A particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the $\mathrm{n}^{\text {th }}$ power of $R$. If the period of rotation of the particle is T , then :

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

(B) $T \propto R^{3 / 2}$ for any $n$.

(C) $T \propto R^{\frac{n}{2}+1}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Gravitation

A particle is moving with a uniform speed in a circular orbit of radius

R

in a central force inversely proportional to the

n^{th}

power of

R

. If the period of rotation of the particle is T , then :

Answer: (D)

A particle in a circular orbit needs a net inward force to supply its centripetal acceleration, so here the central force

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

plays exactly that role, giving

\frac{m v ^{2}}{R} = \frac{k}{R ^{n}}

. Solving for the speed,

v^{2} = \frac{k}{m} R^{1 - n}

, so

v \propto R^{(1 - n) /2}

. The period is the circumference divided by the speed,

T = \frac{2 π R}{v} \propto \frac{R}{R ^{(1 - n) /2}} = R^{1 - (1 - n) /2} = R^{(n + 1) /2}

, so

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

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option