The value of acceleration due to gravity at height h from Earth’s surface will become half it’s value on the surface if ( $R{\text{ }} = {\text{ Radius of Earth}}$ )
A) $h = R$
B) $h = 2R$
C) $h = (\sqrt 2 - 1)R$
D) $h = (\sqrt 2 + 1)R$

Hint: We have to find the height at which the gravity will be half of that on the surface of Earth. So we will use the formula of Force of interaction between two bodies having mass. Then, we will find the value of gravity at Earth’s surface then we will use the given height in the formula we will obtain.

Complete step by step solution:
We know that the force of attraction $F$ between two bodies having masses ${m_1}$ and ${m_2}$ separated by a distance $r$ is given by
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where $G$ is universal gravitational constant.
Let the mass of Earth be $M$ a point object of mass $m$ at surface of Earth of radius $R$ ,
Their force of attraction $F$ will be given by,
$F = \dfrac{{GMm}}{{{R^2}}}$
This force of attraction will be $F = mg$ , where $g$ is gravity so we get,
$mg = \dfrac{{GMm}}{{{R^2}}}$
On further simplification, we get,
$g = \dfrac{{GM}}{{{R^2}}}$
So we get equation as,
$g{R^2} = GM$ -------(1)
Let at height $h$ , distance from centre from Earth will be $R + h$ , gravity will be halved than that of surface, so we get,
$\dfrac{g}{2}{(R + h)^2} = GM$ ----------(2)
From equation $1\& 2$ we get,
$\dfrac{g}{2}{(R + h)^2} = g{R^2}$
On simplifying we get,
${(R + h)^2} = 2{R^2}$
Doing squareroot on both sides we get,
$R + h = \sqrt 2 R$
On simplifying this, we get,
$h = (\sqrt 2 - 1)R$

So the correct answer is option (C).

Note: Since the gravity at obtained height becomes half, it means any person will feel his weight half of his weight at the surface. When this height is enough to get out of the atmosphere, the weight decreases to zero, that’s why we feel zero weight in space.