problem12_75

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.75: a) The semimajor axis is the average of the perigee and apogee distances, m. 10 58 . 8 )) ( ) (( 6 a E p E 2 1 × = + + + = h R h R a From Eq. (12.19) with the mass of the earth, the period of the orbit is s, 10 91 . 7 2 3 2 3 × = π = E GM a T a little more than two hours. b) See Problem 12.74; . 53 . 1 p a p = = r r a υ c) The equation that represents conservation of energy (apart from a common factor of the mass of the spacecraft) is , 2 1 2 1 2 1 a E 2 p 2 a p a E 2 a p E 2 p r Gm v r r r Gm v r Gm v - = - = - where conservation of angular momentum has been used to eliminate . of favor is p a v v Solving for 2 p v and simplifying, ( 29 , s m 10 71 . 7 2 2 2 7 p p a E 2 p × = + = a r r r r Gm v from which s. m 10 51 . 5 and s m 10 43 . 8 3 3 p × = × = a v v d) The escape speed for a given
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Unformatted text preview: distance is , 2 r GM v e = and so the difference between escape speed and p v is, after some algebra, [ ] ⋅ +-=-) ( 1 / 1 1 2 p p p e a E r r r Gm v v Using the given values for the radii gives s. m 10 41 . 2 3 p e × =-v v The similar calculation at apogee give s, m 10 26 . 3 3 a e × =-v v so it is more efficient to fire the rockets at perigee. Note that in the above, the escape speed e v is different at the two points, s. m 10 77 . 8 and s m 10 09 . 1 3 ae 4 pe × = × = v v...
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## This document was uploaded on 02/05/2008.

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