Question

The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:

• A. g/4
• B. 4g
• C. g/2
• D. 2g

Answer 1

The Correct Option is B

To solve the problem, we need to understand how acceleration due to gravity, \(g\), depends on the dimensions and mass of the Earth. The formula for gravitational acceleration at the surface of a planet is given by:

\(g = \frac{G \cdot M}{R^2}\) 

where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the Earth.
• \(R\) is the radius of the Earth.

According to the problem, the diameter of the Earth is reduced to half, which means the radius will also be reduced to half. So, the new radius \(R_{new} = \frac{R}{2}\).

Since the mass of the Earth remains constant, we can substitute \(R_{new}\) into the formula for gravity:

\(g_{new} = \frac{G \cdot M}{(R_{new})^2} = \frac{G \cdot M}{(\frac{R}{2})^2}\)

This simplifies to:

\(g_{new} = \frac{G \cdot M}{\frac{R^2}{4}} = \frac{4 \cdot G \cdot M}{R^2}\)

The above expression shows that the new acceleration due to gravity becomes:

\(g_{new} = 4g\)

Therefore, if the diameter of the Earth is reduced to half and the mass remains constant, the acceleration due to gravity on the surface of the Earth would be \(4g\), which corresponds to the correct answer: 4g.

Answer 2

The acceleration due to gravity on the surface of the earth is given by:

\(g = \frac{GM}{R^2}\)

where \( G \) is the gravitational constant, \( M \) is the mass of the earth, and \( R \) is the radius of the earth.

If the diameter of the earth is reduced to half, the radius \( R \) will also be reduced to half, becoming \( \frac{R}{2} \). Substituting \( R' = \frac{R}{2} \) into the formula for \( g \), we get:

\(g' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = 4 \cdot \frac{GM}{R^2} = 4g\)
Thus, the new acceleration due to gravity on the surface of the earth would be \( 4g \).