PRAYOGX
ADV-2026-P2-PHY-Q04

Orbit Wobble Lab — how fast does a nudged orbit wobble?

JEE Advanced 2026 · Paper 2 · Physics · Q.4 · single correct option (+3 / 0 / −1) · Gravitation

01The question

Q.4 A particle of mass m, and angular momentum ℓ is moving in a circular orbit of radius r0 under the influence of an attractive force F(r) = −kr2r̂. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance δr ≪ r0, due to which its radial distance varies periodically. The corresponding time period is:
  1. (A)2πℓ3mk2
  2. (B)2π√mk
  3. (C)2πℓ33mk2
  4. (D)2πℓ35mk2

02The experiment

Target:period of the radial oscillation Tr

Ready.

10.0 kg m² s⁻¹
50 N m²
1.00 kg
0.05

03Detailed solution

Answer:  T = 2πℓ3/(mk2)

  1. Radial motion with ℓ fixed. With the angular momentum conserved, the radial motion is that of a particle in the effective potential Ueff(r) = ℓ2/(2mr2) − k/r: the first term is the centrifugal barrier, the second the attraction.
  2. The circular orbit. It sits at the bottom of the well: dUeff/dr = −ℓ2/(mr3) + k/r2 = 0 gives r0 = ℓ2/(mk), the same as mv2/r = k/r2 with ℓ = mvr.
  3. Stiffness of the well. Ueff″(r0) = 3ℓ2/(mr04) − 2k/r03. Using ℓ2 = mkr0, this is 3k/r03 − 2k/r03 = k/r03.
  4. Small oscillation. A small displacement δr oscillates with ωr2 = Ueff″/m = k/(mr03), so T = 2π√(mr03/k).
  5. In terms of ℓ. With r03 = ℓ6/(m3k3): T = 2π√(ℓ6/(m2k4)) = 2πℓ3/(mk2): option (A).
  6. A check. The orbital period is 2πr0/v = 2πmr02/ℓ = 2πℓ3/(mk2) too. The radial wobble repeats exactly once per revolution, so the nudged orbit closes on itself: it is an ellipse with the centre at a focus, as Kepler's first law says.
  7. The other options. (B) 2π√(m/k) has the wrong units here (k is in N m², so m/k is in s2 m−3); it is the spring formula. (C) and (D) are a third and a fifth of the orbital period; the experiment shows the wobble takes one full revolution, not a third or a fifth of one.

Understand the concept

An orbit can be split into a turning motion and an in-and-out motion. Angular momentum fixes the turning part, and what is left behaves like a ball rolling in a valley, the effective potential. The circular orbit sits at the bottom of the valley; a small nudge makes the ball roll back and forth. For the inverse-square force the valley's stiffness is such that one back-and-forth takes exactly one revolution, so the farthest point comes back at the same place every time and the orbit is a closed ellipse. For any other power law the two periods differ and the farthest point slowly turns around the centre, tracing a rosette: switch the force law in the experiment to see it. Any inverse-square attraction behaves this way: a charged puck around a charged sphere (as in the experiment, where one orbit takes about two and a half seconds) or a planet around the Sun.

Key takeaways

  • Effective potential: Ueff = ℓ2/(2mr2) + U(r); circular orbit at its minimum.
  • Small radial oscillations: ω2 = Ueff″(r0)/m.
  • Inverse-square force: r0 = ℓ2/(mk) and Tradial = Torbit = 2πℓ3/(mk2): the orbit closes.
  • Check the units of every option: 2π√(m/k) is a time only when k is a spring constant.

Verification: an independent derivation; a symbolic second derivative of the effective potential; a high-accuracy numerical integration of the orbit, independent of the page, whose radial period approaches the formula as the nudge shrinks; the Kepler ellipse check (the exact period for a finite nudge); the units and the values of every option; and the page's own engine, run by tests/verify_p2phyq04.py. The MathonGo solution was read first only for the approach. The official final answer key (JEE Advanced 2026, Paper 2), read last, gives (A), in agreement. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

How to use this simulation

  1. Press ▶ START EXPERIMENT. The stopwatch starts; the puck circles the charged sphere once at r0, then it is nudged outward by δr, keeping its angular momentum.
  2. Everything runs in real time: one second on the stopwatch is one second for the puck. Watch the distance r rise and fall; the farthest points are marked on the table and the graph, and the stopwatch shows each lap.
  3. The period is measured between successive farthest points and compared with the four options; the one that matches is found.
  4. Then change ℓ, k, m or the nudge, or try What if the force were… K/r1.5 or K/r2.5 to see the orbit stop closing.
  5. Replay runs it again. Reset clears it. Classroom fills the screen for a projector.