Question

If the radius of the Earth decreased by $10\%$ the mass remaining unchanged what will happen to the acceleration due to Gravity?
A) Decrease by $19\%$
B) Increase by $19\%$
C) Decrease by more than $19\%$
D) Increase by more than $19\%$

Answer 1

In order to solve this Question, Use the formula of Acceleration due to Gravity ${\text{‘g’}}$,which tells us about the relation between Mass $\left( {\text{M}} \right)$, Radius $\left( {\text{R}} \right)$ & Gravitational constant $\left( {\text{G}} \right).$

Formula used:
${\text{g = }}\dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}}$
Here ${\text{G = }}$ gravitational constant.
${\text{R}} \to$ Radius of Earth
${\text{M}} \to$ Mass of the Earth

Complete step by step Answer:
According to the given question, the radius of earth is decreasing by $10\%$,without changing the mass.
So, We get a New radius ${\text{R’}}$ which is equal to i. e. {\text{R’ = }}\left( {{\text{100% - 10% }}} \right){\text{R}}
Where ${\text{R’}}$ is the Radius of earth when it is decreased by $10\%$ hence it is the new Radius.
& ${\text{R}}$ is the old Radius of Earth.
So,
${\text{g = }}\dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}}$ is the “Acceleration due to gravity ’’.
&
${\text{g’}}$ is the new acceleration due to gravity and is equal to
${\text{g’ = }}\dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}}$ $--- \left( 1 \right)$
Where ${\text{R’}}$ is the new radius, &
${\text{R’}}$ is equal to $90\%$ of ${\text{R}}$
Putting the value of ${\text{R’}}$in Equation $\left( 1 \right)$
${\text{g’ = }}\dfrac{{{\text{GM}}}}{{{{\left( {\dfrac{{{\text{90}}}}{{{\text{100}}}}} \right)}^{\text{2}}}{{\text{R}}^{\text{2}}}}} \Rightarrow {\text{g’ = }}\dfrac{{{\text{100}}}}{{{\text{81}}}}{\text{g}}$
Further,
$\dfrac{{g’}}{g} = \dfrac{{100}}{{81}}$
$\dfrac{{g’}}{g} - 1 = \dfrac{{100}}{{81}} - 1$
$\dfrac{{{\text{g’ - g}}}}{{\text{g}}}{\text{ = }}\dfrac{{{\text{19}}}}{{{\text{81}}}}$
$\dfrac{{{{\Delta g}}}}{{{g}}}{\text{ = }}\dfrac{{{{19}}}}{{{{81}}}}$ $----- \left( 2 \right)$
Where ${{\Delta g}}$ is a change in acceleration due to gravity.
Now, the percentage change is obtained by multiplying $100$ both side in Equation $\left( 2 \right)$
$\dfrac{{{{\Delta g}}}}{{{g}}}{{ \times 100 = }}\dfrac{19}{81}{{ \times 100}}$
So, percentage change in ${\text{g}}$ is equal to
= \dfrac{{19}}{{81}} \times 100 \\\ = 23\% \\\

So, the correct option is (D). i. e. Increases by more than $19\%$.

Additional Information:
Magnitude of g:
At Equator $\left( {\text{g}} \right){\text{ = 978}}{\text{.0316}}\dfrac{{{\text{cm}}}}{{{\text{se}}{{\text{c}}^{\text{2}}}}}$
At poles $\left( {\text{g}} \right){\text{ = 983}}{\text{.152}}\dfrac{{{\text{cm}}}}{{{\text{se}}{{\text{c}}^{\text{2}}}}}$
Normal value of $\left( {\text{g}} \right){\text{ = 980}}\dfrac{{{\text{cm}}}}{{{\text{se}}{{\text{c}}^{\text{2}}}}}$
In Geophysics we get the unit as where
${\text{1}}\dfrac{{{\text{cm}}}}{{{\text{se}}{{\text{c}}^{\text{2}}}}}{\text{ = 1Gal}}$

Note: The acceleration due to gravity is constant for all bodies. It is independent of the masses of the individual bodies. That is why in a free fall the bodies undergo similar conditions irrespective of their masses. However, the acceleration due to gravity depends on the mass of the planet or satellite. Thus, it will have different values for different planets and satellites.