Question

The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of planet 1.
What is the free-fall acceleration on planet 2? Express your answer using two significant figures. m/s^2.

Answer 1

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2