Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 8 - Exercises and Problems - Page 142: 41


$M=2.61\times 10^{41}Kg$

Work Step by Step

As we know that $\frac{GMm}{r^2}=\frac{mv^2}{r}$ This can be rearranged as: $M=\frac{v^2 r}{G}$ As $v=\frac{2\pi r}{T}$ $\implies M=\frac{(2\pi r)^2 r}{GT^2}$ We plug in the known values to obtain: $M=\frac{2(3.1416(2.6\times 10^{20}))^2(2.6\times 10^{20})}{6.67\times 10^{-11}(200)^2}$ $M=2.61\times 10^{41}Kg$
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