Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Problems - Page 134: 53

Answer

Astronaut's apparent weight: $58.8N$ away from the moon With acceleration: $76.2N$ toward the moon

Work Step by Step

$F_G=G\frac{m_1m_2}{r^2}$ $F_{AW}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{75kg\times7.35\times10^{22}kg}{(2500\times10^3m)^2}=58.8N$ (away from the moon) $\sum F=mg-N$ $N=mg_M-ma$ $N=58.8N-75kg\times (1.8\frac{m}{s^2})=-76.2N$ (toward the moon)
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