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.]

Gravitation PYQ — JEE Advanced 2024 · Shift 2 · Q36 with Solution

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Correct answer: A

Given circular motion under gravity alone

For a circular orbit of radius $r_0$ around mass $M$ , the gravitational force provides the centripetal force:

\frac{m v_0^2}{r_0} = \frac{G M m}{r_0^2}

So,

v_0^2 = \frac{G M}{r_0}

The angular speed is

\omega_0 = \frac{v_0}{r_0}, \qquad \omega_0^2 = \frac{G M}{r_0^3}

Hence,

T_0 = \frac{2\pi}{\omega_0}, \qquad T_0^2 = \frac{4\pi^2 r_0^3}{G M}

Additional central potential

The extra potential energy is

V_c(r) = \frac{m\alpha}{r^3}

The corresponding force is

F_c(r) = -\frac{dV_c}{dr}

Differentiate:

\frac{d}{dr}\left(\frac{m\alpha}{r^3}\right) = -\frac{3m\alpha}{r^4}

Therefore,

F_c(r) = -\left(-\frac{3m\alpha}{r^4}\right) = \frac{3m\alpha}{r^4}

This force is radially outward.

So the net inward force available for centripetal motion at radius $r_0$ is

\frac{G M m}{r_0^2} - \frac{3m\alpha}{r_0^4}

New circular orbit condition

Since the particle still moves in the same circular orbit of radius $r_0$ , but with new speed $v_1$ and period $T_1$ ,

\frac{m v_1^2}{r_0} = \frac{G M m}{r_0^2} - \frac{3m\alpha}{r_0^4}

Cancelling $m$ :

v_1^2 = \frac{G M}{r_0} - \frac{3\alpha}{r_0^3}

Thus,

\omega_1^2 = \frac{v_1^2}{r_0^2} = \frac{G M}{r_0^3} - \frac{3\alpha}{r_0^5}

Since $T = \frac{2\pi}{\omega}$ ,

T_1^2 = \frac{4\pi^2}{\omega_1^2}

So,

T_1^2 = \frac{4\pi^2}{\dfrac{G M}{r_0^3} - \dfrac{3\alpha}{r_0^5}}

= \frac{4\pi^2 r_0^5}{G M r_0^2 - 3\alpha}

Also,

T_0^2 = \frac{4\pi^2 r_0^3}{G M}

--- 4. **Compute the required expression** We need

\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \frac{T_0^2}{T_1^2}

Now,

\frac{T_0^2}{T_1^2} = \frac{\omega_1^2}{\omega_0^2}

because $T^2 \propto \frac{1}{\omega^2}$. Hence,

\frac{T_0^2}{T_1^2} = \frac{\dfrac{G M}{r_0^3} - \dfrac{3\alpha}{r_0^5}}{\dfrac{G M}{r_0^3}} = 1 - \frac{3\alpha}{G M r_0^2}

Therefore,

\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \left(1 - \frac{3\alpha}{G M r_0^2}\right) = \frac{3\alpha}{G M r_0^2}

--- 5. **Match with options**

\boxed{\frac{T_1^2-T_0^2}{T_1^2} = \frac{3\alpha}{G M r_0^2}}

So the correct option is:

\boxed{\text{A}}

More from Gravitation

View written solutionFree

Correct answer: A

Given circular motion under gravity alone

For a circular orbit of radius $r_0$ around mass $M$ , the gravitational force provides the centripetal force:

\frac{m v_0^2}{r_0} = \frac{G M m}{r_0^2}

So,

v_0^2 = \frac{G M}{r_0}

The angular speed is

\omega_0 = \frac{v_0}{r_0}, \qquad \omega_0^2 = \frac{G M}{r_0^3}

Hence,

T_0 = \frac{2\pi}{\omega_0}, \qquad T_0^2 = \frac{4\pi^2 r_0^3}{G M}

Additional central potential

The extra potential energy is

V_c(r) = \frac{m\alpha}{r^3}

The corresponding force is

F_c(r) = -\frac{dV_c}{dr}

Differentiate:

\frac{d}{dr}\left(\frac{m\alpha}{r^3}\right) = -\frac{3m\alpha}{r^4}

Therefore,

F_c(r) = -\left(-\frac{3m\alpha}{r^4}\right) = \frac{3m\alpha}{r^4}

This force is radially outward.

So the net inward force available for centripetal motion at radius $r_0$ is

\frac{G M m}{r_0^2} - \frac{3m\alpha}{r_0^4}

New circular orbit condition

Since the particle still moves in the same circular orbit of radius $r_0$ , but with new speed $v_1$ and period $T_1$ ,

\frac{m v_1^2}{r_0} = \frac{G M m}{r_0^2} - \frac{3m\alpha}{r_0^4}

Cancelling $m$ :

v_1^2 = \frac{G M}{r_0} - \frac{3\alpha}{r_0^3}

Thus,

\omega_1^2 = \frac{v_1^2}{r_0^2} = \frac{G M}{r_0^3} - \frac{3\alpha}{r_0^5}

Since $T = \frac{2\pi}{\omega}$ ,

T_1^2 = \frac{4\pi^2}{\omega_1^2}

So,

T_1^2 = \frac{4\pi^2}{\dfrac{G M}{r_0^3} - \dfrac{3\alpha}{r_0^5}}

= \frac{4\pi^2 r_0^5}{G M r_0^2 - 3\alpha}

Also,

T_0^2 = \frac{4\pi^2 r_0^3}{G M}

--- 4. **Compute the required expression** We need

\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \frac{T_0^2}{T_1^2}

Now,

\frac{T_0^2}{T_1^2} = \frac{\omega_1^2}{\omega_0^2}

because $T^2 \propto \frac{1}{\omega^2}$. Hence,

\frac{T_0^2}{T_1^2} = \frac{\dfrac{G M}{r_0^3} - \dfrac{3\alpha}{r_0^5}}{\dfrac{G M}{r_0^3}} = 1 - \frac{3\alpha}{G M r_0^2}

Therefore,

\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \left(1 - \frac{3\alpha}{G M r_0^2}\right) = \frac{3\alpha}{G M r_0^2}

--- 5. **Match with options**

\boxed{\frac{T_1^2-T_0^2}{T_1^2} = \frac{3\alpha}{G M r_0^2}}

So the correct option is:

\boxed{\text{A}}

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