Question

The height at which the weight of a body becomes$\dfrac{1}{16}th$, its weight on the surface of the earth (radius R), is
A. 3R
B. 4R
C. 5R
D. 15R

Answer 1

The question is based on the concept of gravity. This problem can be solved considering the two cases involving the height difference and the related weight difference. As the mass remains constant, using the parameters, weight, height and the radius, the calculation is carried out.
Formula used:
$g=\dfrac{GM}{{{R}^{2}}}$

Complete answer:
Using the gravity equation, we have,
$g=\dfrac{GM}{{{R}^{2}}}$
Where g is the acceleration due to gravity, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.

As, the mass is a universal constant, thus the mass of the body remains constant irrespective of the height reached by that body.
Let us consider the two different cases.
Case I: When the body is on the ground, that is, the height is zero.
Case II: When the body is at some height from the ground, that is, the height is ‘h’.

We know the relation between the weight, mass and the acceleration due to gravity, that is,
$w=mg$
Where w is the weight, m is the mass and g is the acceleration due to gravity.

As, we have considered two different cases, so we get two different equations, with the mass being constant.

& {{w}_{1}}=m{{g}_{1}} \\\ & \Rightarrow m=\dfrac{{{w}_{1}}}{{{g}_{1}}} \\\ \end{aligned}$$……. (1) Similarly, $$\begin{aligned} & {{w}_{2}}=m{{g}_{2}} \\\ & \Rightarrow m=\dfrac{{{w}_{2}}}{{{g}_{2}}} \\\ \end{aligned}$$……. (2) Equate the equations (1) and (2), as the LHS of both the equations are the same. So, we get, $$\dfrac{{{w}_{1}}}{{{g}_{1}}}=\dfrac{{{w}_{2}}}{{{g}_{2}}}$$ ……. (3) From here, we will proceed with the gravity equation. So, we have, Case I: When the body is on the ground, that is, the height is zero. Case II: When the body is at some height from the ground, that is, the height is ‘h’. $$\begin{aligned} & {{g}_{1}}=\dfrac{GM}{{{R}^{2}}} \\\ & {{g}_{2}}=\dfrac{GM}{{{(R+h)}^{2}}} \\\ \end{aligned}$$ Use the above equation values in the equation (3) $$\begin{aligned} & \dfrac{{{w}_{1}}}{{{w}_{2}}}=\dfrac{{{g}_{1}}}{{{g}_{2}}} \\\ & \Rightarrow \dfrac{{{w}_{1}}}{{{w}_{2}}}=\dfrac{\dfrac{GM}{{{R}^{2}}}}{\dfrac{GM}{{{(R+h)}^{2}}}} \\\ \end{aligned}$$ Further, solve the above equation. $$\begin{aligned} & \dfrac{{{w}_{1}}}{{{w}_{2}}}=\dfrac{{{g}_{1}}}{{{g}_{2}}} \\\ & \dfrac{{{w}_{1}}}{{{w}_{2}}}=\dfrac{{{(R+h)}^{2}}}{{{R}^{2}}} \\\ \end{aligned}$$ Now substitute the values of the weight in the above equation. $$\begin{aligned} & \dfrac{{{w}_{1}}}{\dfrac{1}{16}{{w}_{1}}}=\dfrac{{{(R+h)}^{2}}}{{{R}^{2}}} \\\ & \Rightarrow 16=\dfrac{{{(R+h)}^{2}}}{{{R}^{2}}} \\\ \end{aligned}$$ Take the square root on both sides. So, we get, $$\begin{aligned} & \sqrt{16}=\sqrt{\dfrac{{{(R+h)}^{2}}}{{{R}^{2}}}} \\\ & \Rightarrow 4=\dfrac{R+h}{R} \\\ \end{aligned}$$ Further solve the above equation. $$\begin{aligned} & 1+\dfrac{h}{R}=4 \\\ & \Rightarrow h=3R \\\ \end{aligned}$$ The height at which the weight of a body becomes$$\dfrac{1}{16}th$$, its weight on the surface of the earth (radius R), is 3R. **So, the correct answer is “Option A”.** **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The terms, ‘g’ and ‘G’ are different, ‘g’ is the gravitational acceleration whose value equals $$9.8\,{N}/{kg}\;$$, whereas, ‘G’ is the gravitational constant whose value equals $$6.67\times {{10}^{-11}}{N{{m}^{2}}}/{k{{g}^{2}}}\;$$, but, we use the term/abbreviation ‘g’ to represent the gravitational constant.