JEE MAIN - Physics (2004 - No. 50)
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
$${R^n}$$
$${R^{\left( {{{n - 1} \over 2}} \right)}}$$
$${R^{\left( {{{n + 1} \over 2}} \right)}}$$
$${R^{\left( {{{n - 2} \over 2}} \right)}}$$
Explanation
For moving a planet around the sun in the circular orbit,
The necessary centripetal force = Gravitational force exerted on it
$$\therefore$$ $${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$$
$$ \Rightarrow $$ $$v = \sqrt {{{GM} \over {{R^{n - 1}}}}} $$
We know, $$T = {{2\pi R} \over v}$$
$$ = 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}} $$
= $$2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}} $$
= $$2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$$
$$\therefore$$ $$T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$$
The necessary centripetal force = Gravitational force exerted on it
$$\therefore$$ $${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$$
$$ \Rightarrow $$ $$v = \sqrt {{{GM} \over {{R^{n - 1}}}}} $$
We know, $$T = {{2\pi R} \over v}$$
$$ = 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}} $$
= $$2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}} $$
= $$2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$$
$$\therefore$$ $$T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$$
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