Answer
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Hint: Express the new mass and radius of the earth as per given in the question. Recall the expression for the acceleration due to gravity and express the new acceleration due to gravity of the earth with new mass and radius. Calculate the decrease in the acceleration due to gravity.
Formula used:
Acceleration due to gravity, \[g = \dfrac{{GM}}{{{R^2}}}\]
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
Complete step by step answer:
We know the expression for the acceleration due to gravity of the planet,
\[g = \dfrac{{GM}}{{{R^2}}}\] ……. (1)
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
We have given that the mass of the earth has decreased by 25%. Therefore, the new mass of the earth is,
\[M' = M - 25\% M = M - \dfrac{M}{4}\]
\[ \Rightarrow M' = \dfrac{3}{4}M\] …… (2)
Also, the radius of the earth increases by 50%. Therefore, the new radius of the earth is,
\[R' = R + 50\% R = R + \dfrac{R}{2}\]
\[ \Rightarrow R' = \dfrac{3}{2}R\] …… (3)
Now, let us express the new acceleration due to gravity of the earth with new mass and radius as follows,
\[g' = \dfrac{{GM'}}{{{{R'}^2}}}\]
Using equation (2) and (3) in the above equation, we get,
\[g' = \dfrac{{G\left( {\dfrac{3}{4}M} \right)}}{{{{\left( {\dfrac{3}{2}R} \right)}^2}}}\]
\[ \Rightarrow g' = \dfrac{1}{3}\dfrac{{GM}}{{{R^2}}}\]
Using equation (1) in the above equation, we get,
\[g' = \dfrac{g}{3}\]
\[ \Rightarrow g' = 0.33\,g\]
Now, the net decrease in the acceleration due to gravity of the earth is,
\[\Delta g = g - g'\]
\[ \Rightarrow \Delta g = g - 0.33g\]
\[ \therefore \Delta g = 67\% \]
Therefore, the acceleration of the gravity of the earth decreases by 67%.
So, the correct answer is option C.
Note: While determining the decrease in the acceleration due to gravity, students must subtract the new acceleration due to gravity from the initial acceleration due to gravity. Note that the acceleration due to gravity is inversely proportional to the square of the radius of the planet and not just the radius. The universal gravitational constant remains constant even if there is change in the radius and change in mass of the planet or any other heavenly object.
Formula used:
Acceleration due to gravity, \[g = \dfrac{{GM}}{{{R^2}}}\]
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
Complete step by step answer:
We know the expression for the acceleration due to gravity of the planet,
\[g = \dfrac{{GM}}{{{R^2}}}\] ……. (1)
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
We have given that the mass of the earth has decreased by 25%. Therefore, the new mass of the earth is,
\[M' = M - 25\% M = M - \dfrac{M}{4}\]
\[ \Rightarrow M' = \dfrac{3}{4}M\] …… (2)
Also, the radius of the earth increases by 50%. Therefore, the new radius of the earth is,
\[R' = R + 50\% R = R + \dfrac{R}{2}\]
\[ \Rightarrow R' = \dfrac{3}{2}R\] …… (3)
Now, let us express the new acceleration due to gravity of the earth with new mass and radius as follows,
\[g' = \dfrac{{GM'}}{{{{R'}^2}}}\]
Using equation (2) and (3) in the above equation, we get,
\[g' = \dfrac{{G\left( {\dfrac{3}{4}M} \right)}}{{{{\left( {\dfrac{3}{2}R} \right)}^2}}}\]
\[ \Rightarrow g' = \dfrac{1}{3}\dfrac{{GM}}{{{R^2}}}\]
Using equation (1) in the above equation, we get,
\[g' = \dfrac{g}{3}\]
\[ \Rightarrow g' = 0.33\,g\]
Now, the net decrease in the acceleration due to gravity of the earth is,
\[\Delta g = g - g'\]
\[ \Rightarrow \Delta g = g - 0.33g\]
\[ \therefore \Delta g = 67\% \]
Therefore, the acceleration of the gravity of the earth decreases by 67%.
So, the correct answer is option C.
Note: While determining the decrease in the acceleration due to gravity, students must subtract the new acceleration due to gravity from the initial acceleration due to gravity. Note that the acceleration due to gravity is inversely proportional to the square of the radius of the planet and not just the radius. The universal gravitational constant remains constant even if there is change in the radius and change in mass of the planet or any other heavenly object.
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