How much percentage change in acceleration due to gravity on an object occurs if object is initially at ?2R? distance above the surface of the earth and falls through ?R? distance? (Here, R is the radius of earth) options:- (1)decrease by 300% (2)increase by 300% (3)decrease be 125% (4)increase by 125%

Dear Student,

As we know that , the variation of the gravitational acceleration is vary as per the rise of height or down side.

$$\begin{gathered} g\text{'}= \frac{GM}{{(R+h)}^2} \\ h=2R \\ g\text{'}= \frac{GM}{{(R+2R)}^2} = \frac{GM}{9R^2} \end{gathered}$$

given that h1= 2R and h2= R

$$\begin{gathered} g\text{'}\text{'} = \frac{GM}{{(R+h)}^2} \\ g\text{'}\text{'} =\frac{GM}{{(R+R)}^2} \\ g\text{'}\text{'} =\frac{GM}{4R^2} \end{gathered}$$

Hence, the required percentage

 $$\begin{gathered} =\frac{g\text{'}-g\text{'}\text{'}}{g\text{'}}\times 100 \\ =\frac{{\displaystyle \frac{g}{4}}-{\displaystyle \frac{g}{9}}}{{\displaystyle \frac{g}{9}}}\times 100 \\ =\frac{5}{4}\times 100 =125\% \end{gathered}$$

Hence, g will increase 125%.

So, from the above, it is clear that if object is moving from any height to near to the surface of the earth

then acceleration due to gravity is getting increase.

Regards