Question

A man weight $W$ on the surface of the earth and his weight at a height $R$ from the surface of the earth is ( $R$ is the radius of the earth)
(A)${\displaystyle \frac{W}{4}}$
(B)${\displaystyle \frac{W}{2}}$
(C)$W$
(D)$4W$

Answer 1

Hint: The weight of a body is the force acting on that body due to gravitational pull. S.I. unit of weight is newton $(N)$ . The mass of a body always remains constant but the weight of a body changes with the change in gravitational force acting on it.
Formula used:
$W=mg$
$W={\displaystyle \frac{GMm}{D^2}}$​
where $W$ is the gravitational force acting between two bodies, $m$ is mass of one of the body, $M$ is mass of another body, $G$ is the universal gravitational constant, and $D$ is the distance between the center of the two bodies.

Complete Step-by-step solution:
It is given that the weight of the man on the surface of the earth is $W$.
Hence the mass $(m)$ of man will be ${\displaystyle \frac{W}{g}}$.
i.e. $\Rightarrow m={\displaystyle \frac{W}{g}}$
where $g$is the acceleration due to gravity near the earth’s surface which is equal to ${\displaystyle \frac{GM}{R^2}}$.
From Newton's law of gravitation,
$W={\displaystyle \frac{GMm}{D^2}}$​
Where $W$ is the gravitational force acting between two bodies, $m$ is the mass of one of the body, $M$ is the mass of another body, $G$is the universal gravitational constant, and $D$ is the distance between the center of the two bodies.
 In our solution with regard to the question, it can be written as,
$\Rightarrow W={\displaystyle \frac{GMm}{R^2}}$​
where $W$is the weight of man at the surface of the earth, $m$ is the mass of man, $M$ is the mass of earth, $G$is the universal gravitational constant, and $R$is the radius of the earth.
 Let the weight of man at a height $R$from the surface of the earth i.e. at a distance of $2R$from the center of the earth be $W_0$.
So,
$\Rightarrow W_0={\displaystyle \frac{GMm}{{(R+R)}^2}}$
$\Rightarrow W_0={\displaystyle \frac{GMm}{{(2R)}^2}}={\displaystyle \frac{GMm}{4R^2}}$
As we have seen earlier, $W={\displaystyle \frac{GMm}{R^2}}$.
Hence,
$\Rightarrow W_0={\displaystyle \frac{W}{4}}$

Therefore the correct answer to our question is (A) ${\displaystyle \frac{W}{4}}$.

Note:
In many questions with an increase in height, we did not consider the weight of the body is changing as the change of weight with small variation in height (in comparison to radius of the earth) will be very small and thus can be neglected. But in this question change in height was not small and was comparable to the radius of the earth thus the change produced can’t be neglected.