Question

If $R$ is the radius of the earth, then the value of acceleration due to gravity at a height $h$ from the surface of the earth will become half its value on the surface of the earth if:
$$\begin{array}{rl}& A.h=2R\\ & B.h=R\\ & C.h=(\sqrt{2}+1)R\\ & D.h=(\sqrt{2}-1)R\end{array}$$

Answer 1

Hint: To find the relationship between $R$ and $h$, we need to calculate the acceleration due to gravitation at $R$ and $h$, using the general formula. Since a relationship between calculating the acceleration due to gravitation at $R$ and $h$ is given, we can equate them to find the required relation.

Formula used:
$g={\displaystyle \frac{Gm}{R^2}}$ and $g\prime ={\displaystyle \frac{Gm}{(R+h{)}^2}}$

Complete step-by-step answer:
We know that according to Newton-Kepler law of gravitation, the force due to gravitation is given as $F=G{\displaystyle \frac{m_1{m}_2}{r^2}}$, where $G$ is the gravitational constant, $m_1,m_2$ is the masses of the body, and $r$ is the distance between the two bodies.
We know that the acceleration due to gravitation on the surface of the earth is given as $g={\displaystyle \frac{Gm}{R^2}}$ and the acceleration due to gravitation at a height $h$ from the surface is given by $g\prime ={\displaystyle \frac{Gm}{(R+h{)}^2}}$ , where $G$ is the gravitational constant, $m$ is the mass of the body, $R$ is the radius of the earth and $h$ is the height from the surface of the earth .
Here, since the mass of the body is very small as compared to the mass of the earth, to calculate the force on the body, we can ignore the value of mass of the earth.
Since it is given that, the value of acceleration due to gravity at a height $h$ from the surface of the earth is half its value on the surface of the earth , i.e. $g\prime ={\displaystyle \frac{g}{2}}$
Then taking the ratio between $g$ and $g\prime .$ we get,
${\displaystyle \frac{g\prime }{g}}={\displaystyle \frac{{\displaystyle \frac{g}{2}}}{g}}={\displaystyle \frac{{\displaystyle \frac{Gm}{(R+h{)}^2}}}{{\displaystyle \frac{Gm}{R^2}}}}$
Assuming the mass of the body is same, and reducing, we get,
${\displaystyle \frac{1}{2}}={\displaystyle \frac{R^2}{(R+h{)}^2}}$
$\Rightarrow \sqrt{2}={\displaystyle \frac{R+h}{R}}$
$\Rightarrow \sqrt{2}=1+{\displaystyle \frac{h}{R}}$
$\Rightarrow h=R(\sqrt{2}-1)$

Note: The mass of the earth is larger than the mass of the object, it can be neglected. Also note that using the Newton-Kepler law of gravitation, the acceleration due to gravitation at a height $h$ from the surface is given by $g\prime ={\displaystyle \frac{Gm}{(R+h{)}^2}}$.