JEE Advanced 2024 · Paper 2 · Physics Q2

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PhysicsSingle correct+3 / -1Question 19 of 51

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 $V_c(r) = m\alpha/r^3$ , 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 $V_c(r)$ , 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.]

\dfrac{3\alpha}{GMr_0^2}

\dfrac{\alpha}{2GMr_0^2}

\dfrac{\alpha}{GMr_0^2}

\dfrac{2\alpha}{GMr_0^2}

Answer and worked solution

Official answer: A

The extra potential gives a radial force
$F_c=-\frac{dV_c}{dr}=\frac{3m\alpha}{r^4},$
which points outward (repulsive).

Orbit equations.

Before: $m\omega_0^2r_0=\dfrac{GMm}{r_0^2}$ , so $\omega_0^2=\dfrac{GM}{r_0^3}$ .
After: $m\omega_1^2r_0=\dfrac{GMm}{r_0^2}-\dfrac{3m\alpha}{r_0^4}$ , so $\omega_1^2=\dfrac{GM}{r_0^3}-\dfrac{3\alpha}{r_0^5}$ .

Since $T\propto\frac1\omega$ :
$\frac{T_1^2-T_0^2}{T_1^2}=1-\frac{\omega_1^2}{\omega_0^2}=\frac{3\alpha/r_0^5}{GM/r_0^3}=\frac{3\alpha}{GMr_0^2}.$

Topics:Satellites and orbital motion Potential energy and conservative forces Circular motion dynamics

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JEE Advanced 2024 · Paper 2, Physics question 2