Question

A satellite of mass m is revolving in a circular orbit of radius r around the earth. Its angular momentum w.r.t. the center of its orbit is
(M = mass of the earth, G = gravitational constant)

A.)${\left( {GMmr} \right)^{\dfrac{1}{2}}}$
B.)${\left( {GM{m^2}r} \right)^{\dfrac{1}{2}}}$
C.)${\left( {GM{m^2}{r^2}} \right)^{\dfrac{1}{2}}}$
D.)${\left( {G{M^2}{m^2}r} \right)^{\dfrac{1}{2}}}$

Answer 1

Hint- For any circulatory motion there is always a force called centrifugal force ,which acts constantly on any object of its revolution along a circular path with constant speed .In this case there is gravitational force acting in the satellite for its revolution, which makes it possible to rotate .

Complete step-by-step answer:
Angular momentum- A rigid object's angular momentum is defined as the product of both the moment of inertia and angular velocity.

Angular Momentum = (moment of inertia) (angular velocity)

$L = I\omega$

Where

L = angular momentum
I = moment of inertia $\left( {kg{m^2}/s} \right)$
$\omega$ = angular velocity $(radians/s)$

Given

Mass of the object is m
Radius of the orbit is r

Since, the object is revolving in a circular orbit around the earth; the force acting on the object is the gravitational force of the earth which is balanced by the centrifugal force.

$\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$

Solving the above equation for the value of v, we get

$v = \sqrt {\dfrac{{GM}}{r}}$

Also, the angular momentum of satellite

$L = mvr$

Substituting the value of v from the above equation we get

$L = m{\left( {\dfrac{{GM}}{r}} \right)^{\dfrac{1}{2}}}r \\\ L = {\left( {GM{m^2}r} \right)^{\dfrac{1}{2}}} \\\$

Hence, the correct option is B.

Note-This centripetal force is supplied by gravity - the force that universally acts at a distance between any two objects that have mass. Were it not for this impact, the satellite should travel at the same speed and in the same direction. It would follow its inertial, straight-line path. Like any projectile, gravity alone influences the satellite's trajectory such that it always falls below its straight-line, inertial path.