A 213.7 kg satellite is in a circular orbit of 22,236 miles (35,768 km) in radius. The force keeping the satellite in orbit is 44.1 N. What is the velocity (speed) of the satellite

v = 3.08 k ms^-1

Explanation:

As we know that,

where r = R + h = 35768 + 6400 = 42168 km = 4.2168 x 10^7 m

G = 6.673 x 10^-11 N m^-2 kg^-2 and M = 6.0 x 10^24 kg

By substituting the required values, we get...

v = 6.673 x 10^-11 x 6.0 x 10^24 / 4.2168 x 10^7 = 9.494 x 10^6 ms^-1 = 3081.2 ms^-1 = 3.08 k ms^-1

i think the answer is a and b

i wish you luck