Question

"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above the surface of the earth, the force of gravity then becomes what fraction of my present weight?"

Answer 1

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

$F_s=\frac{G\times M\times m}{d^2}=W$

On substituting the of d,

$W=\frac{\text{GMm}}{4000^2}$

At a height of 8000 miles from the surface, the gravitational force is equal to,

$F_a=\frac{GMm}{12000^2}$

On dividing the above two equations,

$\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}$

Therefore,

$F_a=\frac{1}{9}W$

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.