Let 10 re be the distance of the satellite a from the center of the earth, where re is the radius of the earth. what is the gravitational acceleration due to the earth at satellite a?

a = 0.098 m/s²

Explanation:

The satellite, at any distance from the center of the Earth, is subject to the attractive force due to the Earth, according to the Newton´s Universal Law of gravitation, as follows:

Fg = G*ms*mE / (rse)²

According to Newton´s 2nd Law, neglecting any other force acted upon the satellite, we can write the following equation:

Fg = ms*a = G*ms*mE / (rse)²

⇒ a = G*mE / (rse)² (1)

As the distance between the satellite and the center of the Earth is 10 times the radius of the Earth, replacing this value in (1), we have:

a = G*mE / (10*RE)² = G*mE/(RE)² * (1/100)

but G*mE/(RE)², is just g, the acceleration due to gravity on the surface of the earth, so the gravitational acceleration due to Earth at satellite A, is as follows:

a = g*(1/100) = 0.01*g = 0.098 m/s²