at what altitude above the earth’s surface must they be at?

Question

If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight to drop to 800 N? Explain.

Answer 1

Using the definition of weight as being the magnitude of the gravitational pull,

this magnitude is proportional to ~ 1/r², r the distance between the object and the Earth (or another object). The further they are, the lower the weight will be. Do you double the distance, the weight of the object will be 4 times lower.

The weight of a person on the surface of the earth is approximately:

W_s = GMm/R²,

with G the gravitational constant, M mass of the Earth, m the mass of the person and R the radius of the Earth.

We know this equals to 882 N.

The weight of a person on a certain height 'h' above the Earth's surface is:

W_h = GMm/(R+h)²

Your question now is to find the value of 'h' such that W_h = 800 N knowing that W_s = 882 N.

This is now an algebra problem. I am not going to give you all the steps, you should try it out yourself.

You don't need the values of G, M and m to solve this. R = 6.371 x 10⁶ m

Substituting W_s in W_h:

W_h = W_sR² / (R + h)²

Solving this for h gives you: h = 3.19 x 10⁵ m