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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Gravitation

JEE Advanced 2024 · Paper 2 · Physics · Gravitation

A particle of mass m is under the influence of the gravitational...

A particle of mass mm is under the influence of the gravitational field of a body of mass M(≫m)M(\gg m). The particle is moving in a circular orbit of radius r0r_0 with time period T0T_0 around the mass MM. Then, the particle is subjected to an additional central force, corresponding to the potential energy Vc(r)=mα/r3V_{\mathrm{c}}(r)=m \alpha / r^3, where α\alpha is a positive constant of suitable dimensions and risthedistancefromthecenteroftheorbit.Iftheparticlemovesinthesamecircularorbitofradiusr is the distance from the center of the orbit. If the particle moves in the same circular orbit of radiusr_0inthecombinedgravitationalpotentialduetoin the combined gravitational potential due toM and V_{\mathrm{c}}(r),butwithanewtimeperiod, but with a new time periodT_1,then(T12−T02)/T12, then \left(T_1^2-T_0^2\right) / T_1^2 is given by [ GG is the gravitational constant.]

Options

  1. A3αGMr02\frac{3 \alpha}{G M r_0^2}
  2. Bα2GMr02\frac{\alpha}{2 G M r_0^2}
  3. CαGMr02\frac{\alpha}{G M r_0^2}
  4. D2αGMr02\frac{2 \alpha}{G M r_0^2}

Correct answer

A. 3αGMr02\frac{3 \alpha}{G M r_0^2}

Step-by-step solution

F1=GMmr02 F2=GMmr02−3 mαr04ω12ω02=F2 F1=GMr02−3αr04GMr02T02 T12=1−3αGMr02T12−T02 T12=3αGMr02\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{\mathrm{r}_0^2} \\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{\mathrm{r}_0^2}-\frac{3 \mathrm{~m} \alpha}{\mathrm{r}_0^4} \\ & \frac{\omega_1^2}{\omega_0^2}=\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\frac{\mathrm{GM}}{\mathrm{r}_0^2}-\frac{3 \alpha}{\mathrm{r}_0^4}}{\frac{\mathrm{GM}}{\mathrm{r}_0^2}} \\ & \frac{\mathrm{T}_0^2}{\mathrm{~T}_1^2}=1-\frac{3 \alpha}{\mathrm{GMr}_0^2} \\ & \frac{\mathrm{T}_1^2-\mathrm{T}_0^2}{\mathrm{~T}_1^2}=\frac{3 \alpha}{\mathrm{GMr}_0^2}\end{aligned}
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