61. (a) We use the law of periods: T2= (4π2/GM)r3, where Mis the mass of the Sun (1.99 ×1030kg) and ris the radius of the orbit. The radius of the orbit is twice the radius of Earth’s orbit: r= 2re= 2(150 ×109m) = 300 ×109m. Thus, 23293711323044(30010 m)8.9610 s.(6.6710m /s kg)(1.9910 kg)rTGMππ−×===××⋅×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.