ch13-p061 - 61. (a) We use the law of periods: T2 =...

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61. (a) We use the law of periods: T 2 = (4 π 2 / GM ) r 3 , where M is the mass of the Sun (1.99 × 10 30 kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of Earth’s orbit: r = 2 r e = 2(150 × 10 9 m) = 300 × 10 9 m. Thus, 23 2 9 3 7 11 3 2 30 4 4 (300 10 m) 8.96 10 s. (6.67 10 m /s kg)(1.99 10 kg) r T GM ππ × == = × ×⋅ × Dividing by (365 d/y) (24 h/d) (60 min/h) (60 s/min), we obtain T = 2.8 y. (b) The kinetic energy of any asteroid or planet in a circular orbit of radius
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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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