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