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Question: If the radius of earth is decreased by \(10\% \), the mass remaining unchanged, what will happen to ...

If the radius of earth is decreased by 10%10\% , the mass remaining unchanged, what will happen to the acceleration of gravity?
A) Decreased by 19%19\%
B) Increased by 19%19\%
C) Decreased by more than 19%19\%
D) Increased by more than 19%19\%

Explanation

Solution

Hint: Use the relation of acceleration of gravity and radius of earth to deduce the final expression. The acceleration of gravity is inversely proportional to the square of the radius of the earth.

Complete step by step answer:
Let a body of mass mmon the surface on earth. We take this mass to deduce the expression between acceleration of gravity and the radius of earth RR.
We know that, when a body is present near the earth’s surface, a force of gravity which is equal to mgmgacts on this body. Also, we know the force of gravitation between two bodies kept at a distance from each other is GMmr2G\dfrac{{Mm}}{{{r^2}}}(as per Newton’s universal law of gravitation).
On equating both the forces we get,
mg=GMmr2mg = G\dfrac{{Mm}}{{{r^2}}}
g=GMr2g = G\dfrac{M}{{{r^2}}}
Where,
MMis the mass of earth.
GG is the universal gravitational constant.
rr is the radius of earth.
It is given in the question that the radius is decreased by 10%10\% ,
The final radius becomes,
R=r10%rR = r - 10\% r
This equation means that the total radius is now 10%10\% less than the original radius.
R=r10100rR = r - \dfrac{{10}}{{100}}r
R=9r10R = \dfrac{{9r}}{{10}}
The new acceleration due to gravity is,
g=GMR2g' = G\dfrac{M}{{{R^2}}}
g=GM(9r10)2g' = G\dfrac{M}{{{{(\dfrac{{9r}}{{10}})}^2}}}
g=10081GMR2g' = \dfrac{{100}}{{81}}G\dfrac{M}{{{R^2}}}
Now, change in acceleration due to gravity is given by,
100GM81r2GMr2GMr2×100\dfrac{{\dfrac{{100GM}}{{81{r^2}}} - \dfrac{{GM}}{{{r^2}}}}}{{\dfrac{{GM}}{{{r^2}}}}} \times 100
23%23\%
So the acceleration due to gravity is increased by 23%23\% .
D) is correct.

Note: We calculated acceleration due to gravity on the surface of earth. Remember that the value of acceleration due to gravity decreases with the height or depth from the surface of earth.
Acceleration due to gravity at height hh from the surface =g(1+hR)2 = g{\left( {1 + \dfrac{h}{R}} \right)^{ - 2}}
Acceleration due to gravity at depth dd from the surface=g(1dR) = g\left( {1 - \dfrac{d}{R}} \right)
Here RR is the radius of earth.