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Question: When the radius of Earth is reduced by \(1\% \) without changing the mass, then the acceleration due...

When the radius of Earth is reduced by 1%1\% without changing the mass, then the acceleration due to gravity will.
A) Increase by 2%2\%
B) Decrease by 1.5%1.5\%
C) Increase by 1%1\%
D) Decrease by 1%1\%

Explanation

Solution

For this problem, we need to find the change in acceleration due to gravity due to change in the radius of Earth. Use the formula for Newton’s law of gravitation. Find the acceleration due to gravity when the radius of Earth was not reduced then find the acceleration due to gravity when the radius of Earth is reduced by 1%1\% . Use formula to find the percentage change.

Complete step by step solution: From Newton’s law of gravitation, we have:
g=GMR2g = \dfrac{{GM}}{{{R^2}}} --equation 1

Here, gg is the acceleration due to gravity.
GG is gravitational constant and it has magnitude of 6.67×10−11 N m2 kg−26.67 \times {10^{ - 11}}\,N\,{m^2}\,k{g^{ - 2}}
MM is mass of Earth, mass of Earth is given to be constant.
RR is the radius of Earth.
As per the given question, there is 1%1\% decrease in the radius of the Earth. As a result, the new radius of Earth is R′=0.99RR' = 0.99R . We need to find the acceleration due to gravity g′g' for this changed radius. It will be given as:

g′=GMR′2g' = \dfrac{{GM}}{{R{'^2}}}

Substituting R′=0.99RR' = 0.99R in the above equation, we get

g′=GM(0.99R)2g' = \dfrac{{GM}}{{{{\left( {0.99R} \right)}^2}}}
⇒g′=1.02×GMR2\Rightarrow g' = 1.02 \times \dfrac{{GM}}{{{R^2}}}

From equation 1 we have, g=GMR2g = \dfrac{{GM}}{{{R^2}}} , put this value in above equation, we get

g′=1.02×gg' = 1.02 \times g

Percentage change will be given as g′−gg×100\dfrac{{g' - g}}{g} \times 100

⇒g′−gg×100=1.02g−gg×100 \Rightarrow \dfrac{{g' - g}}{g} \times 100 = \dfrac{{1.02g - g}}{g} \times 100
⇒g′−gg×100=0.02×100\Rightarrow \dfrac{{g' - g}}{g} \times 100 = 0.02 \times 100
=2%= 2\%

Therefore, there is 2%2\% increase in the acceleration due to gravity as the radius of Earth is reduced by 1%1\% without changing the mass.

Thus, option A is the correct option.

Note: We are given that the mass of the Earth does not change. The acceleration due to gravity is inversely proportional to the square of the radius of the Earth. If the radius increases, the acceleration due to gravity will decrease and vice-versa. Acceleration due to gravity is constant of bodies on the surface of a particular planet/satellite. The acceleration due to gravity is different for different planets/satellites only because it depends on the radius of the planet/satellite.