Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 12 - Gravity - Problems and Conceptual Exercises - Page 412: 62

Answer

a) $F=16N$ (Note that we assumed a mass of 70kg; other assumptions are possible.) b) $2.14\times 10^8m$

Work Step by Step

(a) The required force can be determined as $F=\frac{4GmM_a}{r^3}$ We plug in the known values to obtain: $F=4(6.67\times 10^{-11}N.m^2/Kg^2)\frac{(70Kg)(10^6\times 2.00\times 10^{30}Kg)(1.8m)}{(10^6mi\times 1609m/mi)^3}$ $F=16N$ (Note that we assumed a mass of 70kg; other assumptions are possible.) (b) We can find the required distance as follows: $r=(\frac{4GmM_a}{F})^{\frac{1}{3}}$ $\implies r=(\frac{4GmM_a}{10mg})^{\frac{1}{3}}=(\frac{4GM_a}{10g})^{\frac{1}{3}}$ We plug in the known values to obtain: $r=[\frac{4(6.67\times 10^{-11})(2.00\times 10^{36})(1.8)}{10(9.81)}]^{\frac{1}{3}}=2.14\times 10^8m$
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