The orbital angular momentum of a satellite is $\mathrm{L}$, when it is revolving in a circular orbit at height $\mathrm{h}$ 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

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

$\frac{m{v}^2}{r}=\frac{GMm}{r^2} ...\left(1\right)$ ($r=R+h$ is the distance from the centre of the Earth)

Simplify equation (1) to obtain the linear speed of the satellite.

$v=\sqrt{\frac{GM}{r}}$

Thus, the angular momentum of the satellite is given by

$\begin{array}{rcl}L& =& mvr\\ & =& mr\sqrt{\frac{GM}{r}}\\ & =& m\sqrt{GMr} ...\left(2\right)\end{array}$

Similarly, for distance $r\text{'}$ from the centre of the Earth, the angular momentum is given by

$L\text{'}=m\sqrt{GMr\text{'}} ...\left(3\right)$

Divide equation (3) by equation (2) to obtain the required angular momentum.

$\begin{array}{rcl}\frac{L\text{'}}{L}& =& \frac{m\sqrt{GMr\text{'}}}{m\sqrt{GMr}}\\ & =& \sqrt{\frac{r\text{'}}{r}} ...\left(4\right)\end{array}$

Now, $r^{\text{'}}=r+8r=9r$

Substitute the known values of the parameters into equation (4) to obtain the required angular momentum.

$\begin{array}{rcl}\frac{L\text{'}}{L}& =& \sqrt{9}\\ & \Rightarrow & L\text{'}=3L\end{array}$