   SEARCH HOME Math Central Quandaries & Queries  ROHAN, a student: AT WHAT HEIGHT FROM THE SURFACE OF EARTH,THE WEIGHT OF THE BODY BECOMES HALF? (RADIUS OF THE EARTH = 6400) Hi Rohan.

The formula for gravitational force F is given by

F = GMm/R2

where G is the gravitational constant (good for the whole universe), M is the mass of one object (say the Earth), m is the mass of the other object (say, you) and R is the distance between the centers of the two objects.

So for your question, G, m and M don't change, just F and R.

If we say R is the radius of the earth and h is the height about the surface of the earth at which the force is half the original, then we have this:

(1/2)GMm/R2 = GMm/(h+R)2

And we try to solve this now:
1/(2R2) = 1/(h2 + 2Rh + R2)
2R2 = h2 + 2Rh + R2
h2 + 2Rh - R2 = 0

This is a quadratic. So we can use the quadratic formula to calculate it:

h = [ -b ± √(b2 - 4ac) ] / (2a)

where a, b and c are the co-efficients in ah^2 + bh + c = 0, so in our case a = 1, b = 2R, c = -R^2

h = [-2R ± √(4R2 + 4R2) ] / 2
h = [-1 ± √(2)] R

Now clearly of these two answers, the negative value of h is not all that useful; we want the height above ground, not the height below ground. So the answer is h = [-1 + √(2) ] R

According to Wikipedia, the mean radius R of the earth is 6,371 km. Thus h = [-1 + 1.41421] 6371 = 2639 km.

So if you are 2639 km above earth's surface, the gravitation force that the planet exerts on you is half what you experience on the surface.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.