Question

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 𝑉c(𝑟) = 𝑚𝛼/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 𝑉c(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. \(\frac{3\alpha}{ 𝐺𝑀𝑟_0^2}\)
• B. \(\frac{\alpha}{ 2𝐺𝑀𝑟_0^2}\)
• C. \(\frac{\alpha}{ 𝐺𝑀𝑟_0^2}\)
• D. \(\frac{2\alpha}{𝐺𝑀𝑟_0^2}\)

Answer 1

The Correct Option is A

To solve the problem, consider the forces and time periods involved in circular motion under combined potentials.

1. Initial circular orbit under gravitational force alone:
Gravitational force: \[ F_g = \frac{G M m}{r_0^2} \] Time period of orbit: \[ T_0 = 2 \pi \sqrt{\frac{r_0^3}{G M}} \] Angular velocity: \[ \omega_0 = \frac{2 \pi}{T_0} = \sqrt{\frac{G M}{r_0^3}} \]

2. Additional central force from potential:
Given potential: \[ V_c(r) = \frac{m \alpha}{r^3} \] Force from this potential (radial inward, since \(\alpha > 0\)): \[ F_c = - \frac{dV_c}{dr} = - m \alpha \frac{d}{dr}\left(\frac{1}{r^3}\right) = - m \alpha (-3 r^{-4}) = \frac{3 m \alpha}{r^4} \]

3. Total radial force for circular motion at \(r_0\):
\[ F_{\text{total}} = F_g + F_c = \frac{G M m}{r_0^2} + \frac{3 m \alpha}{r_0^4} \] This force provides the centripetal force for circular motion: \[ m \omega_1^2 r_0 = F_{\text{total}} = m \left(\frac{G M}{r_0^2} + \frac{3 \alpha}{r_0^4}\right) \] \[ \omega_1^2 = \frac{G M}{r_0^3} + \frac{3 \alpha}{r_0^5} \]

4. Relate new time period \(T_1\) to \(\omega_1\):
\[ T_1 = \frac{2 \pi}{\omega_1} \implies \omega_1 = \frac{2 \pi}{T_1} \] Square both sides: \[ \omega_1^2 = \frac{4 \pi^2}{T_1^2} \] Similarly, \[ \omega_0^2 = \frac{4 \pi^2}{T_0^2} = \frac{G M}{r_0^3} \]

5. Substitute \(\omega_1^2\):
\[ \frac{4 \pi^2}{T_1^2} = \frac{G M}{r_0^3} + \frac{3 \alpha}{r_0^5} \] Rearranged: \[ \frac{4 \pi^2}{T_1^2} - \frac{4 \pi^2}{T_0^2} = \frac{3 \alpha}{r_0^5} \] Divide both sides by \(4 \pi^2\): \[ \frac{1}{T_1^2} - \frac{1}{T_0^2} = \frac{3 \alpha}{4 \pi^2 r_0^5} \]

6. Express \(\frac{T_1^2 - T_0^2}{T_1^2}\):
\[ \frac{T_1^2 - T_0^2}{T_1^2} = 1 - \frac{T_0^2}{T_1^2} = 1 - \frac{1/T_1^2}{1/T_0^2} = 1 - \frac{\frac{1}{T_1^2}}{\frac{1}{T_0^2}} = 1 - \frac{1/T_1^2}{1/T_0^2} \] \[ = \frac{\frac{1}{T_0^2} - \frac{1}{T_1^2}}{\frac{1}{T_0^2}} = \frac{\frac{3 \alpha}{4 \pi^2 r_0^5}}{\frac{1}{T_0^2}} = 3 \alpha \cdot \frac{T_0^2}{4 \pi^2 r_0^5} \]

Using \(T_0^2 = \frac{4 \pi^2 r_0^3}{G M}\), substitute:

\[ \frac{T_1^2 - T_0^2}{T_1^2} = 3 \alpha \cdot \frac{\frac{4 \pi^2 r_0^3}{G M}}{4 \pi^2 r_0^5} = 3 \alpha \cdot \frac{r_0^3}{G M r_0^5} = \frac{3 \alpha}{G M r_0^2} \]

Final Answer:
\[ \boxed{\frac{3 \alpha}{G M r_0^2}} \]

Answer 2

1. Starting with the given equation:
We are given the equation that relates the gravitational force and the centripetal force:

$ \frac{Gmm}{r_0^2} - \frac{3\alpha m}{r_0^4} = \frac{mv^2}{r_0} $

2. Rearranging the expression:
From this, we can solve for the expression for \( v^2 \):

$ v^2 = \frac{Gm}{r_0^2} - \frac{3\alpha}{r_0^3} $

3. Expression for the period \( T \):
The formula for the period \( T \) of revolution is:

$ T = \frac{2\pi r_0}{v} $

4. Substituting \( v^2 \):
Substitute the expression for \( v^2 \) into the equation for \( T \):

$ T = \frac{2\pi r_0}{\sqrt{\frac{Gm}{r_0^2} - \frac{3\alpha}{r_0^3}}} $

5. Squaring both sides:
Squaring both sides of the equation:

$ T^2 = \frac{4\pi^2 r_0^3}{Gm - 3\alpha} $

6. Using the ratio \( \frac{T_3^2 - T_2^2}{T_1^2} \):
Now, using the ratio \( \frac{T_3^2 - T_2^2}{T_1^2} \), we get:

$ \frac{T_3^2 - T_2^2}{T_2^2} = 1 - \frac{T_1^2}{T_2^2} $

7. Substituting the expression for \( T^2 \):
Substituting the expression for \( T^2 \) into the ratio:

$ = 1 - \frac{4\pi^2 r_0^3}{Gm} \times \frac{r_0^2}{4\pi^2 r_0^2} $

8. Simplifying the expression:
Simplifying the expression further:

$ = 1 - 1 + \frac{3\alpha}{Gm r_0^2} $

9. Final result:
The final result is:

$ = \frac{3\alpha}{GMr_0^2} $