A particle of mass m is under the influence of the gravitational field of a body of mass M ( m). T

Q: Why is the new force repulsive?

The potential energy is given as $V_c(r) = \frac{m\alpha}{r^3}$. Since force is the negative gradient of potential energy ($F = -\frac{dV}{dr}$), differentiating a positive $1/r^3$ term yields a positive force. In central force problems, a positive force indicates repulsion (directed away from the center).

Q: Why does the radius $r_0$ remain constant?

The problem explicitly states, "If the particle moves in the same circular orbit of radius $r_0$". This means the particle adjusts its velocity (and hence its time period) so that the new net force perfectly provides the required centripetal force for that exact same radius.

Q: Can we use conservation of energy here?

While energy is conserved, it's not the most direct path to find the time period. The time period is directly linked to the angular velocity, which is determined by the radial force balance (centripetal force). Using dynamics is much faster here.

Q: What happens if $\alpha$ is negative?

If $\alpha$ were negative, the additional force would be attractive (inward). This would increase the net inward force, requiring the particle to move faster to stay in the same orbit, thereby decreasing its time period $T_1$.

Q: Why did we calculate $T_0^2$ initially?

We need to find the expression $\frac{T_1^2 - T_0^2}{T_1^2}$. By finding $T_0^2$ in terms of the given constants ($G, M, r_0$), we can easily substitute it later to eliminate the $4\pi^2$ and $r_0$ terms and arrive at the final answer.

Question

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 $V_c(r) = \frac{m\alpha}{r^3}$, 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 $V_c(r)$, but with a new time period $T_1$, then $\frac{T_1^2 - T_0^2}{T_1^2}$ is given by

[G is the gravitational constant.]

Options:

(A) $\frac{3\alpha}{GMr_0^2}$

(B) $\frac{\alpha}{2GMr_0^2}$

(C) $\frac{\alpha}{GMr_0^2}$

(D) $\frac{2\alpha}{GMr_0^2}$

Sigma Percentile · Accepted Answer

### The Gravitational Dance

Imagine a massive body of mass $M$ sitting quietly at the center of the universe. A much smaller particle of mass $m$ is caught in its gravitational embrace, revolving around it in a perfect circular orbit of radius $r_0$.

Initially, what keeps this particle from flying off into deep space? It's the gravitational force acting as the centripetal force. We can express this mathematically by balancing the forces:

$$ \frac{GMm}{r_0^2} = m\omega_0^2 r_0 $$

We know that the angular velocity $\omega_0$ is intimately connected to the time period $T_0$ by the relation $\omega_0 = \frac{2\pi}{T_0}$. Substituting this into our force equation gives us a beautiful expression for the square of the initial time period:

$$ \frac{GM}{r_0^3} = \frac{4\pi^2}{T_0^2} \implies T_0^2 = \frac{4\pi^2 r_0^3}{GM} $$

Keep this equation safe; it will be our master key later on.

### The Plot Twist: A New Potential

Now, the universe throws a curveball. The particle is suddenly subjected to an additional central force. We aren't given the force directly, but rather its potential energy:

$$ V_c(r) = \frac{m\alpha}{r^3} $$

To find the force, we must remember the fundamental relationship between conservative forces and potential energy: force is the negative gradient of potential energy ($F = -\frac{dV}{dr}$). Let's differentiate our new potential:

$$ F_c = -\frac{d}{dr}\left(\frac{m\alpha}{r^3}ight) = \frac{3m\alpha}{r^4} $$

Notice the positive sign! In the language of central forces, a positive sign means the force is directed radially outwards. It is a **repulsive force**, actively trying to push the particle away from the central mass $M$.

### The Battle of Forces

The particle is now caught in a tug-of-war. Gravity pulls it inward, while this new mysterious force pushes it outward. However, the problem states a crucial constraint: **the particle continues to move in the exact same circular orbit of radius $r_0$**.

For the radius to remain constant, the particle must adjust its speed so that the new net inward force perfectly provides the required centripetal force. Let's write the new force balance equation:

$$ F_{	ext{net}} = F_g - F_c = m\omega_1^2 r_0 $$

Substituting our known force expressions, we get:

$$ \frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4} = m\omega_1^2 r_0 $$

### The Mathematics of Time

Let's clean up this equation by dividing everything by $mr_0$ and replacing the new angular velocity $\omega_1$ with $\frac{2\pi}{T_1}$, where $T_1$ is the new time period:

$$ \frac{GM}{r_0^3} - \frac{3\alpha}{r_0^5} = \frac{4\pi^2}{T_1^2} $$

Look closely at the first term on the left side: $\frac{GM}{r_0^3}$. Does it look familiar? It is exactly the term we found in our very first step! We can replace it with $\frac{4\pi^2}{T_0^2}$:

$$ \frac{4\pi^2}{T_0^2} - \frac{3\alpha}{r_0^5} = \frac{4\pi^2}{T_1^2} $$

Now we have a direct bridge connecting the new time period $T_1$ to the old time period $T_0$. Let's divide the entire equation by $4\pi^2$ to isolate the time periods:

$$ \frac{1}{T_0^2} - \frac{3\alpha}{4\pi^2 r_0^5} = \frac{1}{T_1^2} $$

### The Grand Finale

We are asked to find the value of the expression $\frac{T_1^2 - T_0^2}{T_1^2}$. Let's rearrange our equation to match this form:

$$ \frac{1}{T_1^2} - \frac{1}{T_0^2} = -\frac{3\alpha}{4\pi^2 r_0^5} $$

$$ \frac{T_0^2 - T_1^2}{T_1^2 T_0^2} = -\frac{3\alpha}{4\pi^2 r_0^5} $$

$$ \frac{T_1^2 - T_0^2}{T_1^2} = T_0^2 \left(\frac{3\alpha}{4\pi^2 r_0^5}ight) $$

We are almost there! Remember that master key we saved in the first step? It's time to use it. Substitute $T_0^2 = \frac{4\pi^2 r_0^3}{GM}$ into our equation:

$$ \frac{T_1^2 - T_0^2}{T_1^2} = \left(\frac{4\pi^2 r_0^3}{GM}ight) \left(\frac{3\alpha}{4\pi^2 r_0^5}ight) $$

The $4\pi^2$ terms cancel out beautifully, and $r_0^3$ divides $r_0^5$ to leave $r_0^2$ in the denominator. The dust settles, revealing our final, elegant answer:

$$ \frac{T_1^2 - T_0^2}{T_1^2} = \frac{3\alpha}{GMr_0^2} $$

Because the net inward force decreased due to the repulsive potential, the particle had to slow down to maintain its orbit, resulting in a longer time period $T_1$. This is a masterful demonstration of how forces dictate the rhythm of orbital mechanics!

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Visualized Solution

$I ni t ia l S e t u p$

$I ni t ia l F or ce B a l an ce$

$I ni t ia l T im e P er i o d T_{0}$

$A dd i t i o na l P o t e n t ia l E n er g y$

$A dd i t i o na l C e n t r a l F or ce$

$N e w F or ce B a l an ce$

$S u b s t i t u t in g V a l u es$

$I n t r o d u c in g N e w T im e P er i o d T_{1}$

$C o nn ec t in g T_{1} an d T_{0}$

$A l g e b r ai c R e a r r an g e m e n t$

$F ina l C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Orbital Motion of a Satellite

The Gravitational Dance

The Plot Twist

The Battle of Forces

The Mathematics of Time

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Force Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Time Period T0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Additional Potential Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Additional Central Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Force Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Introducing New Time Period T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Connecting T1​ and T0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Rearrangement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion