Question

At what height, is the value of g half that on the surface of the earth? (R = radius of the earth).
A) $0.414\;R$
B) $R$
C) $2R$
D) $3.5R$

Answer 1

The law of gravitation is used to determine the value of g on the surface of earth or from the different heights from the earth. It is also used to give the relation between the gravitational of different heights and the radius of the earth.

Complete step by step answer:
Suppose that the acceleration due to gravity on earth is $g$, the acceleration due to gravity on height is $g{'}$ and the height is h.

We know that the gravitational force is given by $F = \dfrac{{GmM}}{{{r^2}}}$
Here, the gravitational constant is $G$, the mass of the earth is $M$, the mass of the body is $m$ and the distance between the body and the core of the earth is $r$.

The acceleration of gravity is given by $g = \dfrac{F}{m}$
We will now substitute the known values in the above equation of acceleration due to gravity.
$\Rightarrow g = \dfrac{{\dfrac{{GmM}}{{{r^2}}}}}{m}$
$\Rightarrow g = \dfrac{{GM}}{{{r^2}}}$

For earth the acceleration of gravity is given by $g = \dfrac{{GM}}{{{R^2}}}$.
Here, the radius of the earth is $R$.
The acceleration due to gravity for a object at height h form the earth is given by $g{'} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
According to the question, the relation between the acceleration due to gravity on the surface of the earth and the acceleration due to gravity at a distance h from the surface of the earth is $g{'} = \dfrac{g}{2}$.
We will now substitute the known values in the above equation of acceleration due to gravity.
$\Rightarrow \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}} = \dfrac{1}{2}\dfrac{{GM}}{{{R^2}}}$
$\Rightarrow 2{R^2} = {\left( {R + h} \right)^2}$
On simplification,
$\Rightarrow {\left( {\sqrt 2 R} \right)^2} = {\left( {R + h} \right)^2}$
$\Rightarrow \sqrt 2 R = R + h$
On further simplification,
$\Rightarrow h = \sqrt 2 R - R$
$\Rightarrow h= \left( {1.414 - 1} \right)R$
$\Rightarrow h = 0.414R$

$\therefore$ At a height of $0.414R$ the value of $g$ half that on the surface of the earth and hence, option (A) is correct.

Note:
Make sure not to get confused between the gravitational force and acceleration of gravity. And also take square root correctly after cross multiplication otherwise answers will not match with the correct one.