Question

A man weighs 60 kg at earth’s surface. At what height above the earth’s surface his weight becomes 30 kg? (radius of earth $= 6400km$)
(A) 1624 km
(B) 2424 km
(C) 2624 km
(D) 2826 km

Answer 1

The weight of a person near the gravitational field of a celestial body (such as planets) is inversely proportional to the square of the distance from the centre of the body. The distance at the surface of the earth would be the radius of the earth.
Formula used: In this solution we will be using the following formulae;
$W = \dfrac{k}{{{r^2}}}$ where $W$ is the weight of an object, and $r$ is the distance from the centre of the earth.

Complete step by step solution:
Generally, the weight of a person or any object for that matter is dependent on the location of the object. How close or how far is it from the gravitational field causing the weight? In terms of distance from the field, the weight of a body is inversely proportional to the square of the distance to the centre of the body, i.e. in mathematical terms
$W \propto \dfrac{1}{{{r^2}}}$
$\Rightarrow W = \dfrac{k}{{{r^2}}}$
Hence, it can be written as
${W_1}r_1^2 = {W_2}{r_2}^2$
Now, at the surface of the earth,
${r_1} = R = 6400km$ and ${W_1} = 60kg$
It is 30 kg at an unknown distance from the surface of the earth. Hence
${r_2} = R + h = \left( {6400 + h} \right)km$ (since the distance from the centre would be the radius of the earth plus the actual height above the earth surface)
Then, by inserting the known values into ${W_1}r_1^2 = {W_2}{r_2}^2$
$60 \times {6400^2} = 30 \times {r_2}^2$
$\Rightarrow {r_2}^2 = \dfrac{{60 \times {{6400}^2}}}{{30}}$
By computation, we have
${r_2} = \sqrt {81920000} = 9051km$
But ${r_2} = R + h = \left( {6400 + h} \right)km$
Then
$h = {r_2} - R = 9051 - 6400 = 2651km$
The solution is very close to option C.

Hence, option (C) is the correct answer.

Note: For clarity, although the unit kg is used in quantifying the weight, the quantity of interest is still the weight. Note that the mass of an object does not change with location, hence in actual terms, the mass would still be 60 kg irrespective of height above the surface.