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 answer: the orbital speed of a satellite 6.16 x 108 m above the surface of the earth is a. 800m/s
