where Ris the radius of the particle, and 2ARπ=is the cross-sectional area. On the other hand, the gravitational force on the particle is given by Newton’s law of gravitation (Eq. 13-1): 33222(4/3)43SSSgGM mGMRGMRFrrrρππ===, where 3mRρπ=is the mass of the particle. When the two forces balance, the particle travels in a straight path. The condition that rgFF=implies 23443PRGM Rrcr=, which can be solved to give 26833113230733(3.9 10W)1616 (3 10 m/s)(3.5 10 kg/m )(6.67 10m /kg s )(1.99 10 kg)1.7 10 m .SSPRcGMπρ−−××××⋅×=×(b) Since gFvaries with 3RandrFvaries with 2,Rif the radius
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.