Question

The decrease in the value of $g$ on going to a height $\dfrac{R}{2}$ above the earth’s surface will be:
(A) $\dfrac{g}{2}$
(B) $\dfrac{{5g}}{9}$
(C) $\dfrac{{4g}}{9}$
(D) $\dfrac{g}{3}$

Answer 1

The acceleration due to gravity is inversely proportional to the square of the radius of the earth. If the gravity is measured at a height then the value of distance will be increased. Thus, the acceleration due to gravity will be decreased.

Complete solution:
Acceleration due to gravity is the rate of change in velocity of a free-falling body under the influence of gravity. The numerical value seems to be a contestant on the surface of earth. And can be called acceleration due to gravity.
The expression for the acceleration due to gravity is given as,
$g = \dfrac{{GM}}{{{R^2}}}...................\left( 1 \right)$
Where, $G$ is the gravitational constant, $M$ is the mass of earth and $R$ is the radius of earth.
When the acceleration due to gravity is calculated at a height of $h$ , then the distance is taken as, $R + h$ .
Then the equation $\left( 1 \right)$ changes to,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
If we are considering $h = \dfrac{R}{2}$. That is if the height taken is half the radius of earth.
Then, we can write the equation as,
$\Rightarrow {g_h} = \dfrac{{GM}}{{{{\left( {R + \dfrac{R}{2}} \right)}^2}}} \\\ \Rightarrow \dfrac{{GM}}{{{R^2} + {R^2} + \dfrac{{{R^2}}}{4}}} \\\ \Rightarrow \dfrac{{GM}}{{\dfrac{{9{R^2}}}{4}}} \\\ \Rightarrow \dfrac{{4GM}}{{9{R^2}}}$
Substitute the equation (1) in the above equation.
$\Rightarrow {g_h} = \dfrac{{4g}}{9}$
In order to find the decrease in the value of acceleration due to gravity, subtract the above expression from the acceleration due to gravity.
Therefore, $g - {g_h}$
Substitute the value in above expression,
$\Rightarrow g - \dfrac{{4g}}{9} = \dfrac{{5g}}{9}$
The decrease in the value of $g$ on going to a height above the earth’s surface will be $\dfrac{{5g}}{9}$.

The answer is option B.

Note: It is clear that the acceleration due to gravity will decrease when we go up to the surface of earth. At the surface of earth, the acceleration due to gravity is constant. The acceleration due to gravity is directly proportional to the gravitational constant.