A particle of mass m, and angular momentum l is moving in a circular orbit of radius r0 under the in

Q: Why do we use the centrifugal force $\frac{\ell^2}{mr^3}$ instead of just $m\omega^2 r$?

Because the problem states that the angular momentum $\ell$ is kept unchanged during the displacement, not the angular velocity $\omega$. Using $\ell$ ensures we are analyzing the system under the correct conserved quantity.

Q: Can we solve this using the energy method instead of forces?

Absolutely! You can write the effective potential energy $U_{\text{eff}}(r) = \frac{\ell^2}{2mr^2} - \frac{k}{r}$. The second derivative of this potential at the minimum $r_0$ gives the effective spring constant $K_{\text{eff}} = U_{\text{eff}}''(r_0)$, which leads to the exact same result.

Q: Why did we ignore the higher-order terms in the binomial expansion?

The problem specifies that the displacement $x$ is much smaller than $r_0$. In such cases, terms like $x^2/r_0^2$ and higher become infinitesimally small and can be safely neglected, leaving us with a linear restoring force.

Q: Does this radial oscillation affect the circular motion?

The particle continues to revolve around the center, but its radius now slightly oscillates in and out. The resulting path is no longer a perfect circle but a slightly perturbed, rosette-like shape.

Q: What if the attractive force was proportional to $1/r^3$ instead of $1/r^2$?

If the force was $1/r^3$, the effective spring constant would turn out to be zero (or negative), meaning the circular orbit would be unstable. Any slight nudge would cause the particle to either spiral into the center or escape to infinity!

Question

A particle of mass $m$, and angular momentum $\ell$ is moving in a circular orbit of radius $r_0$ under the influence of an attractive force $\vec{F}(r) = -\frac{k}{r^2}\hat{r}$. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance $\delta r \ll r_0$, due to which its radial distance varies periodically. The corresponding time period is:

Options:

(A) $\frac{2\pi\ell^3}{mk^2}$

(B) $2\pi\sqrt{\frac{m}{k}}$

(C) $\frac{2\pi\ell^3}{3mk^2}$

(D) $\frac{2\pi\ell^3}{5mk^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle gracefully moving in a perfect circular orbit. The only reason it maintains this path is because a central attractive force is constantly pulling it inward, acting as an invisible tether.

The force given is $\vec{F}(r) = -\frac{k}{r^2}\hat{r}$. This is an inverse-square law force, much like gravity or electrostatics.

### The Master Equation for Equilibrium

For the particle to stay in this stable circular orbit of radius $r_0$, the inward attractive force must perfectly balance the outward centrifugal tendency.

In a rotating frame of reference, we can express this balance as:

$\frac{k}{r_0^2} = \frac{mv_0^2}{r_0}$

We are told that the angular momentum $\ell$ remains unchanged. We know that $\ell = m v_0 r_0$, which gives us $v_0 = \frac{\ell}{m r_0}$.

Substituting this into our force balance equation:

$\frac{k}{r_0^2} = \frac{m}{r_0} \left( \frac{\ell}{m r_0} ight)^2 = \frac{\ell^2}{m r_0^3}$

From this, we can isolate the equilibrium radius $r_0$:

$r_0 = \frac{\ell^2}{m k}$

### Introducing the Disturbance

Here is where the physics gets really interesting. We introduce a tiny radial disturbance, nudging the particle slightly away from its perfect circle to a new radius $r = r_0 + x$, where $x \ll r_0$.

Because it is a stable equilibrium, it will try to return, creating an oscillation. To analyze this wobble, we calculate the net effective force at this new radius.

The effective radial force is the difference between the centrifugal force and the attractive force:

$F_{	ext{eff}} = \frac{\ell^2}{m r^3} - \frac{k}{r^2}$

### The Mathematics of the Wobble

Let's substitute $r = r_0 + x$ into our effective force equation:

$F_{	ext{eff}} = \frac{\ell^2}{m (r_0 + x)^3} - \frac{k}{(r_0 + x)^2}$

Since the displacement $x$ is very small, we can factor out $r_0$ and use the binomial approximation $(1 + \epsilon)^n \approx 1 + n\epsilon$:

$F_{	ext{eff}} = \frac{\ell^2}{m r_0^3} \left(1 + \frac{x}{r_0}ight)^{-3} - \frac{k}{r_0^2} \left(1 + \frac{x}{r_0}ight)^{-2}$

$F_{	ext{eff}} \approx \frac{\ell^2}{m r_0^3} \left(1 - \frac{3x}{r_0}ight) - \frac{k}{r_0^2} \left(1 - \frac{2x}{r_0}ight)$

Recall our equilibrium condition: $\frac{\ell^2}{m r_0^3} = \frac{k}{r_0^2}$. We can substitute this into the first term:

$F_{	ext{eff}} \approx \frac{k}{r_0^2} \left( 1 - \frac{3x}{r_0} - 1 + \frac{2x}{r_0} ight)$

$F_{	ext{eff}} \approx -\frac{k}{r_0^3} x$

Look at the structure of this resulting force! It is directly proportional to the displacement $x$, and it acts in the opposite direction. This is the undeniable signature of **Simple Harmonic Motion**.

### Final Calculation

Our effective spring constant is $K_{	ext{eff}} = \frac{k}{r_0^3}$.

Now, we simply plug it into the classic time period formula for SHM:

$T = 2\pi \sqrt{\frac{m}{K_{	ext{eff}}}} = 2\pi \sqrt{\frac{m r_0^3}{k}}$

Finally, we substitute our earlier expression for the stable radius $r_0 = \frac{\ell^2}{m k}$:

$T = 2\pi \sqrt{\frac{m}{k} \left( \frac{\ell^2}{m k} ight)^3} = 2\pi \sqrt{\frac{\ell^6}{m^2 k^4}}$

The algebra elegantly collapses into our final answer:

**$T = \frac{2\pi \ell^3}{m k^2}$**

This explains why planets do not just crash or escape when slightly nudged by the gravity of other celestial bodies!

Select Answer:

Visualized Solution

Initial Setup & Force $F (r)$

Equilibrium Condition

Calculating Radius $r_{0}$

Effective Force $F_{eff}$

Binomial Approximation

Restoring Force & SHM

Time Period $T$

Conclusion

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

The Master Equation for Equilibrium

Introducing the Disturbance

The Mathematics of the Wobble

Final Calculation

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