Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size. If the acceleration due to gravity on the surface of the earth is \(g\) and that on the surface of the new planet is \(g'\), then

The acceleration due to gravity on the surface of a planet is given by \(g = \dfrac{GM}{R^2}\).

Substituting the mass of the planet \(M = \dfrac{4}{3}\pi R^3 \rho\), we get:

\(g = \dfrac{G}{R^2} \left(\dfrac{4}{3}\pi R^3 \rho\right) = \dfrac{4}{3}\pi G R \rho\)

This shows that \(g \propto R \rho\).

For the new planet, the density is the same as Earth (\(\rho' = \rho\)) and the size (radius) is twice that of Earth (\(R' = 2R\)).

Therefore, the acceleration due to gravity on the new planet is:

\(g' = \dfrac{4}{3}\pi G R' \rho'\)

\(g' = \dfrac{4}{3}\pi G (2R) \rho = 2 \left(\dfrac{4}{3}\pi G R \rho\right)\)

\(g' = 2g\)

Answer: \(g' = 2g\)