Question

the mass of earth is 5.972 * 1024 kg, and the radius of earth is 6,371 km. a 750 kg satellite orbits 35,800 km above earth in a perfectly circular orbit. if the gravitational force is acting as the centripetal force to keep the satellite in orbit, what is the tangential velocity of the satellite in its orbit?

Answer 1

The speed or the tangential velocity of the satellite to keep it in its orbit is  3,073.5 meters per second.

The gravitational force between two objects is given by:

F = GMm/r²

Where:

G = constant of gravity = 6.674 x 10⁻¹¹ N.m²/kg²

M, m = mass of each object

r = distance between objects

Parameters given in the problem:

M = 5.972 x 10²⁴ kg

m = 750 kg

r = 6371 + 35,800 = 42,171 km = 42,171,000 m

Hence,

F = 6.674 x 10⁻¹¹ x 5.972 x 10²⁴ x 750 / 42,171,000²

   = 168.1 N

This is equal to centripetal force:

F = m  . v² / r

168.1  = 750 .  v² /  42,171,000

v² = 9,446,304

v = 3,073.5 m/s

Hence, the tangential velocity of the satellite to keep it in its orbit is  3,073.5 m/s

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