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: Understanding the Question:
A particle of mass \(m\) (assumed) moves in a circular orbit of radius \(r\) under a central force \(F = -k/r\) (attractive, magnitude \(k/r\)). We need to find how the period \(T\) depends on \(r\).

Step 2: Key Formula or Approach:
For circular motion, the centripetal force is provided by the central force: \( \frac{mv^2}{r} = \frac{k}{r} \). The period \(T = \frac{2\pi r}{v}\).

Step 3: Detailed Explanation:
From \(\frac{mv^2}{r} = \frac{k}{r}\) we cancel \(r\) (provided \(r \neq 0\)): \[ m v^2 = k \quad \Rightarrow \quad v = \sqrt{\frac{k}{m}}. \] Thus the orbital speed \(v\) is independent of \(r\). The period is: \[ T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{k/m}} = \left( \frac{2\pi}{\sqrt{k/m}} \right) r. \] Hence \(T \propto r\).

Step 4: Final Answer:
The periodic time is proportional to \(r\), which corresponds to option (C).