Question

Find the height from the earth’s surface where $g$ will be $25\%$ of its value on the surface of earth $(R = 6400km)$. Find the percentage increase in the value of $g$ at a depth $h$ from the surface of earth.

Answer 1

Acceleration due to gravity depends on the distance between the centre of the earth and the object. Derive a formula for the value of acceleration due to gravity at different altitudes. In order to do this, use Newton's law of gravitation. Now, find the value of $g$ at height $h$ and then apply the given condition and afterwards find the percentage change in the value of $g$ when one goes from the surface to a depth of height $h$.

Complete step by step answer:
(a) The gravitational force at the surface of the earth is given by $F = \dfrac{{GMm}}{{{R^2}}}$
Here $M$is the mass of earth, $m$ is the mass of a particle and $R$ is the radius of the earth. From here, we can get $g$ as $g = \dfrac{{GM}}{{{R^2}}}$.
Now, to get the value of $g$at a height $h$,
$g$ can be given as
$g' = \dfrac{{GM}}{{{r^2}}} \\\ \implies g' = \dfrac{{GM}}{{{{(R + h)}^2}}} \\\ \implies g' = \dfrac{{GM}}{{{R^2}}}{\left( {\dfrac{R}{{R + h}}} \right)^2} \\\ \implies g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2} \\\$
Given that the value of $g$ at height $h$should be $25\%$ of the value of $g$ at surface, mathematically we have,
$g' = \dfrac{{25}}{{100}}g \\\ \therefore g' = \dfrac{g}{4} \\\$
Now substituting value of $g'$ in the above equation, we get
\dfrac{g}{4} = g{\left( {\dfrac{R}{{R + h}}} \right)^2} \\\
\implies \dfrac{1}{4} = {\left( {\dfrac{R}{{R + h}}} \right)^2} \\\
\implies \dfrac{R}{{R + h}} = \dfrac{1}{{ 2 }} \\\
\implies h = R \\\
\dfrac{g}{4} = g{\left( {\dfrac{R}{{R + h}}} \right)^2} \\\
\implies \dfrac{1}{4} = {\left( {\dfrac{R}{{R + h}}} \right)^2} \\\
\implies \dfrac{R}{{R + h}} = \dfrac{1}{2} \\\
$\therefore h = R \\\$
Therefore, the height from the earth’s surface where $g$ will be $25\%$ of its value on the surface of earth is $6400km$.
(b) The value of $g$ inside the earth at a distance $R - h$ is given as
g' = \dfrac{{GMr}}{{{R^3}}} \\\
\implies g' = \dfrac{{GM}}{{{R^2}}}\dfrac{r}{R} \\\
\implies g' = \dfrac{{gr}}{R} \\\
g' = \dfrac{{GMr}}{{{R^3}}} \\\
\implies g' = \dfrac{{GM}}{{{R^2}}}\dfrac{r}{R} \\\
\implies g' = \dfrac{{gr}}{R} \\\
We have $r = R - h$
$g' = \dfrac{g}{R}\left( {R - h} \right) \\\ \therefore g' = g\left( {1 - \dfrac{h}{R}} \right) \\\$
Now, change in percentage in the value of $g$ as it goes from $g$ to $g'$ is given by
$\dfrac{{g' - g}}{g} \times 100 = \dfrac{{g\left( {1 - \dfrac{h}{R}} \right) - g}}{g} \times 100 \\\ \implies \dfrac{{g' - g}}{g} \times 100 = - \dfrac{{100h}}{R} \\\$

Therefore, the percentage increase in the value of $g$ at a depth $h$ from the surface of earth is $- \dfrac{{100h}}{R}$.

Note:
As we have seen that the acceleration due to gravity changes as you go at different altitudes, we can classify it into three cases. First case will be when the acceleration due to gravity is to be found below the surface of the earth, second case will be at the surface and the third case will be at a height greater than radius of the earth, that is above the surface of the earth.