Earth has a mass of 5.97 x 1024 kg, and a mean radius of 6.38 x 106 m. what is the orbital speed of a satellite 6.16 x 108 m above the surface of the earth? round the answer off to the nearest whole number. (g = 6.67 x 10-11 n·m2/kg2) a. 800m/s c. 80,000m/s b. 8000m/s d. none of the above.

The first thing you should know is that the gravitational force is:
 F = GMm / r ^ 2
 Then, by definition, the force is equal to
 F = m * a = mv ^ 2 / r
 We match both expressions:
 GMm / r ^ 2 = mv ^ 2 / r
 Let's clear the speed:
 v = (GM / r) ^ (1/2)
 On the other hand:
 r = R + h
 Substituting:
 v = (GM / (R + h)) ^ (1/2)
 Where,
 G = 6.67x10 ^ -11 n · m2 / kg2
 M = 5.98x10 ^ 24 kg
 R = 6.38x10 ^ 6 m
 h = 6.16 x 10 ^ 8 m
 Substituting the values:
 v = ((6.67 * 10 ^ -11) * (5.98 * 10 ^ 24) / ((6.38 * 10 ^ 6) + (6.16 * 10 ^ 8))) ^ (1/2)
 v = 800m / s
 
 the orbital speed of a satellite 6.16 x 108 m above the surface of the earth is
 a. 800m/s
 

The answer is b) 32 your welcome