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boyakko [2]
1 year ago
5

"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 th

e surface of the earth, the force of gravity then becomes what fraction of my present weight?"
Physics
1 answer:
lord [1]1 year ago
5 0

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.

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D/H =15

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