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

Time Period Calculation Under Additional Force 

The additional force due to the potential is given by:

\[ F = -\frac{dV}{dr} = -\frac{3\alpha m}{r^4} \]

The net centripetal force acting on the particle is:

\[ F_{\text{net}} = \frac{GMm}{r^2} + \frac{3\alpha m}{r^4} \]

Step 1: Time Period of the Orbit Under the Additional Force

The modified time period is:

\[ T_1 = \frac{2\pi r_0}{v} \] where \( v \) is: \[ v = \sqrt{\frac{GM}{r_0} + \frac{3\alpha}{r_0^3}} \] Thus: \[ T_1 = \frac{4\pi^2 r_0^3}{GM + 3\alpha r_0^3} \]

Step 2: Original Time Period

The original time period without the additional force is:

\[ T_0 = \frac{4\pi^2 r_0^3}{GM} \]

Step 3: Relationship Between \( T_1 \) and \( T_0 \)

The difference in time periods can be expressed as:

\[ \frac{T_1 - T_0}{T_1} \] Substituting \( T_0 \) and \( T_1 \): \[ \frac{T_1 - T_0}{T_1} = -1 - \frac{T_0}{T_1} \] Expanding and simplifying: \[ \frac{T_1 - T_0}{T_0} = \frac{T_1 - T_0}{T_1} + \frac{T_0}{T_1} \] This simplifies to: \[ \frac{T_1 - T_0}{T_0} = \frac{3\alpha}{GM r_0^2} \]

Final Expression:

The difference in the time periods is given by:

\[ \frac{T_1 - T_0}{T_0} = \frac{3\alpha}{GM r_0^2} \]

Quick Tip:

• \( \frac{T_1 - T_0}{T_0} = \frac{3\alpha}{GM r_0^2} \)