Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Exercises and Problems - Page 166: 90


$743\times10^6 \ m$

Work Step by Step

The distance between the two is an average of $779 \times 10^9 m$. The sun weighs $1.989\times 10^{30} \ kg $, and Jupiter weighs $1.898 \times 10^{27}\ kg $. Thus, we use the equation for center of mass to find: $\frac{(1.898 \times 10^{27}\ kg)(779 \times 10^9 m)}{(1.989\times 10^{30})+(1.898 \times 10^{27}\ kg)}=743\times10^6 \ m$
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