Text Material Preview
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436 PROBLEM 12.86 A space vehicle is in a circular orbit of 2200-km radius around the moon. To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it passes through A. Knowing that the mass of the moon is 73.49 2110 kg,× determine (a) the speed of the vehicle as it approaches B on the elliptic path, (b) the amount by which its speed should be reduced as it approaches B to insert it into the smaller circular orbit. SOLUTION For a circular orbit, 2 :n n v F ma F m r Σ = = Eq. (12.28): 2 Mm F G r = Then 2 2 Mm v G m rr = or 2 GM v r = Then 12 3 2 21 2 circ 3 66.73 10 m /kg s 73.49 10 kg ( ) 2200 10 m Av −× ⋅ × ×= × or circ( ) 1493.0 m/sAv = and 12 3 2 21 2 circ 3 66.73 10 m /kg s 73.49 10 kg ( ) 2080 10 m Bv −× ⋅ × ×= × or circ( ) 1535.5 m/sBv = (a) We have circ( ) ( ) (1493.0 26.3) m/s 1466.7 m/s A TR A Av v v= + Δ = − = Conservation of angular momentum requires that ( ) ( )A A TR B B TRr m v r m v= or 2200 km ( ) 1466.7 m/s 2080 km 1551.3 m/s B TRv = × = or ( ) 1551 m/sB TRv = (b) Now circ( ) ( )B B TR Bv v v= + Δ or (1535.5 1551.3) m/sBvΔ = − or 15.8 m/sBvΔ = −
