JEE Main 2024 · 1 February · evening · Single correct · +4 / −1

A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R^(-3 / 2) then choose the correct option :

  1. (a) T² ∝ R^(7 / 2)
  2. (b) T² ∝ R³
  3. (c) T² ∝ R^(5 / 2)
  4. (d) T² ∝ R^(3 / 2)
Answer and solution

Answer (c) T² ∝ R^(5 / 2)

Solution

Physics sir · 3 steps

Gravity here supplies the centripetal force, so set the given force equal to mω² R and solve for T.

  1. Force balance. In a circular orbit the attraction towards the star provides the centripetal force. Write the given force as F = kR^(-3/2), where k is a constant.

    kR^(-3/2) = mω² R = m(2π/T)² R

  2. Collect powers. Rearranging gives T² = (4π² m)/k · R/R^(-3/2) = (4π² m)/k R^(1 + 3/2). Everything in front of R is constant.
  3. So. T² ∝ R^(5/2), which is option (c).

The idea to keep. If the central force varies as R⁻ⁿ, then T² ∝ Rⁿ⁺¹. For ordinary gravity n = 2 gives Kepler's T² ∝ R³; change n and the power changes with it.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 160 times in JEE Main since 2020. This question is from 2024.