Question

A planet has its mass twice of the earth and its
radius is 3 times of the radius of earth. If the
acceleration due gravity on earth is g, the
acceleration due to gravity on the surface of the
planet will be​

Answer 1

Given :

• Mass of the Planet X = Twice the Mass of Earth.

• Radius of the Planet X = Thrice the Radius of Earth.

• Acceleration due to gravity on the surface of the planet = g m/s².

To find :

The Acceleration due to gravity on the Planet X.

Solution :

Let the mass of earth be m Kg and the Radius of Earth be r m.

So According to the given information , we get the Mass and Radius of Planet X as 2m and 3r ,respectively.

Now to find the acceleration due to gravity on that planet with respect to the acceleration due to gravity on earth (in terme of g).

To find the acceleration due to gravity on earth :

We know the formula for Gravitational Acceleration :

$\underline{\boxed{\bf{g = \dfrac{GM}{R^{2}}}}}$

Where :-

• g = Acceleration due to gravity
• G = Universal Gravitational constant
• R = Distance

Now , using the formula and substituting the values in it, we get :

$:\implies \bf{g_{1} = \dfrac{GM}{R^{2}}}$

$:\implies \bf{g_{1} = \dfrac{Gm}{r^{2}}}$

$\therefore \bf{g_{1} = \dfrac{Gm}{r^{2}}}$

Hence, the acceleration due to gravity on the planet earth is $\bf{\dfrac{Gm}{r^{2}}}$

To find the acceleration due to gravity on Planet X :

We know the formula for Gravitational Acceleration :

$\underline{\boxed{\bf{g = \dfrac{GM}{R^{2}}}}}$

Where :-

• g = Acceleration due to gravity
• G = Universal Gravitational constant
• R = Distance

Now , using the formula and substituting the values in it, we get :

$:\implies \bf{g_{2} = \dfrac{G \times 2m}{(3m)^{2}}}$

$:\implies \bf{g_{2} = \dfrac{2Gm}{9r^{2}}}$

$\therefore \bf{g_{2} = \dfrac{2Gm}{9r^{2}}}$

Hence, the acceleration due to gravity on the planet X is $\bf{\dfrac{2Gm}{3r^{2}}}$

Now , by dividing the Acceleration on earth by Acceleration on planet X.

$:\implies \bf{\dfrac{g_{1}}{g_{2}}}$

$:\implies \bf{\dfrac{g_{1}}{g_{2}} = \dfrac{\dfrac{Gm}{r^{2}}}{\dfrac{2Gm}{3r^{2}}}}$

$:\implies \bf{\dfrac{g_{1}}{g_{2}} = \dfrac{Gm}{r^{2}} \times \dfrac{3r^{2}}{2Gm}}$

$:\implies \bf{\dfrac{g_{1}}{g_{2}} = \dfrac{\not{G}\not{m}}{\not{r^{2}}} \times \dfrac{3\not{r^{2}}}{2\not{G}\not{m}}}$

$:\implies \bf{\dfrac{g_{1}}{g_{2}} = \dfrac{3}{2}}$

By Cross-multiplication , we get :

$:\implies \bf{2g_{1} = 3g_{2}}$

$:\implies \bf{2g_{1} = 3g_{2}}$

$:\implies \bf{\dfrac{2g_{1}}{3} = g_{2}}$

$\boxed{\therefore \bf{g_{2} = \dfrac{2g_{1}}{3}}}$

Hence, the acceleration due to gravity on that planet will be ⅔ of the Acceleration due to gravity on the Earth.