JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Gravitation

A particle of mass

m

, and angular momentum

ℓ

is moving in a circular orbit of radius

r_{0}

under the influence of an attractive force

F (r) = - \frac{k}{r ^{2}} \overset{r}{^}

. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance

δ r ≪ r_{0}

, due to which its radial distance varies periodically. The corresponding time period is:

Answer: (A)

Solution

In equilibrium:

\frac{k}{r _{0}^{2}} = \frac{m v _{0}^{2}}{r _{0}}

and

ℓ = m v_{0} r_{0}

, giving

r_{0} = \frac{ℓ ^{2}}{mk}

.

For a small radial displacement

d r

, conservation of angular momentum gives

v = v_{0} (1 - \frac{d r}{r _{0}})

.

The restoring force is

F_{R} = - \frac{k}{r _{0}^{3}} d r

, giving acceleration

a = - \frac{k}{m r _{0}^{3}} d r

.

This is SHM with

ω = \frac{k}{m r _{0}^{3}}

.

Time period:

T = \frac{2 π}{ω} = 2 π \frac{m r _{0}^{3}}{k} = 2 π \frac{m}{k} \cdot \frac{ℓ ^{6}}{m ^{3} k ^{3}} = \frac{2 π ℓ ^{3}}{m k ^{2}}

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option