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: 34

Answer

$1.98 m/s^2$

Work Step by Step

Mass per unit volume is the definition of density. Assuming that the density of Europa is the same as of the Earth's, the radius of Europa can be calculated. $$\rho_{Europa} = \rho_{Earth}$$ $$\frac{m_{Europa}}{\frac{4}{3}\pi r^3_{Europa}}=\frac{m_{Earth}}{\frac{4}{3}\pi r^3_{Earth}}$$ $$r_{Europa}=r_{Earth}(\frac{m_{Europa}}{m_{Earth}})^{\frac{1}{3}}$$ $$g_{Europa}=\frac{Gm_{Europa}}{r^2_{Europa}}=\frac{Gm_{Europa}}{(r_{Earth}(\frac{m_{Europa}}{m_{Earth}})^{\frac{1}{3}})^2}=\frac{Gm^{1/3}_{Europa}m^{2/3}_{Earth}}{r^2_{Earth}}=\frac{Gm_{Earth}}{r^2_{Earth}}(\frac{m^{1/3}_{Europa}}{m^{1/3}_{Earth}})=g_{Earth}(\frac{m_{Europa}}{m_{Earth}})^{\frac{1}{3}}$$ $$=9.8m/s^2 (\frac{4.9x10^{22}kg}{5.98x10^{24}kg})^{\frac{1}{3}}=1.98m/s^2$$
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