Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 8 - Exercises and Problems - Page 147: 41

Answer

$35.9\space GJ$ Since this is a positive value the spacecraft is not bound to the sun.

Work Step by Step

Here we use the equations of kinetic & potential energy to find the total energy consisting of the new horizons spacecraft. Total energy (E) $=\frac{1}{2}mV^{2}-\frac{GM_{S}}{m}$ Where, m - mass of the spacecraft, V - speed of the spacecraft, $M_{S}$ - mass of the sun, R - distance to the spacecraft from the sun. $E= 450\space kg\left \{ \frac{1}{2}(51000\times\frac{1000}{3600}m/s)^{2}-\frac{6.67\times10^{-11}Nm^{2}/kg^{2}\times1.99\times10^{30}kg}{6.5\times10^{12}m}\right \}$ $E= 35.9\times10^{9}J= 35.9\space GJ$
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