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# 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) $\dfrac{g}{3}$

Last updated date: 20th Jun 2024
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Hint: 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 $\dfrac{{5g}}{9}$.