Question

If radius of earth is R then the height h at which the value of g becomes one fourth is:
A) $2 \mathrm{R}$
B) $3 \mathrm{R}$
C) $\mathrm{R}$
D) $4 \mathrm{R}$

Answer 1

Hint: The Earth’s gravity is denoted by g which is the total acceleration that is distributed to objects due to the lump sum effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the rotation of Earth's).

Complete step by step answer:
We can denote the gravitation of earth as,
$\mathrm{g}=\dfrac{G M}{R^{2}}$
We can assume that at some height $\mathrm{H}$, we can consider the acceleration due to gravity to be $\text{g }\!\!'\!\!\text{ }$
Now, we can denote the given situation in the following formula,
$\Rightarrow \mathrm{g}^{\prime}=\dfrac{G M}{(R+H)^{2}}$
In the given question,
$\Rightarrow g’=g / 4$
Therefore, while solving for $\mathrm{H}$, we get,
$\Rightarrow \dfrac{g}{4}=g’$
Therefore,
$\Rightarrow \dfrac{G M}{4 R^{2}}=\dfrac{G M}{(R+H)^{2}}$
Or,$R+H=2 R$
Or, $H=R$

Therefore, the correct answer is option C.

Note: While denoting in terms of SI units, this acceleration is measured in metres per second squared (in symbolic notation, it is denoted as $\mathrm{m} / \mathrm{s}^{2}$ or $\mathrm{m}.\mathrm{s}^{-2}$ ) or similarly in newtons per kilogram (N/kg or $\mathrm{N}.\mathrm{kg}^{-1}$ ). At some distance from the Earth's surface, gravitational acceleration is approximately $9.81 \mathrm{~m} / \mathrm{s}^{2}$. This denotes that, if we ignore the effects of air resistance, the speed of an object falling in free motion will get incremented by almost around 9.81 metres per second per second.