Question

Planet Z-34 has a mass equal to one-third that of Earth and a radius equal to one-third that of Earth. With g representing, as usual, the acceleration due to gravity on the surface of Earth, the acceleration due to gravity on the surface of Z-34 is

Answer 1

The acceleration due to gravity on the surface of Z-34 is equal to
`3g` .

Step-by-step explanation:

Here we know that acceleration due to gravity on a planet of mass M and radius R is given by

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

Given that mass of earth is M and radius is R .

Mass of Z-34 =
`(M)/(3)` ,

Radius of Z-34 =
`(R)/(3)`

G = gravitational constant .

`g_(z)` = acceleration due to gravity on Z-34 ,

`(g)/(g_(z)) = (M)/((M)/(3) ) * (((R)/(3) )/(R))^(2) = (1)/(3) \\ \\g_(z) = 3g`