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Assuming that the mass of Earth is constant when its radius shrinks by 5%, the value of acceleration due to gravity on Earth will be (Assume g = 10 m/s2)
1. 9.81 m/s2
2. 11.08 m/s2
3. 2.81 m/s2
4. 5.67 m/s2

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Correct Answer - Option 2 : 11.08 m/s2

The correct answer is option 2) i.e. 11.08 m/s2


  • Law of Universal Gravitation: It states that all objects attract each other with a force that is proportional to the masses of two objects and inversely proportional to the square of the distance that separates their centres.

It is given mathematically as follows:

\(F = \frac{Gm_1m_2}{R^2}\)

Where m1 and m2 are the mass of two objects, G is the gravitational constant and R is the distance between their centres.

  • From the Law of Universal Gravitation, the gravitational force acting on an object of mass m placed on the surface of Earth is:

\(F = \frac{GMm}{R^2}\)

Where R is the radius of the earth. 

From Newton's second law, F = ma = mg

\(⇒ mg =\frac{GMm}{R^2}\)

⇒ Acceleration due to gravity, \(g =\frac{GM}{R^2}\)


Let M be the mass of Earth, R be its radius and g be the acceleration due to gravity of Earth.

When the radius shrink by 5%, the new radius R' = 95%R = 0.95R

The mass remains the same i.e. M' = M

Initially, \(g =\frac{GM}{R^2} =10 \: m/s^2\)

After shrinking the radius, \(g' =\frac{GM'}{R'^2} \)

\(\Rightarrow g' =\frac{GM}{(0.95R)^2} =\frac{GM}{0.9025R^2}\)

\(\Rightarrow g' = \frac{g}{0.9025} = \frac{10}{0.9025} = 11.08 \:m/s^2\)

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