(a) For a head-on collision, the relative speed of the two objects must be 2v= 5.4 ×104km/h. (b) A perpendicular collision is possible if one satellite is, say, orbiting above the equator and the other is following a longitudinal line. In this case, the relative speed is given by the Pythagorean theorem: 22ν+= 3.8 ×104km/h. 44. From Eq. 13-37, we obtain
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.