Question

The free-fall acceleration at the surface of planet 1 is 14 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is g on planet 2?

Answer 1

7 m/s²

Step-by-step explanation:

Acceleration on the surface of the planet in question is

g = GM/r², where G is the gravitational constant, and M is the mass of the planet and r is the radius of the planet

From the question, we know that

g1 = 14

M2 = 2M1

r2 = 2r1

g2 = GM2/r2²

Substituting for the variables, we have

g2 = G.(2M1) / (2r1)²

g2 = 1/2 * GM1/r1², where GM1/r1² is the stated g in the first part of this solution

g2 = 1/2 * g

g2 = 1/2 * 14

g2 = 7 m/s²