Question

What is the magnitude of g at a height above Earth's surface where free fall acceleration equals 6.5m/s^2?

Answer 1

Assuming you are seeking the altitude where gravity = g = 6.5m/s^2:

As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the expression g = GM/r^2 = µ/r^2 where GM = µ = the gravitational constant of the body (G = the Universal Gravitational Constant and M = the mass of the body) and r = the distance from the center of the body to the point in question.

Therefore, r = sqrt[µ/g]
= sqrt[3.986365x10^14/6.5
= sqrt[6.132869x10^13] = 7,831,263m
= 7831.263km.

The distance above the surface is therefore h = ~(7831 - 6378) = 1453km.