A satellite of mass m and radius R is moving in a circular orbit of radius r around a planet of mass M.

Options

A. The magnitude of its angular momentum with respect to the centre of the orbit is `m sqrt(GMr)`, where G is the gravitation constant and the direction of L is perpendicular to the plane of the orbit
B. The magnitude of its angualr momentum is `mR sqrt(2gr)` where is the acceleration due to gravity on the surface of the planet
C. The direction of angular momentum is parallel to the plane of the orbit
D. The direction of angular momentum is inclined to the plane of the orbit

Answer

A

Explanation

The magnitude of angular momentum (L) of a satellite, moving in a circular orbit of radius r about a planet of mass m is given by
`L = I omega = mr^(2).(v)/(r )= mvr`
But the orbital velocity `v = sqrt((GM)/(r ))`
`therefore L = mvr = m sqrt((GM)/(r )).r = m sqrt(GMr)`
This is option (a).
The direction of angular momentum is perpendicular to the plane of the orbit. Thus (a) is the correct option.