Question

The height at which the weight of a body becomes $\frac{1}{16}th$ its weight on the surface of (radius$R$) is

• A. $3R$
• B. $4R$
• C. $5R$
• D. $15R$

Answer 1

The correct option is A

$3R$

Step 1: Given that:

The weight of a body becomes$\frac{1}{16}th$.

Know that, when the body is on the ground, that is the height is zero and when the body is at some height from the ground that is height$=h$

Step 2: Formula used of gravity:

Therefore the gravity, $g=\frac{GM}{R^2}$

Where$M$ is the earth's mass, $R$ is its radius, $G$ is the gravitational constant, and$g$ is the acceleration caused by gravity.

Know that the relation between the weight, mass, and the acceleration due to gravity,

$w=mg$ where$w$ is weight, $m$is mass and $g$ is the acceleration due to gravity.

Suppose that two different equations with the mass$\left(m\right)$ being constant.

$$\begin{gathered} w_1=m{g}_1 \\ m=\frac{w_1}{g_1}.......................\left(1\right) \\ {w}_2=m{g}_2 \\ m=\frac{w_2}{g_2}.........................\left(2\right) \end{gathered}$$

Therefore from the equation $\left(1\right)\mathrm{and}\left(2\right)$

$\frac{w_1}{g_1}=\frac{w_2}{g_2}.........................\left(3\right)$

Step 3: Use the condition and write the equation:

When the body is on the ground, that is the height is zero and when the body is at some height from the ground that is height is $“h”$.

Therefore,

$$\begin{gathered} g_1=\frac{GM}{R^2} \\ {g}_2=\frac{GM}{{\left(R+h\right)}^2} \end{gathered}$$

Now from the equation $\left(3\right)$.

$$\begin{gathered} \frac{w_1}{w_2}=\frac{{\displaystyle \frac{GM}{R^2}}}{{\displaystyle \frac{GM}{{\left(R+h\right)}^2}}} \\ \frac{w_1}{w_2}=\frac{{\displaystyle {\left(R+h\right)}^2}}{{\displaystyle R^2}}..........................\left(4\right) \end{gathered}$$

Step 4: Substitute the value of the weight in the equation$\left(4\right)$.

$$\begin{gathered} \frac{w_1}{{\displaystyle \frac{1}{16}}{w}_1}=\frac{{\left(R+h\right)}^2}{R^2} \\ \frac{R+h}{R}=4 \\ R+h=4R \\ h=3R \end{gathered}$$

Therefore the height at which the weight of a body becomes $\frac{1}{16}th$ its weight on the surface of (radius$R$) is$3R$.

Hence, the correct option is $\left(A\right)$.