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=(G* M* 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=(G* M* 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=(GMm)/(12000^2)`

On dividing the above two equations,

`(F_a)/(W)=\frac{4000^2^{}}{12000^2}=(1)/(9)`

Therefore,

`F_a=(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.