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

Published by Pearson

# Chapter 8 - Section 8.4 - Gravitational Energy - Example - Page 136: 8.5

#### Answer

$6.9\ Mm$

#### Work Step by Step

We know that kinetic energy plus potential energy is constant. Using this fact and solving for r, we find: $r = (\frac{1}{R_E}-\frac{v_0^2}{2GM_E})^{-1}=(\frac{1}{6.37 \times 10^6}-\frac{3100^2}{2(6.67\times10^{-11}\times5.97 \times10^{24})})^{-1} = 6.9 \ Mm$

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