Question

How high above the earth's surface would the acceleration of a free falling object be half its value at earth's surface? Write the answer in terms of Earth's radius r.

Answer 1

The gravitational force varies as the Square of the distance between the centers of the two masses (in this case the distance between the center of the Earth and the center of the mass we are discussing).  

Therefore, being at a distance of twice the Earth's radius, the gravitational force would be one-quarter.

At an altitude h above the surface of the Earth, [math]g_2 = \dfrac {g_1}{2}[/math]

or, [math]\dfrac{GM}{{R+h}^2} = \dfrac{\frac{GM}{R^2}}{2}[/math]

or, [math]{R+h}^2 = 2{R^2}[/math]

And math]2*\dfrac{M}{(R+h)^2} - 2*\dfrac{M_s}{(R_s+x)^2