Question

At what altitude is the acceleration due to gravity one-half that on Earth’s surface?

Answer 1

Well, you need the following information:\[M_E = 5.9722\times10^{24}\]where Me represents the average mass of the earth.\[R_E=6.371\times10^6\]and Re represents the average radius of the earth (this means we're ON the surface). And of course we need the gravitational constant,\[G = 6.67\times10^{-11}\]According to Newton's formula, we have,\[F=\frac{ G* M_E }{ R_E^2}\]if you input those values you get F=9.81 which is the average acceleration on the surface. Now to get the result you need, you just substitute F=(9.81)/2 and solve for Re. Can you do that? (NOTE: I am working in the metric system)