Physics problem.............?

http://en.wikipedia.org/wiki/Gravitation

http://en.wikipedia.org/wiki/Earth%27s_gravity

http://en.wikipedia.org/wiki/Earth_radius

1st please note:

Gravity decreases with altitude, since greater altitude means greater distance from the Earth's center. All other things being equal, an increase in altitude from sea level to the top of Mount Everest (8,850 metres) causes a weight decrease of about 0.28%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object's buoyancy.[3]) It is a common misconception that astronauts in orbit are weightless because they have flown high enough to "escape" the Earth's gravity. In fact, at an altitude of 400 kilometres (250 miles), equivalent to a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the Earth's surface, and weightlessness actually occurs because orbiting objects are in free-fall.

g = (G * M_earth) / r^2

to solve this we need to modify r so that g becomes 1/4 it's original value!

Let r_start = Earth_radius and g_start = gravity at Earth_surface...

g_start = (G * M_earth) / r_start^2

making r_final = (r_start)*2 gives...

g_final = (G * M_earth) / ((r_start)*2)^2

g_final = (G * M_earth) / ((r_start)^2 * 2^2)

g_final = ((G * M_earth) / (r_start)^2) / (2^2)

g_final = g_start / (2^2) = g_start / 4

answer is height above surface equals 1 "Earth radius".

When height equals 2 "Earth radius"...

g_final becomes 1/(3^2) the value of the gravity experienced at Earth's surface.

Good luck.