Calculate the force of gravity on a spacecraft 12800 km above the earth's surface if its mass is 700 kg.

Answer 1

The acceleration of gravity (g') is reduced from g = 9.81 m/s^2 (the value at the earth's surface) by an inverse-square-law factor
[Re/(12800 + Re)]^2

Re is the radius of the Earth, which is about 6400 km. Therefore the value of g' is about (1/3)^2 = 1/9 of the value at the Earth's surface.

The weight (force of gravity) of the 700 kg object at that altitude is
W = M g', even though it will appear to be "weightless" inside the spacecraft, because the spacecraft moves with it.

Answer 2

Inverse square law factor: [Re/m+Re]^2
Re=6400km
Plugging in your values you get 1/3

W=mg
700*9.81 = 6867

(1/3)(6867) = 2289N is your answer