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 $L'$ is:

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