Question

An Earth satellite of mass 'm' revolves at a height 'h' from the surface of the Earth. If 'R' is the radius and 'g' is the acceleration due to gravity at the surface of the Earth, then the velocity of the satellite is given by- (Give a detailed procedure of getting the answer as well as the correct option)-

a) $\frac{Gh}{R}$

b) $\sqrt{ \frac{Gh}{R}}$

c) $\sqrt{ \frac{gR^2}{R+h}}$

d) $\sqrt{ \frac{g(R-h)^2}{R+h}}$

Answer 1

Mass of Earth = M
acceleration due to gravity on surface of Earth = g
mass of satellite = m
altitude of satellite above Earth's surface = h
Earth's radius  = R
distance of satellite from center of Earth = R + h

For any satellite to go around in a circular orbit, there should be a centripetal force that makes it go around in a circle.  For a satellite, this force is supplied by the gravitational force of attraction (Newton's law).

 let v = linear velocity of satellite along the circular orbit.

$F=\frac{GMm}{(R+h)^2}=\frac{mv^2}{(R+h)}\\\\= >\ \ v^2=\frac{GM}{R+h}\\\\On\ Earth's\ surface weight=gravitational\ force\ due\ to\ Earth\\\\mg=\frac{GMm}{R^2}\\\\= > \ \ g=\frac{GM}{R^2},\ \ \ = > \ \ GM=gR^2\\\\Hence,\ v^2=\frac{gR^2}{R+h}\\\\v=\sqrt{\frac{gR^2}{R+h}}$

Option C.

Actually the orbit is an elliptical orbit with very small deviation from circle.  However, for a simple solution, we assume it as a circular orbit.

Answer 2

Heya.....

======= Answer =======

Option C is correct ...

____________ Solution ___________

For any satellite to go around in a circular orbit , there should be a centripetal force that makes it to go around...

That is :-

let y = linear velocity of satellite along moving around

F = GMm/(R+h)^2

= mv^2/(R+H)

= v^2 = GM/R+H

Earth face weight :-

mg = GMm/R^2

g = GM/R^2

GM = gR^2

v^2 = √gR^2/R+h

Thank you