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 -

<p>If we take the velocity (v) of the satellite to be $v = \sqrt{GM/r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is:</p> <p>$ L = mvr = m \sqrt{GMr} $</p> <p>If the distance (r) from the Earth&#39;s center is increased by a factor of 8, the new angular momentum $L&#39;$ is:</p> <p>$ L&#39; = m \sqrt{GM(8r)} = m \sqrt{8GMr} = 2 \sqrt{2} L $</p> <p>However, the distance from the earth&#39;s surface is given, so we have to take into account the radius of the Earth ($R$) in our calculations. When the height from the earth&#39;s surface is increased eight times, the distance from the earth&#39;s center is $r = R + 8h$, which is approximately $9R$ (because the height of the satellite above the Earth is generally much less than the radius of the Earth).</p> <p>So, the new angular momentum (L&#39;&#39;) is:</p> <p>$ L&#39;&#39; = m \sqrt{GM(9R)} = 3 \sqrt{GM(R)} = 3L $</p>