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