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 $\mathrm{r}_{0}$ 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 $\left(T_{1}^{2}-T_{0}^{2}\right) / T_{1}^{2}$ is given by [ $G$ is the gravitational constant.]

(A) $\frac{3 \alpha}{G M r_{0}^{2}}$

(B) $\frac{\alpha}{2 G M r_{0}^{2}}$

(C) $\frac{\alpha}{G M r_{0}^{2}}$

(D) $\frac{2 \alpha}{G M r_{0}^{2}}$

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Gravitation

A particle of mass

m

is under the influence of the gravitational field of a body of mass

M (≫ 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 α / r^{3}

, where

α

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.]

Answer: (A)

Initially, the gravitational force provides centripetal force:

F_{1} = \frac{GM m}{r _{0}^{2}}

.

With the additional potential

V_{c} (r) = m α / r^{3}

, the additional force is

F_{add} = - \frac{d V _{c}}{d r} = \frac{3 m α}{r ^{4}}

(radially inward). The total centripetal force becomes

F_{2} = \frac{GM m}{r _{0}^{2}} - \frac{3 m α}{r _{0}^{4}}

.

Since

ω^{2} r_{0} = F / m

, we have:

\frac{ω _{1}^{2}}{ω _{0}^{2}} = \frac{F _{2}}{F _{1}} = \frac{\frac{GM}{r _{0}^{2}} - \frac{3 α}{r _{0}^{4}}}{\frac{GM}{r _{0}^{2}}} = 1 - \frac{3 α}{GM r _{0}^{2}}

Since

ω = 2 π / T

, we get

\frac{T _{0}^{2}}{T _{1}^{2}} = 1 - \frac{3 α}{GM r _{0}^{2}}

.

Therefore,

\frac{T _{1}^{2} - T _{0}^{2}}{T _{1}^{2}} = \frac{3 α}{GM r _{0}^{2}}

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Single correct option