The orbital angular momentum of a satellite is L , when it is...

For a satellite revolving around the Earth in a circular orbit, it can be written that

$\dfrac{m{v}^{2}}{r} = \dfrac{GMm}{{r}^{2}} . . .$\left(1\right)$($r = R + h$is the distance from the centre of the Earth)</p><p>Simplify equation (1) to obtain the linear speed of the satellite.</p><p>$v = \sqrt{\dfrac{GM}{r}}$</p><p>Thus, the angular momentum of the satellite is given by</p><p>$\begin{cases}L & = & mvr \\ & = & mr\sqrt{\dfrac{GM}{r}} \\ & = & m\sqrt{GMr} . . . \left(2\right)\end{cases}$</p><p>Similarly, for distance$r '$from the centre of the Earth, the angular momentum is given by</p><p>$L ' = m\sqrt{GMr '} . . . \left(3\right)$</p><p>Divide equation (3) by equation (2) to obtain the required angular momentum.</p><p>$\begin{cases}\dfrac{L '}{L} & = & \dfrac{m\sqrt{GMr '}}{m\sqrt{GMr}} \\ & = & \sqrt{\dfrac{r '}{r}} . . . \left(4\right)\end{cases}$</p><p>Now,${r}^{'} = r + 8r = 9r$</p><p>Substitute the known values of the parameters into equation (4) to obtain the required angular momentum.</p><p>$\begin{cases}\dfrac{L '}{L} & = & \sqrt{9} \\ & ⇒ & L ' = 3L\end{cases}