A particle of mass $m$ is under the influence of the gravitational field of a body of mass $M(\gg m)$. The…

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 $V_{\mathrm{c}}(r)=m \alpha / r^3$, 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 $V_{\mathrm{c}}(r)$, but with a new time period $T_1$, then $\left(T_1^2-T_0^2\right) / T_1^2$ is given by [ $G$ is the gravitational constant.]
  1. $\frac{3 \alpha}{G M r_0^2}$
  2. $\frac{\alpha}{2 G M r_0^2}$
  3. $\frac{\alpha}{G M r_0^2}$
  4. $\frac{2 \alpha}{G M r_0^2}$

Solution

$\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{\mathrm{r}_0^2} \\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{\mathrm{r}_0^2}-\frac{3 \mathrm{~m} \alpha}{\mathrm{r}_0^4} \\ & \frac{\omega_1^2}{\omega_0^2}=\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\frac{\mathrm{GM}}{\mathrm{r}_0^2}-\frac{3 \alpha}{\mathrm{r}_0^4}}{\frac{\mathrm{GM}}{\mathrm{r}_0^2}} \\ & \frac{\mathrm{T}_0^2}{\mathrm{~T}_1^2}=1-\frac{3 \alpha}{\mathrm{GMr}_0^2} \\ & \frac{\mathrm{T}_1^2-\mathrm{T}_0^2}{\mathrm{~T}_1^2}=\frac{3 \alpha}{\mathrm{GMr}_0^2}\end{aligned}$

Asked in: JEE Advanced 2024 (Paper 2)

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