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 - General Problems: 87

Answer

$1.21\times10^6m$

Work Step by Step

Find the 'new' Earth radius by setting the acceleration due to gravity at the Sun's surface equal to the acceleration due to gravity at the 'new' Earth's surface. $$g_{Earth new}=g_{Sun}$$ $$\frac{GM_{Earth}}{r^2_{Earthnew}}=\frac{GM_{Sun}}{r^2_{Sun}}$$ $$r_{Earthnew}=r_{Sun}\sqrt{\frac{M_{Earth}}{M_{Sun}}}=(6.96\times10^8m)\sqrt{\frac{5.98\times10^{24}kg}{1.99\times10^{30}kg}}=1.21\times10^6m$$
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