Question

1. A moon of mass \(m\) orbits around a non-rotating planet of mass \(M\) with orbital angular velocity \(\Omega\). The moon also rotates about its own axis with angular velocity \(\omega\). The axis of rotation of the moon is perpendicular to the plane of the orbit. Let \(I\) be the moment of inertia of the moon about its own axis. You can assume \(m<<M\)so that the center of

mass of the system is at the center of the planet.

(a) What is the total angular momentum \(L\) of the system about the center of the planet? What is the total energy \(E ?\)

(b) In general, the two angular velocities \(\omega\) and \(\Omega\) are unequal. Suppose there is a mechanism which reduces \(E\) if \(\omega \neq \Omega\), but conserves angular momentum. Show that it is possible to obtain a stable configuration with \(\omega=\Omega\) if the final separation \(D\) between the moon and planet satisfies \(D>\sqrt{3 I / m}\). You can assume that the orbit is circular during the evolution.

[Hint: Consider \(E\) as a function of \(\omega\). For a configuration to be stable, \(E\) must be a minimum.]

Answer 1

(a) Total angular momentum L is given by LE IR + m woz ( When D can be calculated as - Yan >I V=WD moon - circular barn •for circular motion Centripetal foud - Gravitational force so, mw²D² = GMm D = GM 33 Herce L=IRt mw GM And total energy of system &= notational Energy of planet + + P + KE of moont pe of system е E I² + I w²0² Mem D putting all the values ! E: /I ² + √ mw? (ampt G M m / w2 1/2 I 2? + * - m ( MW) 22 Gr (M) w 3 m I m (GMW ) GIM