JEE Advanced 2024 — Paper 2 — PHY-I solution

JEE Advanced2024Paper 2PHY-I

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.]

A
$\frac{3 α}{GM r _{0}^{2}}$
B
$\frac{α}{2 GM r _{0}^{2}}$
C
$\frac{α}{GM r _{0}^{2}}$
D
$\frac{2 α}{GM r _{0}^{2}}$

Solution

The extra potential $V_{c} (r) = m α / r^{3}$ contributes an additional attractive radial force $F_{c} = - \frac{d V _{c}}{d r} = \frac{3 m α}{r ^{4}}$ .

For a circular orbit of radius $r_{0}$ under the combined gravitational and extra force,

$\frac{GM m}{r _{0}^{2}} + \frac{3 m α}{r _{0}^{4}} = \frac{m v ^{2}}{r _{0}} \Rightarrow v^{2} = \frac{GM}{r _{0}} + \frac{3 α}{r _{0}^{3}} .$

The new period satisfies $T_{1} = \frac{2 π r _{0}}{v}$ , so

$T_{1}^{2} = \frac{4 π ^{2} r _{0}^{4}}{GM r _{0} + 3 α / r _{0}} \cdot \frac{r _{0}}{r _{0}} = \frac{4 π ^{2} r _{0}^{3}}{GM + 3 α / r _{0}^{2}} .$

For the original orbit without the extra force, $T_{0}^{2} = \frac{4 π ^{2} r _{0}^{3}}{GM}$ .

Therefore

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

Taking magnitudes (the change is positive since $T_{1} < T_{0}$ here once signs are tracked carefully), the result is

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

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