Question

A particle of mass \(m\) moves in a circular orbit under the central potential field, \(U(r) = -\frac{C}{r}\), where \(C\) is a positive constant. The correct radius – velocity graph of the particle's motion is: 

• A. A
• B. B
• C. C
• D. D

Hint:

For inverse square forces (like gravity or electrostatics), the orbital speed \(v\) is inversely proportional to the square root of the radius \(\sqrt{r}\).

Answer 1

The Correct Option is C

Step 1: Find the force \(F\) from the potential \(U(r)\): \[F = -\frac{dU}{dr} = -\frac{d}{dr}\left(-\frac{C}{r}\right) = -\frac{C}{r^2}\] The magnitude of the force is \(\frac{C}{r^2}\).
Step 2: For circular motion, centripetal force is provided by this field: \[\frac{mv^2}{r} = \frac{C}{r^2} \implies v^2 = \frac{C}{mr} \implies r = \frac{C}{mv^2}\]
Step 3: Since \(r \propto \frac{1}{v^2}\), the graph of \(r\) vs \(v\) is a hyperbola-like curve where \(r\) decreases as \(v\) increases.