# Question 25b65

Feb 23, 2018

Option (3) only if ${\omega}_{o} = {\omega}_{s} \setminus i . e . ,$
Orbital angular velocity is equal to spin angular velocity.

#### Explanation:

Let the spherical satellite of mass $m$ be spinning about an axis passing through its center with angular velocity ${\omega}_{s}$.
Its spin angular momentum ${L}_{s}$ is is given as

${L}_{s} = I \omega$
where $I$ is its moment of inertia and for a sphere $= \frac{2}{5} m {a}^{2}$

$\therefore {L}_{s} = \frac{2}{5} m {a}^{2} {\omega}_{s}$ .........(1)

Orbital angular momentum ${\vec{L}}_{o}$ of satellite, considered a point mass, revolving in a circle of radius $r$ is given as the cross product of the position vector $\vec{r}$ and its linear momentum $\vec{p} = m \vec{v}$.

vecL = vecr × vecp#

Let ${\omega}_{o}$ be orbital angular velocity. We have $v = b {\omega}_{o}$.
Orbital velocity is always perpendicular to the radius vector hence $\sin \theta$ of the cross product is $= 1$, and $r = b$, we get

${L}_{o} = {b}^{2} m {\omega}_{o}$ .........(2)

Ratio of orbital angular momentum to spin angular momentum is

${L}_{o} / {L}_{s} = \frac{{b}^{2} m {\omega}_{o}}{\frac{2}{5} m {a}^{2} {\omega}_{s}}$
${L}_{o} / {L}_{s} = \frac{5 {b}^{2} {\omega}_{o}}{2 {a}^{2} {\omega}_{s}}$