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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems: 12

Answer

The free-fall acceleration on the newly discovered planet is $12.25~m/s^2$

Work Step by Step

We can write an expression for the free-fall acceleration on the earth. $g = \frac{G~M}{R^2} = 9.80~m/s^2$ We can find the free-fall acceleration $g'$ on the newly discovered planet. $g' = \frac{G~(5M)}{(2R)^2}$ $g' = (\frac{5}{4})~(\frac{G~M}{R^2})$ $g' = (\frac{5}{4})~(9.80~m/s^2)$ $g' = 12.25~m/s^2$ The free-fall acceleration on the newly discovered planet is $12.25~m/s^2$.
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