JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Gravitation

A light planet is revolving around a massive star in a circular orbit of radius

R

with a period of revolution

T

. If the force of attraction between planet and star is proportional to

R^{- 3/2}

then choose the correct option :

(A) $T^{2} \propto R^{5/2}$ ✓ Correct
(B) $T^{2} \propto R^{7/2}$
(C) $T^{2} \propto R^{3/2}$
(D) $T^{2} \propto R^{3}$

Answer: (A)

Solution

For a circular orbit, the attractive force supplies the centripetal force:

\frac{GM m}{R ^{3/2}} = m ω^{2} R

, which gives

ω^{2} \propto R^{- 5/2}

. Since

T = \frac{2 π}{ω}

T^{2} = \frac{4 π ^{2}}{ω ^{2}} \propto R^{5/2}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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