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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?

Respuesta :

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

[tex]g_m = \frac{GM}{R^2}[/tex]

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

[tex]g = \frac{GM}{(3R+R)^2}[/tex]

now using this

[tex]g = \frac{GM}{16R^2}[/tex]

[tex]g = \frac{g_m}{16}[/tex]

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

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