Hw 7 Solutions

# Hw 7 Solutions - P1 The period T and orbit radius r are...

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P1.- The period T and orbit radius r are related by the law of periods: T 2 = (4 π 2 / GM ) r 3 , where M is the mass of Mars. The period is 7 h 39 min, which is 2.754 × 10 4 s. We solve for M : ( ) 2 3 2 6 3 23 2 2 11 3 2 4 4 4 (9.4 10 m) 6.5 10 kg. (6.67 10 m /s kg) 2.754 10 s r M GT π π × = = = × × × P2.- Kepler’s law of periods, expressed as a ratio, is 3 2 2 3 (1.52) 1y M M M E E a T T a T = = where we have substituted the mean-distance (from Sun) ratio for the semimajor axis ratio. This yields T M = 1.87 y. The value in Appendix C (1.88 y) is quite close, and the small apparent discrepancy is not significant, since a more precise value for the semimajor axis ratio is a M / a E = 1.523 which does lead to T M = 1.88 y using Kepler’s law. A question can be raised regarding the use of a ratio of mean distances for the ratio of semimajor axes, but this requires a more lengthy discussion of what is meant by a mean distance” than is appropriate here. P3.- (a) If r is the radius of the orbit then the magnitude of the gravitational force acting on the satellite is given by GMm/r 2 , where M is the mass of Earth and m is the mass of the satellite. The magnitude of the acceleration of the satellite is given by v 2 /r , where v is its speed. Newton’s second law yields GMm/r 2 = mv 2 /r . Since the radius of Earth is 6.37 × 10 6 m the orbit radius is r = (6.37 × 10 6 m + 160 × 10 3 m) = 6.53 × 10 6 m. The solution for v is 11 3 2 24 3 6 (6.67 10 m /s kg)(5.98 10 kg) 7.82 10 m/s. 6.53 10 m GM v r × × = = = × × (b) Since the circumference of the circular orbit is 2 π r , the period is 6 3 3 2 2 (6.53 10 m) 5.25 10 s. 7.82 10 m/s r T v π π × = = = × × This is equivalent to 87.5 min

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P4.- . (a) The period is T = 27(3600) = 97200 s, and we are asked to assume that the orbit is circular (of radius r = 100000 m). Kepler’s law of periods provides us with an approximation to the asteroid’s mass: ( ) 2 3 2 16 4 (97200) 100000 6.3
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