Question 41: Gravitation

Question

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.]

Accepted Answer

To find the effect of an additional central force on the time period of a particle in a circular orbit, consider the following:
Initial Gravitational Force ($F_1$):
The force due to gravity for a particle in a circular orbit of radius $r_0$ is given by:
$    F_1 = \frac{GMm}{r_0^2}    $
where $ G $ is the gravitational constant, $ M $ is the mass of the central body, and $ m $ is the mass of the particle.
Modified Force with Additional Potential ($F_2$):
When an additional force derived from the potential energy $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $ acts on the particle, the effective central force becomes:
$    F_2 = \frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4}    $
Relation of Angular Velocities:
The ratio of squares of the angular velocities ($\omega_0$ and $\omega_1$) before and after the additional force is:
$    \frac{\omega_1^2}{\omega_0^2} = \frac{F_2}{F_1} = \frac{\frac{GM}{r_0^2} - \frac{3\alpha}{r_0^4}}{\frac{GM}{r_0^2}}    $
Relating Time Periods:
Since the angular velocity is inversely proportional to the time period, $\frac{\omega_1^2}{\omega_0^2} = \frac{T_0^2}{T_1^2}$. Thus, we have:
$    \frac{T_0^2}{T_1^2} = 1 - \frac{3\alpha}{GMr_0^2}    $
Find the Change in Time Period:
The expression for the change in the square of the time periods is:
$    \frac{T_1^2 - T_0^2}{T_1^2} = \frac{3\alpha}{GMr_0^2}    $
This shows how the additional central force, represented by the potential $ V_{\mathrm{c}}(r) = \frac{m\alpha}{r^3} $, affects the particleÃ¢ÂÂs orbital period when combined with the gravitational field of the mass $ M $.

Answer

$\frac{3 \alpha}{G M r_0^2}$

Answer

$\frac{\alpha}{2 G M r_0^2}$

Answer

$\frac{\alpha}{G M r_0^2}$

Answer

$\frac{2 \alpha}{G M r_0^2}$

Gravitation

Choose the correct answer: