Question

Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is ${g^1}$ then that would be the relation between them?

Answer 1

Basically there are four fundamental forces in nature. They are gravitational force, electrostatic force, strong nuclear force, weak nuclear force. Here the force in action is gravitational force and it is a long range force and the weakest of the four fundamental forces.
Formula used:
\eqalign{ & g = \dfrac{{GM}}{{{R^2}}} \cr & g = \dfrac{{4\pi G\rho r}}{3} \cr}

Complete answer:
Gravitational force will be acting between any two masses at a particular distance. Since the magnitude of that force is very small nobody feels it. There are some similarities between the gravitational force and electrostatic force. gravitational force acts between the two masses whereas electrostatic force acts between the two charges. Both follow the inverse square law which means that the force will be inversely proportional to the square of distance between them.
So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body.
We have the expression for the force between the earth of mass ‘M’ and the body of mass ‘m’ which are separated at the distance ‘r’ from the center of the earth. If we consider the surface of earth then $r = R$ i.e radius of earth
That expression is
$F = \dfrac{{GMm}}{{{R^2}}}$
Where ‘G’ is the gravitational constant
If we divide that force with mass of a body we will get acceleration due to gravity at that place so acceleration due to gravity on earth’s surface (g) will be
$g = \dfrac{{GM}}{{{R^2}}}$
If we convert the above expression in terms of density of planet then we will get
$g = \dfrac{{4\pi G\rho r}}{3}$
Where $\rho$ is the density of the planet and $r = R$ is radius of the planet in the given problem.
Given the density of two planets are the same and the new planet is 3 times bigger than the earth in size.
That means radius of new planet is 3 times bigger than the earth’s radius
${R^1} = 3R$
\eqalign{ & \Rightarrow g = \dfrac{{4\pi G\rho r}}{3} \cr & \Rightarrow g\alpha r \cr & \Rightarrow \dfrac{{{g^1}}}{g} = \dfrac{{{R^1}}}{R} \cr & \Rightarrow \dfrac{{{g^1}}}{g} = \dfrac{{3R}}{R} \cr & \Rightarrow {g^1} = 3g \cr}
Hence acceleration due to gravity on the surface of the new planet is thrice that of the earth’s.

Note:
The formula we had used i.e the acceleration due to gravity proportional to radius one is valid if densities are same or else it will be proportional to product of density and the radius.