JEE Main 2023 — Gravitation Question with Solution

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:

$L=mvr=m\sqrt{GMr}$

If the distance (r) from the Earth's center is increased by a factor of 8, the new angular momentum $L^{\prime }$ is:

$L^{\prime }=m\sqrt{GM(8r)}=m\sqrt{8GMr}=2\sqrt{2}L$

However, the distance from the earth'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's surface is increased eight times, the distance from the earth'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).

So, the new angular momentum (L'') is:

$L^{\prime \prime }=m\sqrt{GM(9R)}=3\sqrt{GM(R)}=3L$