Question 22: Gravitation

Question

The radius of earth is $R$ and acceleration due to gravity on its surface is $g$. The height at which the acceleration due to gravity becomes $\frac{g}{8}$ is:

Accepted Answer

We use the inverseÃ¢ÂÂsquare law for gravity. At a height $h$ above EarthÃ¢ÂÂs surface the acceleration is
$ g(h) \;=\;\frac{GM}{(R+h)^2} $
and at the surface
$ g \;=\;\frac{GM}{R^2}\,.   $
So
$ \frac{g(h)}{g} \;=\;\frac{R^2}{(R+h)^2}      \;=\;\frac1{8}\,. $
Solving
$ (R+h)^2 \;=\;8\,R^2 \quad\Longrightarrow\quad R+h \;=\;\sqrt{8}\,R \;=\;2\sqrt2\,R \quad\Longrightarrow\quad h \;=\;(2\sqrt2-1)\,R. $
Answer: $(2\sqrt2-1)\,R$. (Option B)

Answer

$2 R$

Answer

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

Answer

$\sqrt{2} R$

Answer

$2 \sqrt{2} R$

Gravitation

The radius of earth is R and acceleration due to gravity on its surface is g. The height at which the acceleration due to gravity becomes \frac{g}{8} is:

Choose the correct answer: