Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems: 56

Answer

Mercury's distance from the sun at the closest point is $4.60\times 10^{10}~m$

Work Step by Step

We can use conservation of angular momentum to find Mercury's speed at its closest point. Let $L_c$ be the angular momentum at the closest point. Let $L_f$ be the angular momentum at the farthest point. Let $M_m$ be Mercury's mass. $L_c = L_f$ $M_m~v_c~r_c ~sin(90^{\circ})= M_m~v_f~r_f~sin(90^{\circ})$ $r_c = \frac{v_f~r_f}{v_c}$ $r_c = \frac{(38.8~km/s)(6.99\times 10^{10}~m)}{59.0~km/s}$ $r_c = 4.60\times 10^{10}~m$ Mercury's distance from the sun at the closest point is $4.60\times 10^{10}~m$.
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