Question

A particle of mass m and angular momentum L moves in space where its potential energy is U(r) = kr² (k > 0) and r is the radial coordinate.
If the particle moves in a circular orbit, then the radius of the orbit is

• A. \((\frac{L^2}{mk})^{\frac{1}{4}}\)
• B. \((\frac{L^2}{2mk})^{\frac{1}{4}}\)
• C. \((\frac{2L^2}{mk})^{\frac{1}{4}}\)
• D. \((\frac{4L^2}{mk})^{\frac{1}{4}}\)

Answer 1

The Correct Option is B

To find the radius of the circular orbit for a particle with mass \( m \) and angular momentum \( L \) moving under the potential energy \( U(r) = kr^2 \), we proceed as follows:

The force acting on the particle due to the potential energy is given by the negative gradient of \( U(r) \):

\(F(r) = -\frac{dU}{dr} = -\frac{d}{dr}(kr^2) = -2kr\)

For circular motion, the centripetal force required is provided by this force, and is given by \( \frac{L^2}{mr^3} \), the expression for centripetal force considering angular momentum:

\(\frac{L^2}{mr^3} = 2kr\)

Solve for \( r \) to determine the radius of the circular orbit:

\(L^2 = 2kmr^4\)

Rearranging gives:

\(r^4 = \frac{L^2}{2km}\)

Taking the fourth root on both sides:

\(r = \left(\frac{L^2}{2km}\right)^{\frac{1}{4}}\)

This matches the given option \((\frac{L^2}{2mk})^{\frac{1}{4}}\), thereby confirming it as the correct choice.