Question

A particle moves in a circular orbit of radius 'r' under a central attractive force, \( F = -\frac{k}{r} \), where k is a constant. The periodic time of its motion is proportional to

• A. \( r^{\frac{1}{2}} \)
• B. \( r^{\frac{2}{3}} \)
• C. \( r \)
• D. \( r^{\frac{3}{2}} \)

Hint:

For a central force \(F \propto r^n\), the period scales as \(T \propto r^{(2-n)/2}\) when \(n \neq -1\). Here \(n = -1\) gives \(T \propto r^{(2-(-1))/2} = r^{3/2}\)? Wait careful: Actually \(F = -k/r\) means \(n = -1\). But the derivation shows \(v\) constant, so \(T \propto r\). Double-check: \(mv^2/r = k/r \Rightarrow v^2 = k/m\) constant. Yes. So the general formula \(T \propto r^{(1-n)/2}\)? Let's derive: \(mv^2/r = k r^n\) gives \(v^2 \propto r^{n+1}\), then \(T = 2\pi r/v \propto r / r^{(n+1)/2} = r^{(1-n)/2}\). For \(n=-1\), \(T \propto r^{(1-(-1))/2} = r^{1}\). Correct. So tip: memorize \(T \propto r^{(1-n)/2}\) for \(F \propto r^n\).

Answer 1

The Correct Option is C

Step 1: Set the scene.
A particle of mass $m$ goes around a circle of radius $r$. The attractive force pulling it in has magnitude $\dfrac{k}{r}$. We want how the period $T$ depends on $r$.

Step 2: Balance the forces.
For circular motion, the inward force provides the centripetal force: $\dfrac{mv^2}{r} = \dfrac{k}{r}$.

Step 3: Solve for the speed.
The $r$ on both sides cancels, leaving $mv^2 = k$, so $v = \sqrt{\dfrac{k}{m}}$.

Step 4: Notice something nice.
The speed $v$ does not depend on $r$ at all. It is the same on every orbit.

Step 5: Write the period.
The time for one loop is the circumference over the speed: $T = \dfrac{2\pi r}{v}$.

Step 6: Read off the dependence.
Since $v$ is constant, $T = \left(\dfrac{2\pi}{v}\right) r$, so $T \propto r$, which is option (C).
\[ \boxed{T \propto r} \]