JEE Advanced 2026 · Paper 2 · Physics Q4

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PhysicsSingle correct+3 / -1Question 22 of 54

A particle of mass $m$ , and angular momentum $\ell$ is moving in a circular orbit of radius $r_0$ under the influence of an attractive force $\vec F(r)=-\dfrac{k}{r^2}\hat r$ . Keeping its angular momentum unchanged, the particle is displaced radially by a small distance $\delta r\ll r_0$ , due to which its radial distance varies periodically. The corresponding time period is:

\dfrac{2\pi\ell^3}{mk^2}

2\pi\sqrt{\dfrac mk}

\dfrac{2\pi\ell^3}{3mk^2}

\dfrac{2\pi\ell^3}{5mk^2}

Answer and worked solution

Official answer: A

Circular orbit. $\dfrac{k}{r_0^2}=\dfrac{mv^2}{r_0}$ and $\ell=mvr_0$ give $r_0=\dfrac{\ell^2}{mk}$ .

Radial oscillation. With $U_{\text{eff}}=-\dfrac kr+\dfrac{\ell^2}{2mr^2}$ ,
$U_{\text{eff}}''(r_0)=-\frac{2k}{r_0^3}+\frac{3\ell^2}{mr_0^4}=\frac{k}{r_0^3}\quad(\text{using }\ell^2=mkr_0).$
So $\omega_r^2=\dfrac{k}{mr_0^3}=\dfrac{m^2k^4}{\ell^6}$ , and
$T=\frac{2\pi}{\omega_r}=\frac{2\pi\ell^3}{mk^2}.$
(For an inverse-square force the radial period equals the orbital period.)

Topics:Satellites and orbital motion Simple harmonic motion Angular momentum and its conservation

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