A newly discovered planet has twice the mass and three times theradius of the earth. What is the free-fall acceleration at itssurface, in terms of the free-fall acceleration g at the surface ofthe earth?A. 2/9gB. 2/3gC. 3/4gD. 4/3g

A) 2/9 g

Explanation:

Acceleration due to gravity (free fall acceleration) of the earth is given as:

g = GM/R² (1)

Where G = gravitational constant

M = mass of earth

R = radius of earth

For the newly discovered planet having twice the mass of earth and thrice it's radius, free fall acceleration will be:

a = (G*2M)/(3R)²

a = 2MG/9R² (2)

Making G subject of formula in (1),

G = R²g/M

Putting this into (2):

a = (2 * M * R² * g)/(9 * R² * M)

=> a = 2g/9

i don't know that i agree with that but what's the question ?

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.