Question

Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is given by g = Gm/R2 . What is the acceleration of a satellite moving in a circular orbit round the earth of radius 2R​

Answer 1

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

$W=F_G$

$mg = G \dfrac{mM}{R^2}$

which gives us an expression for the acceleration due to gravity g as

$g = G\dfrac{M}{R^2}$

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity $g_h$ at this height is

$mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}$

Simplifying this, we get

$g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g$