Question

Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:
A. $4:1$
B. $2:1$
C. $1:1$
D. $1:2$

Answer 1

To solve this question we have to know what acceleration due to gravity and the formula is. We know that a body falls freely due to gravity. So this is called acceleration due to gravity. This is also called free-fall acceleration. This is equal to $\dfrac{{GM}}{{{R^2}}}$ . we are going to use this formula for this question.

Complete step by step answer:
We know that acceleration of a free fall is equal to g which is the gravitational acceleration. Which is equal to
$\dfrac{{GM}}{{{R^2}}} = \dfrac{{G(\rho \dfrac{4}{3}\pi {R^3})}}{{{R^2}}} \propto \rho R$
We know that the acceleration of both the earth’s surface and the planet’s surface is the same .So, we can write that,
${\rho _e}{R_e} = {\rho _P}{R_P}$
Here we are assuming that ${\rho _e}$ is the density of the earth and ${\rho _P}$ is the density of the planet
Again, ${R_e}/{R_p} = {\rho _P}/{\rho _e} = \dfrac{{{d_e}}}{{{d_p}}} = 1:2$

So the right option will be D.

Note: We know that the value of g is $9.3m/s^2$ but here we do not have to put this value. We can get confused when we should put this value and when we should use the full formula. Here we used the full formula of gravitational acceleration. Here according to the formula we know that M is the mass R is the radius of the planet or the earth.