Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 12 - Gravity - Problems and Conceptual Exercises - Page 413: 79

Answer

$ v=\sqrt{\frac{GM_E}{R_E+h}}$

Work Step by Step

In the given scenario $F_{centripetal}=F_{gravity}$ $\implies m(\frac{v^2}{R_E+h})=G\frac{M_E\space m}{(R_E+h)^2}$ This can be rearranged as: $v^2=\frac{GM_E}{R_E+h}$ $\implies v=\sqrt{\frac{GM_E}{R_E+h}}$
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