A particle of mass m 𝑚 is under the influence of the gravitational field of a body of mass M ( >> m ) 𝑀 ( >> 𝑚 ) . The particle is moving in a circular orbit of radius r 0 𝑟 0 with time period T 0 𝑇 0 around the mass M 𝑀 . Then, the particle is subjected to an additional central force, corresponding to the potential energy V c ( r ) = m α / r 3 𝑉 𝑐 ( 𝑟 ) = 𝑚 𝛼 / 𝑟 3 , where α 𝛼 is a positive constant of suitable dimensions and r 𝑟 is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r 0 𝑟 0 in the combined gravitational potential due to M 𝑀 and V e ( r ) 𝑉 𝑒 ( 𝑟 ) , but with a new time period T 1 𝑇 1 , then ( T 2 1 − T 2 0 ) / T 2 1 ( 𝑇 1 2 − 𝑇 0 2 ) / 𝑇 1 2 is given by [G is the gravitational constant.]

Alright, let's tackle this problem step by step. I'm going to break it down to understand how the time period of the particle's orbit changes when an additional central force is introduced.

Understanding the Problem

We have:

A particle of mass $m$ orbiting a much larger mass $M$ (so $M ≫ m$ , meaning we can ignore the motion of $M$ ).
Initially, the particle is in a circular orbit of radius $r_{0}$ with a time period $T_{0}$ .
The gravitational force provides the necessary centripetal force for this orbit.
Then, an additional central force is introduced, described by the potential energy $V_{c} (r) = \frac{m α}{r^{3}}$ , where $α$ is a positive constant.
With this new force, the particle still moves in a circular orbit of the same radius $r_{0}$ , but now with a new time period $T_{1}$ .
We need to find the expression for $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}}$ in terms of the given constants.

Initial Orbit (Only Gravitational Force)

First, let's recall the initial situation where only gravity is acting.

The gravitational force provides the centripetal force:

F_{g} = \frac{G M m}{r_{0}^{2}} = m ω_{0}^{2} r_{0}

where $ω_{0}$ is the initial angular velocity.

Solving for $ω_{0}$ :

ω_{0}^{2} = \frac{G M}{r_{0}^{3}}

ω_{0} = \sqrt{\frac{G M}{r_{0}^{3}}}

The time period $T_{0}$ is related to $ω_{0}$ by:

T_{0} = \frac{2 π}{ω_{0}} = 2 π \sqrt{\frac{r_{0}^{3}}{G M}}

Adding the Additional Central Force

Now, we add the potential $V_{c} (r) = \frac{m α}{r^{3}}$ .

The force associated with this potential is:

F_{c} = - \frac{d V_{c}}{d r} = - \frac{d}{d r} (\frac{m α}{r^{3}}) = \frac{3 m α}{r^{4}}

Wait, let's double-check the force from the potential. The force is the negative gradient of the potential, so for a central potential $V (r)$ , the radial force is:

F = - \frac{d V}{d r}

Given $V_{c} (r) = \frac{m α}{r^{3}}$ :

F_{c} = - \frac{d}{d r} (\frac{m α}{r^{3}}) = - m α (- 3) r^{- 4} = \frac{3 m α}{r^{4}}

So, the additional force is:

F_{c} = \frac{3 m α}{r^{4}}

This force is outward (positive sign), meaning it's repulsive.

Combined Force in the New Orbit

In the new orbit with the same radius $r_{0}$ , the total centripetal force is now provided by the sum of the gravitational force and this additional force. However, since the additional force is outward, it effectively reduces the required centripetal force.

Wait, no—the centripetal force is the net force pointing inward required to keep the particle in circular motion. The gravitational force is inward, and the additional force is outward, so the net inward force is:

F_{net} = F_{g} - F_{c} = \frac{G M m}{r_{0}^{2}} - \frac{3 m α}{r_{0}^{4}}

This net force provides the centripetal force for the new orbit:

F_{net} = m ω_{1}^{2} r_{0}

\frac{G M m}{r_{0}^{2}} - \frac{3 m α}{r_{0}^{4}} = m ω_{1}^{2} r_{0}

Divide both sides by $m$ :

\frac{G M}{r_{0}^{2}} - \frac{3 α}{r_{0}^{4}} = ω_{1}^{2} r_{0}

ω_{1}^{2} = \frac{G M}{r_{0}^{3}} - \frac{3 α}{r_{0}^{5}}

Relating $ω_{1}$ and $T_{1}$

The new time period $T_{1}$ is:

T_{1} = \frac{2 π}{ω_{1}}

T_{1}^{2} = \frac{4 π^{2}}{ω_{1}^{2}} = \frac{4 π^{2}}{\frac{G M}{r_{0}^{3}} - \frac{3 α}{r_{0}^{5}}}

T_{1}^{2} = \frac{4 π^{2} r_{0}^{3}}{G M - \frac{3 α}{r_{0}^{2}}}

Similarly, the initial $T_{0}^{2}$ was:

T_{0}^{2} = \frac{4 π^{2} r_{0}^{3}}{G M}

Calculating $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}}$

Now, let's compute the given expression:

\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}} = 1 - \frac{T_{0}^{2}}{T_{1}^{2}}

Compute $\frac{T_{0}^{2}}{T_{1}^{2}}$ :

\frac{T_{0}^{2}}{T_{1}^{2}} = \frac{\frac{4 π^{2} r_{0}^{3}}{G M}}{\frac{4 π^{2} r_{0}^{3}}{G M - \frac{3 α}{r_{0}^{2}}}} = \frac{G M - \frac{3 α}{r_{0}^{2}}}{G M} = 1 - \frac{3 α}{G M r_{0}^{2}}

So:

\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}} = 1 - (1 - \frac{3 α}{G M r_{0}^{2}}) = \frac{3 α}{G M r_{0}^{2}}

Final Answer

Thus, the expression $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}}$ is given by:

\frac{3 α}{G M r_{0}^{2}}

Verification

Let me quickly verify the steps to ensure no mistakes were made:

Initial centripetal force: $\frac{G M m}{r_{0}^{2}} = m ω_{0}^{2} r_{0}$ leads to $ω_{0}^{2} = \frac{G M}{r_{0}^{3}}$ .
Additional force from $V_{c} (r)$ : $F_{c} = \frac{3 m α}{r^{4}}$ (outward).
New centripetal force: $\frac{G M m}{r_{0}^{2}} - \frac{3 m α}{r_{0}^{4}} = m ω_{1}^{2} r_{0}$ , so $ω_{1}^{2} = \frac{G M}{r_{0}^{3}} - \frac{3 α}{r_{0}^{5}}$ .
$T_{1}^{2} = \frac{4 π^{2}}{ω_{1}^{2}} = \frac{4 π^{2} r_{0}^{3}}{G M - \frac{3 α}{r_{0}^{2}}}$ .
$\frac{T_{0}^{2}}{T_{1}^{2}} = \frac{G M - \frac{3 α}{r_{0}^{2}}}{G M} = 1 - \frac{3 α}{G M r_{0}^{2}}$ .
Thus, $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}} = \frac{3 α}{G M r_{0}^{2}}$ .

Everything checks out! The final answer is indeed $\frac{3 α}{G M r_{0}^{2}}$ .

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Understanding the Problem

Initial Orbit (Only Gravitational Force)

Adding the Additional Central Force

Combined Force in the New Orbit

Relating $ω_{1}$ and $T_{1}$

Calculating $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}}$

Final Answer

Verification

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Understanding the Problem

Initial Orbit (Only Gravitational Force)

Adding the Additional Central Force

Combined Force in the New Orbit

Relating ω1ω1​ and T1T1​

Calculating T12−T02T12T12​T12​−T02​​

Final Answer

Verification

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Relating $ω_{1}$ and $T_{1}$

Calculating $\frac{T_{1}^{2} - T_{0}^{2}}{T_{1}^{2}}$