Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Problems - Page 768: 33

Answer

$9.0\times10^{13}J$, $9.2\times10^9kg$.

Work Step by Step

Calculate the energy equivalent of the mass from Einstein’s famous equation (numbered 26-7 in the textbook). $$E=mc^2=(0.0010kg)(3.00\times10^8m/s)^2=9.0\times10^{13}J$$ Use this energy is used to increase the gravitational potential energy of a mass M. $$E=Mgh$$ $$M=\frac{E}{hg}=\frac{9.0\times10^{13}J }{(1.0\times10^3 m)(9.80m/s^2)}=9.2\times10^9kg$$
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