Question

The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes $1\% $ of its value at the surface is (Radius of the earth= R)
A. $9R$
B. $10R$
C. $8R$
D. $20R$

Answer 1

Hint: Here to solve this problem we need to know the formula of acceleration due to gravity at height h. In this formula we will get the relation between the acceleration due to gravity on the surface, radius of the earth, height and the acceleration due to gravity at that height. We know the relation between the acceleration due to gravity at height at and the acceleration due to gravity at the surface. Now using this relation in the formula we can find the solution to this problem.

Formula used:
$g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
Where, the acceleration due to gravity at height $h = g’$
The acceleration due to gravity at surface of the earth = $g$
Radius of the earth = $R$
Height = $h$

Complete step by step answer:
As per the problem we have a point vertically above the earth’s surface, at which acceleration due to gravity becomes $1\% $ of its value at the surface.Mathematically we can represent the above give statement as,
$g' = 1\% g = \dfrac{g}{{100}} \ldots \ldots \left( 1 \right)$
We need to calculate the height of the point that is vertically above the earth surface.We know that acceleration due to gravity at height h is given by,
$g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$

From equation (1) we will get the acceleration due to gravity at height h as,
$\dfrac{g}{{100}} = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
Cancelling the common terms we will get,
$\dfrac{1}{{100}} = {\left( {\dfrac{R}{{R + h}}} \right)^2}$
On further solving we will get,
$\dfrac{1}{{10}} = \left( {\dfrac{R}{{R + h}}} \right)$
Rearranging the above equation we will get,
$R + h = 10R \\
\therefore h = 9R$

Therefore the correct option is A.

Note: Remember that acceleration due to gravity is defined as the acceleration of the body that is falling freely under the influence of the earth gravity that is expressed as the rate of increase in velocity per unit time. Note that this is a constant term that is acceleration known as $9.8\,m{s^{ - 1}}$. Also keep in mind that acceleration of gravity varies with height above the surface of the earth.