## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 13 - Newton's Theory of Gravity - Conceptual Questions: 6

#### Answer

The free-fall acceleration on the surface of planet 2 is $10~m/s^2$

#### Work Step by Step

Let $M$ be the mass of planet 1. Let $R$ be the radius of planet 1. We can write an expression for the free-fall acceleration $g_1$ at the surface of planet 1. $g_1 = \frac{G~M}{R^2} = 20~m/s^2$ Note that the mass of planet 2 is $2M$ and the radius of planet 2 is $2R$. We can write an expression for the free-fall acceleration $g_2$ on the surface of planet 2. $g_2 = \frac{G~(2M)}{(2R)^2}$ $g_2 = \frac{G~M}{2~R^2}$ $g_2 = \frac{g_1}{2}$ $g_2 = \frac{20~m/s^2}{2}$ $g_2 = 10~m/s^2$ The free-fall acceleration on the surface of planet 2 is $10~m/s^2$.

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