Question

At which height from the earth’s surface acceleration due to gravity is decreased by $75\%$ of its value at the earth’s surface.

Answer 1

The acceleration due to gravity is inversely proportional to the square of the distance of the body from the centre of the earth. The distance of a body at the surface of the earth to the centre is equal to the radius of the earth.

Formula used: In this solution we will be using the following formulae;
$g = G\dfrac{M}{{{r^2}}}$ where $g$ is the acceleration due to gravity at a point on the earth, $G$ is the universal gravitation constant, $M$ is the mass of the earth, and $r$ is the distance of that point from the centre of the earth.

Complete step by step answer:
To solve the above, we only need to note that the acceleration due to gravity is inversely related to the square of the distance between them. Mathematically written as
$g = G\dfrac{M}{{{r^2}}}$ where $g$ is the acceleration due to gravity at a point on the earth, $G$ is the universal gravitation constant, $M$ is the mass of the earth, and $r$ is the distance of that point from the centre of the earth.
Hence, for object on the surface, we have
${g_s} = G\dfrac{M}{{{R_E}^2}}$ where ${R_E}$ is the radius of the earth
For object at height $h$ we have
${g_h} = G\dfrac{M}{{{{\left( {{R_E} + h} \right)}^2}}}$
Hence,
$\dfrac{{{g_h}}}{{{g_s}}} = G\dfrac{M}{{{{\left( {{R_E} + h} \right)}^2}}} \div G\dfrac{M}{{{R_E}^2}}$
$\Rightarrow \dfrac{{{g_h}}}{{{g_s}}} = G\dfrac{M}{{{{\left( {{R_E} + h} \right)}^2}}} \times \dfrac{{{R_E}^2}}{{GM}} = \dfrac{{{R_E}^2}}{{{{\left( {{R_E} + h} \right)}^2}}}$
Inserting known values, we have
$\dfrac{{25}}{{100}} = \dfrac{{{R_E}^2}}{{{{\left( {{R_E} + h} \right)}^2}}}$
$\Rightarrow {\left( {{R_E} + h} \right)^2} = \dfrac{{100}}{{25}}R_E^2$
By square rooting both sides, we have
${R_E} + h = \dfrac{{10}}{5}{R_E} = 2{R_E}$
$\Rightarrow h = 2{R_E} - {R_E} = {R_E}$
Although not give, if we assume the radius of the earth is known, which is usually 6400 km.
Then, we have
$h = 6400km$

Note: For clarity, note that the acceleration due to gravity reducing by 75 percent means that the acceleration due to gravity at that location is equal to 25 percent of the value of the surface of the earth.
Also, note that to calculate the height above the earth surface, we do not have to know the exact formula for the acceleration due to gravity but only the relationship with the distance. Hence, since we not that
$g \propto \dfrac{1}{{{r^2}}}$
$\Rightarrow g = k\dfrac{1}{{{r^2}}}$
Hence,
$\dfrac{{{g_h}}}{{{g_s}}} = \dfrac{k}{{{{\left( {{R_E} + h} \right)}^2}}} \div \dfrac{k}{{{R_E}^2}}$
$\Rightarrow \dfrac{{{g_h}}}{{{g_s}}} = \dfrac{k}{{{{\left( {{R_E} + h} \right)}^2}}} \times \dfrac{{{R_E}^2}}{k} = \dfrac{{R_E^2}}{{{{\left( {{R_E} + h} \right)}^2}}}$
Which is identical to the expression in the solution.