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

• A. $h=2R$
• B. $h=R$
• C. $h=(\sqrt{2}+1)R$
• D. $h=(\sqrt{2}-1)R$

Answer 1

Answer: (D)

$h=(\sqrt{2}-1)R$

$g^1=\frac{Gm}{(R+h{)}^2}...\left(1\right)$ & at surface $g=\frac{Gm}{R^2}...\left(2\right)$

given that $g^1=g/2$

(1)/(2) we get $\frac{g^1}{g}=\frac{R^2}{(R+h{)}^2}$

Put $g^1=g/2$ we get $\frac{1}{2}=\frac{R^2}{(R+h{)}^2}$

$\Rightarrow \sqrt{2}=\frac{R+h}{R}=1+\frac{h}{R}$

$\Rightarrow \frac{h}{R}=\sqrt{2}-1\Rightarrow h=R(\sqrt{2}-1)$