Question

Find the height above the surface of the earth where the acceleration due to gravity reduces by 1%.

Answer 1

The acceleration due to gravity varies below and above the surface of the earth. It decreases with an increase in the height above the surface. The acceleration due to gravity also depends on the shape of the earth and hence will depend on the radius of the earth.

Formula used:
The acceleration due to gravity above the surface of the earth is given by, $g' = g\left( {1 - \dfrac{{2h}}{{{R_E}}}} \right)$ where $g$ is the acceleration due to gravity on the surface of the earth, $h$ is the height above the surface and ${R_E}$ is the radius of the earth.

Complete step by step answer.
Step 1: Express the relation showing the variation of the acceleration due to gravity with heights above the surface of the earth.
The acceleration due to gravity above the surface of the earth is a function of the height above the surface of the earth.
Let $h$ be the height at which the acceleration due to gravity decreases.
Let $g$’ be the new acceleration due to gravity at $h$ .
It is given that the acceleration due to gravity at the height $h$ reduces by 1%.
i.e., $g' = g - \left( {\dfrac{1}{{100}}g} \right) = 0 \cdot 99g$
Thus the new acceleration due to gravity is $g' = 0 \cdot 99g$ .
The acceleration due to gravity at $h$ is given by, $g' = g\left( {1 - \dfrac{{2h}}{{{R_E}}}} \right)$ --------- (1)
where $g$ is the acceleration due to gravity on the surface of the earth and ${R_E}$ is the radius of the earth.
Substituting for $g' = 0 \cdot 99g$ in equation (1) we get, $0 \cdot 99g = g\left( {1 - \dfrac{{2h}}{{{R_E}}}} \right)$
Cancelling out the similar terms on either side of the above equation we get, $0 \cdot 99 = \left( {1 - \dfrac{{2h}}{{{R_E}}}} \right)$
$\Rightarrow 0 \cdot 99{R_E} = {R_E} - 2h$
On simplifying the above expression we get, $h = \dfrac{{{R_E} - 0 \cdot 99{R_E}}}{2} = 0 \cdot 005{R_E}$
Thus at a height, $h = 0 \cdot 005{R_E}$ the acceleration due to gravity reduces by 1%.

Note: The radius of the earth has different values at the equator and at the poles since earth is slightly flattened at the poles. The commonly known value of the acceleration due to gravity $g = 9 \cdot 8{\text{m}}{{\text{s}}^{ - 2}}$ is the value of acceleration due to gravity on the surface of the earth. Equation (1) suggests that when the height $h = \infty$ then the acceleration due to gravity would be zero.