+1}}}$

$\dpi{100} \therefore \; \; \; T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R^{n+1}}{GM}}=\frac{2\pi }{\sqrt{GM}}R^{\left ( \frac{n+1}{2} \right )}$

$\dpi{100} \therefore \; \; T\; is\; proportional\; to\; R^{\left ( \frac{n+1}{2} \right )}$

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

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

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

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

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

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

Option 1)

Correct

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Incorrect

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