Chapter 8, Problem 69GP

To determine

To Find: The ratio of the moon’s spin angular momentum (about its own axis) to its orbital angular momentum.

Explanation of Solution

Introduction:

Spin is an intrinsic form of angular momentum carried by composite particles and elementary particles, etc. Tidal locking also known as gravitational locking (spin-orbit locking) occurs when an orbiting astronomical body always has same face towards the object it is orbiting.

Assume,

The radius of the Moon, $R$

Distance between Earth and the Moon, $r$

The moon’s spin angular momentum, $L_s$

The moon’s orbital angular momentum, $L_0$

The ratio between the moon’s spin angular momentum to the orbital angular momentum, $=\frac{L_s}{L_o}$

Where,

  $L_s$ =Spin angular momentum

  $L_0$ = Orbital angular momentum

For sphere, $I=\frac{2}{5}m{R}^2$

  $\frac{L_s}{L_o}=\frac{I{\omega }_s}{I{\omega }_0}=\frac{\frac{2}{5}m{R}^2\omega }{m{r}^2\omega }$

  $\frac{L_s}{L_o}=\frac{2}{5}\left(\frac{R^2}{r^2}\right)$

Conclusion: 

The ratio of the Moon’s spin angular momentum to its orbital angular momentum is $\frac{2}{5}\left(\frac{R^2}{r^2}\right)$ .