Question

The acceleration due to gravity becomes $\frac{g}{2}$ at a height equal to ($g$= acceleration due to gravity on the surface of the earth)

• A. $4R$
• B. $\frac{R}{4}$
• C. $2R$
• D. $\frac{R}{2}$

Answer 1

The correct option is B

$\frac{R}{4}$

Step 1: Given data and assumptions

Suppose the height will be- $h$

The acceleration due to gravity at height $\left(h\right)$ is $g\text{'}$.

Then, $g\text{'}=\frac{g}{2}$

Step 2: Formula used:

The acceleration because of gravity is $g=\frac{Gm}{R^2}----\left(1\right)$

Where, $G$ is gravitational constant, $R$ is the radius of the earth, $m$ is the mass of the earth.

The acceleration because of gravity at height $h$ is $g^{\text{'}}=\frac{Gm}{{(R+h)}^2}----\left(2\right)$

Step 2: Calculating the height

On dividing, equation $\left(2\right)$ by equation $\left(1\right)$ we get-

$\frac{g^{\text{'}}}{g}=\frac{R^2}{{(R+h)}^2}$

Substitute $g^{\text{'}}=\frac{g}{2}$,

$\begin{array}{l}\frac{g}{2g}=\frac{R^2\times {\displaystyle \frac{1}{R^2}}}{{(R+h)}^2\times {\displaystyle \frac{1}{R^2}}}\\ \frac{1}{2}=\frac{1}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}\\ \frac{1}{2}={\left(1+\frac{h}{R}\right)}^{-2}\end{array}$

If the value of $h<<R$. So, after expanding the series,

$\begin{array}{l}\frac{1}{2}=1-\frac{2h}{R}\\ \frac{2h}{R}=\frac{1}{2}\\ h=\frac{R}{4}\end{array}$

So, at height $h=\frac{R}{4}$, the acceleration due to gravity becomes $\frac{g}{2}$.

Hence, option B is the correct answer.