Question

A satellite encircles Mars at a distance above its surface equal to 3 times the radius of Mars. If g m is the acceleration due to gravity at the surface of Mars, what is the acceleration due to gravity at the location of the satellite?

Answer 1

Acceleration due to gravity at the surface of Mars is given by the formula

$g_m = \frac{GM}{R^2}$

here

M = mass of Mars

R = radius of Mars

Now if we take a point which is at height 3 times the radius of Mars from its surface

So the acceleration due to gravity is given as

$g = \frac{GM}{(3R+R)^2}$

now using this

$g = \frac{GM}{16R^2}$

$g = \frac{g_m}{16}$

so here the gravity is 16 times lesser than gravity at surface