Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 373: 35


The projectile goes up to a height of 420 km.

Work Step by Step

We first convert the initial speed to units of m/s: $v = (10,000~km/h)(1000~m/km)(1 ~hr/3600~s)$ $v = 2777.8~m/s$ We then use conservation of energy to find the height $h$ of the projectile when the speed decreases to zero. Let $R$ be the earth's radius. Let $M_e$ be the earth's mass. Let $M_p$ be the projectile's mass. $K_f+U_f = K_0+U_0$ $0+U_f = K_0+U_0$ $-\frac{G~M_e~M_p}{R+h} = \frac{1}{2}M_pv^2-\frac{G~M_e~M_p}{R}$ $-2~G~M_e~R = (R)(R+h)(v^2)-(2~G~M_e)(R+h)$ $2~G~M_e~h-R~h~v^2 = R^2~v^2$ $h = \frac{R^2~v^2}{2~G~M_e-R~v^2}$ $h = \frac{(6.38\times 10^6~m)^2(2777.8~m/s)^2}{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)-(6.38\times 10^6~m)(2777.8~m/s)^2}$ $h = 4.20\times 10^5~m$ $h = 420~km$ The projectile goes up to a height of 420 km.
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