University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 30: 1.51

Answer

(a) $r = 1.65\times 10^4~km$ (b) The radius is 2.59 times the Earth's radius.

Work Step by Step

We can use $5.97\times 10^{24}~kg$ as the Earth's mass. We can use $6.38\times 10^6~m$ as the Earth's radius. (a) $\rho = \frac{mass}{volume} = \frac{(5.5)(5.97\times 10^{24}~kg)}{\frac{4}{3}\pi r^3} = 1.76\times 10^3~kg/m^3$ $r^3 = \frac{(3)(5.5)(5.97\times 10^{24}~kg)}{(4\pi)(1.76\times 10^3~kg/m^3)} = 4.454\times 10^{21}~m^3$ $r = 1.65\times 10^7~m = 1.65\times 10^4~km$ (b) We can convert this radius to a multiple of Earth's radius. $r = \frac{1.65\times 10^4~km}{6.38\times 10^3~km} = 2.59$ The radius is 2.59 times the Earth's radius.
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