JEE Advanced · Physics · 9. Gravitation
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 \(V_{\mathrm{c}}(r)=m \alpha / r^3\), 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 \(V_{\mathrm{c}}(r)\), 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 \(\frac{2 \alpha}{G M r_0^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \alpha}{G M r_0^2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{\mathrm{r}_0^2} \\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{\mathrm{r}_0^2}-\frac{3 \mathrm{~m} \alpha}{\mathrm{r}_0^4} \\ & \frac{\omega_1^2}{\omega_0^2}=\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\frac{\mathrm{GM}}{\mathrm{r}_0^2}-\frac{3 \alpha}{\mathrm{r}_0^4}}{\frac{\mathrm{GM}}{\mathrm{r}_0^2}} \\ & \frac{\mathrm{T}_0^2}{\mathrm{~T}_1^2}=1-\frac{3 \alpha}{\mathrm{GMr}_0^2} \\ & \frac{\mathrm{T}_1^2-\mathrm{T}_0^2}{\mathrm{~T}_1^2}=\frac{3 \alpha}{\mathrm{GMr}_0^2}\end{aligned}\)
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