Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 356: 62

Answer

Comet Halley's speed as it crosses Neptune's orbit is 4.48 km/s

Work Step by Step

We can use conservation of energy to find Comet Halley's speed when it crossed Neptune's orbit. Let $M_s$ be the sun's mass and let $M_c$ be Comet Halley's mass. Let $R_1$ be the distance at the closest point and let $R_2$ be the distance at Neptune's orbit. $K_2+U_2 = K_1+U_1$ $\frac{1}{2}M_cv_2^2 = \frac{1}{2}M_c~v_1^2-\frac{G~M_s~M_c}{R_1}-(-\frac{G~M_s~M_c}{R_2})$ $v_2^2 = v_1^2-\frac{2~G~M_s}{R_1}+\frac{2~G~M_s}{R_2}$ $v_2^2 = v_1^2+(2~G~M_s)(\frac{1}{R_2}-\frac{1}{R_1})$ $v_2 = \sqrt{v_1^2+(2~G~M_s)(\frac{1}{R_2}-\frac{1}{R_1})}$ $v_2 = \sqrt{(5.46\times 10^4~m/s)^2+(2)(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)(\frac{1}{4.50\times 10^{12}~m}-\frac{1}{8.79\times 10^{10}~m})}$ $v_2 = 4.48\times 10^3~m/s$ $v_2 = 4.48~km/s$ Comet Halley's speed as it crosses Neptune's orbit is 4.48 km/s
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