Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 13 - Gravitation - Problems - Page 383: 67c

Answer

The altitude is $~~420~km$

Work Step by Step

We can find the original energy of the system: $E = -\frac{GMm}{2r}$ $E = -\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)(220~kg)}{(2)(6.37\times 10^6~m+6.40\times 10^5~m)}$ $E = -6.25\times 10^9~J$ We can find the energy that is lost in 1500 revolutions: $(1500)(1.4\times 10^5~J) = 2.1\times 10^8~J$ We can find the energy in the system after 1500 revolutions: $E = -6.25\times 10^9~J - 2.1\times 10^8~J = -6.46\times 10^9~J$ We can find the orbital radius after 1500 revolutions: $E = -\frac{GMm}{2r}$ $r = -\frac{GMm}{2E}$ $r = -\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)(220~kg)}{(2)(-6.46\times 10^9~J)}$ $r = 6.79\times 10^6~m$ We can find the altitude $h$ above the Earth's surface: $h = 6.79\times 10^6~m-6.37\times 10^6~m$ $h = 4.2\times 10^5~m$ $h = 420~km$ The altitude is $~~420~km$
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