ch13-p068

# ch13-p068 - 68. The orbital radius is r = RE + h = 6370 km...

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(g) To find the period, we use Eq. 13-34 but replace r with a . The result is 23 2 6 3 3 11 3 2 24 44 ( 6 . 6 3 1 0 m ) 5.37 10 s 89.5 min. (6.67 10 m /s kg)(5.98 10 kg) a T GM ππ × == = × ×⋅ × (h) The orbital period T for Picard’s elliptical orbit is shorter than Igor’s by 0 5540 s 5370 s 170 s TTT Δ= −= = . Thus, Picard will arrive back at point P ahead of Igor by 170 s – 90 s = 80 s. 68. The orbital radius is 6 6370 km 400 km 6770 km 6.77 10 m. E rR h =+ = + = = × (a) Using Kepler’s law given in Eq. 13-34, we find the period of the ships to be 2 6 3 3 0 11 3 2 24 ( 6 . 7 7 1 0 m ) 5.54 10 s 92.3 min. (6.67 10 m /s kg)(5.98 10 kg) r T GM × = × × (b) The speed of the ships is 6 32 0 3 0 22 ( 6 . 7 7 1 0 m ) 7.68 10 m/s 5.54 10 s r v T × × . (c) The new kinetic energy is 2 3 2 1 0 0 11 1 (0.99 ) (2000 kg)(0.99) (7.68 10 m/s) 5.78 10 J. 2 Km v m v = × = × (d) Immediately after the burst, the potential energy is the same as it was before the burst.
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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.

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