Question

The free-fall acceleration on the surface of planet A is about one tenth of that on the surface of planet B. If the radius of planet A is a quarter of the radius of planet B (RB/4), find the ratio of their average densities?

Answer 1

`(\rho_A)/(\rho_B) = 0.4`

Step-by-step explanation:

As we know that the acceleration due to gravity on the surface of the planet is given as

`a = (GM)/(R^2)`

so we will have

M = mass of the planet

`M = \rho ((4)/(3) \pi R^3)`

so we have

`a = (G \rho (4)/(3) \pi R^3)/(R^2)`

`a = (4)/(3)\rho \pi G R`

so we have

`(a_A)/(a_B) = (\rho_A R_A)/(\rho_B R_B)`

`(1)/(10) = (\rho_A (R_B/4))/(\rho_B R_B)`

`(1)/(10) = (\rho_A)/(4\rho_B)`

`(\rho_A)/(\rho_B) = 0.4`