## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 8 - Exercises and Problems - Page 141: 12

#### Answer

Please see the work below.

#### Work Step by Step

We know that in circular motion $a_r=\frac{v^2}{r}$ We plug in the known values to obtain: $a_r=\frac{(1.02\times 10^3)^2}{0.3844\times 10^6\times 10^3}$ $a_r=2.7\times 10^{-3}\frac{m}{s^2}$ According to Newton's law of gravitation $F=\frac{GMm}{r^2}$ This can be rearranged as: $\frac{F}{m}=\frac{GM}{r^2}$ $\implies a=\frac{GM}{r^2}$ We plug in the known values to obtain: $a=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{0.3844\times 10^9}$ $a=2.7\times 10^{-3}\frac{m}{s^2}$ Thus, it is verified that both cases give the same answer.

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