JEE Main · 2004 · 24 Apr 2004 · Offline · Gravitation

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Suppose the gravitational force varies inversely as the nth power of distance. Then the time period planet in circular orbit of radius

R

around the sun will be proportional to

(A) $R^{(\frac{n + 1}{2})}$ ✓ Correct
(B) $R^{(\frac{n - 1}{2})}$
(C) $R^{n}$
(D) $R^{(\frac{n - 2}{2})}$

Answer: (A)

Solution

If the attractive force varies as

\frac{k}{R ^{n}}

, then for a circular orbit it must supply the centripetal force,

\frac{k}{R ^{n}} = m ω^{2} R = m \frac{4 π ^{2}}{T ^{2}} R

. Rearranging,

T^{2} = \frac{4 π ^{2} m}{k} R^{n + 1}

, so

T^{2} \propto R^{n + 1}

and hence

T \propto R^{(n + 1) /2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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