JEE Advanced 2026 Paper 2, Physics Section 1 Q4: Satellite, Planetary Motion And Kepler’s Law

JEE Advanced2026Paper 2Physics Section 1

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:

A
$\frac{2 π ℓ ^{3}}{m k ^{2}}$
B
$2 π \frac{m}{k}$
C
$\frac{2 π ℓ ^{3}}{3 m k ^{2}}$
D
$\frac{2 π ℓ ^{3}}{5 m k ^{2}}$

Solution

Step 1 — Circular orbit condition. Centripetal balance gives $\frac{k}{r _{0}^{2}} = \frac{m v _{0}^{2}}{r _{0}}$ , and angular momentum $ℓ = m v_{0} r_{0}$ yields $v_{0} = \frac{ℓ}{m r _{0}}$ .

Combining: $r_{0} = \frac{ℓ ^{2}}{mk}$ .

Step 2 — Effective potential approach. With $ℓ$ fixed, the effective potential is $U_{eff} (r) = - \frac{k}{r} + \frac{ℓ ^{2}}{2 m r ^{2}}$ . Small radial oscillations about $r_{0}$ obey $m \overset{x}{¨} = - U_{eff}^{''} (r_{0}) x$ where $x = δ r$ .

Step 3 — Compute $U_{eff}^{''} (r_{0})$ .

$U_{eff}^{''} (r) = - \frac{2 k}{r ^{3}} + \frac{3 ℓ ^{2}}{m r ^{4}}$ . At $r = r_{0}$ , using $ℓ^{2} = mk r_{0}$ , $U_{eff}^{''} (r_{0}) = - \frac{2 k}{r _{0}^{3}} + \frac{3 k}{r _{0}^{3}} = \frac{k}{r _{0}^{3}}$ .

Step 4 — Period. $ω^{2} = \frac{k}{m r _{0}^{3}}$ , so

$T = 2 π \frac{m r _{0}^{3}}{k} = 2 π \frac{m}{k} \cdot \frac{ℓ ^{6}}{m ^{3} k ^{3}} = \frac{2 π ℓ ^{3}}{m k ^{2}}$ , matching option (A).

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JEE Advanced 2026 Paper 2, Physics Section 1 Q4: Satellite, Planetary Motion And Kepler’s Law

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