JEE Advanced 2024 Paper 2 Q02 Physics Gravitation Newton's Law of Gravitation Hard

JEE Advanced 2024 Paper 2 · Q02 · Newton's Law of Gravitation

A particle of mass $m$ is under the influence of the gravitational field of a body of mass $M\,(\gg m)$. The particle is moving in a circular orbit of radius $r_0$ with time period $T_0$ around the mass $M$. Then, the particle is subjected to an additional central force, corresponding to the potential energy $V_1(r) = m\alpha/r^3$, where $\alpha$ is a positive constant of suitable dimensions and $r$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius $r_0$ in the combined gravitational potential due to $M$ and $V_1(r)$, but with a new time period $T_1$, then $(T_1^2 - T_0^2)/T_1^2$ is given by

[$G$ is the gravitational constant.]

  1. A. $\dfrac{3\alpha}{GMr_0^2}$
  2. B. $\dfrac{\alpha}{2GMr_0^2}$
  3. C. $\dfrac{\alpha}{GMr_0^2}$
  4. D. $\dfrac{2\alpha}{GMr_0^2}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

For circular orbit of radius $r_0$ under gravity alone: $\dfrac{GMm}{r_0^2} = m\omega_0^2 r_0$, giving $\omega_0^2 = \dfrac{GM}{r_0^3}$, so $T_0^2 = \dfrac{4\pi^2 r_0^3}{GM}$.

The additional potential $V_1(r) = m\alpha/r^3$ contributes a radial force $F_1 = -\dfrac{dV_1}{dr} = \dfrac{3m\alpha}{r^4}$ (attractive). At $r = r_0$, the net inward force is $\dfrac{GMm}{r_0^2} + \dfrac{3m\alpha}{r_0^4} = m\omega_1^2 r_0$.

Hence $\omega_1^2 = \dfrac{GM}{r_0^3} + \dfrac{3\alpha}{r_0^5}$, so $T_1^2 = \dfrac{4\pi^2}{\omega_1^2}$.

Then $\dfrac{T_1^2 - T_0^2}{T_1^2} = 1 - \dfrac{T_0^2}{T_1^2} = 1 - \dfrac{\omega_1^2 - 3\alpha/r_0^5\cdot (r_0^3/GM)}{\omega_1^2}\cdot\dfrac{\omega_1^2}{\omega_1^2}$.

Writing $\dfrac{T_1^2 - T_0^2}{T_1^2} = \dfrac{\omega_0^{-2} - \omega_0^{-2}\cdot\omega_0^2/\omega_1^2 \cdot }{}$ — equivalently, since $T^2 \propto 1/\omega^2$:

$\dfrac{T_1^2 - T_0^2}{T_1^2} = 1 - \dfrac{\omega_1^2}{\omega_0^2}\cdot\dfrac{T_1^2}{T_0^2}\cdot\dfrac{T_0^2}{T_1^2}$. Cleaner: $\dfrac{T_0^2}{T_1^2} = \dfrac{\omega_1^2}{\omega_0^2} = 1 + \dfrac{3\alpha}{GM r_0^2}$.

So $\dfrac{T_1^2 - T_0^2}{T_1^2} = 1 - \dfrac{1}{1 + \frac{3\alpha}{GM r_0^2}}$. For $\alpha$ small (weak pert…[truncated]

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