The orbital angular momentum of a satellite is L , when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight tim

Question

The orbital angular momentum of a satellite is L , when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be

TestHub · Accepted Answer

For a satellite revolving around the Earth in a circular orbit, it can be written that
$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}} . . . (1)$ ( $r = R + h$ is the distance from the centre of the Earth)
Simplify equation (1) to obtain the linear speed of the satellite.
$v = \sqrt{\frac{G M}{r}}$
Thus, the angular momentum of the satellite is given by
$\begin{array}{rcl} L & = & m v r \\ = & m r \sqrt{\frac{G M}{r}} \\ = & m \sqrt{G M r} . . . (2) \end{array}$
Similarly, for distance $r'$ from the centre of the Earth, the angular momentum is given by
$L' = m \sqrt{G M r'} . . . (3)$
Divide equation (3) by equation (2) to obtain the required angular momentum.
$\begin{array}{rcl} \frac{L'}{L} & = & \frac{m \sqrt{G M r'}}{m \sqrt{G M r}} \\ = & \sqrt{\frac{r'}{r}} . . . (4) \end{array}$
Now, $r^{'} = r + 8 r = 9 r$
Substitute the known values of the parameters into equation (4) to obtain the required angular momentum.
$\begin{array}{rcl} \frac{L'}{L} & = & \sqrt{9} \\ \Rightarrow & L' = 3 L \end{array}$

Physics - Gravitation Question with Solution | TestHub

Options:

Doubts & Discussion