The orbital angular momentum of a satellite is $$\mathbf{L}$$, 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

For a satellite in circular orbit, the orbital angular momentum is $$L = mvr$$, where $$v$$ is the orbital velocity and $$r$$ is the distance from the Earth's centre.

For a circular orbit, gravitational force provides centripetal force, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

which gives $$v = \sqrt{\frac{GM}{r}}$$.

Substituting into the angular momentum expression:

$$L = mr\sqrt{\frac{GM}{r}} = m\sqrt{GMr}$$

So $$L \propto \sqrt{r}$$. Now, the problem says the distance is increased by eight times to its initial value, meaning the new distance is $$r' = r + 8r = 9r$$ (increased by eight times the original).

The new angular momentum is then

$$\frac{L'}{L} = \sqrt{\frac{9r}{r}} = \sqrt{9} = 3$$

So $$L' = 3L$$. Hence, the correct answer is Option 4.