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Physics 210 Problems - My Solutions Dr. Hulan E. Jack Jr. Chapter 7 P56 Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3 THE PROBLEM STATEMENT Ch 7 P56 Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet. Page 1 of 2 Physics 210 Problems - My Solutions Dr. Hulan E. Jack Jr. Ch 7 P56 Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet. Basic Solution (including BRAINSTORMING-Definitions, concepts , principles and Discussion Given: Find: Show that vescape at surface, R, is proportional to density, . KE(r) + PE(r) = KE(r=infinity) + PE(r=infinity) (1) E (J) vinfinity >0 2 PE = -GMm/r where M is the mass of the planet (parent – this includes stars) and m is the mass of the satellite which includes planets, moons, even smaller stars around more massive stars. M >>m. Etotal >0 1/2mv - GMm/r < 0 1/2mv - GMm/r = 0 Etotal=0 r top Etotal <01/2mv - GMm/r > 0 vinfinity >0 2 r (m) 2 PEgravity=-GMm/r To escape the body of mass m must just reach r= infinity with no energy left. Since PE = 0 when r =infinity, but KE = 0 must also be true. So v=0 at r=infinity. ½ mvescape(r)2 - GMm/r = 0 [v=0] - 0. So, mve(r)2 = 2GMm/r (2) cancel m gives ve(r)2 = 2GM/r Hence ve(r) = sqrt(2GM/r) (3) This is the v that the body of mass m must have at a distance r from the center of, and outside of M. Some definitions: Volume of sphere = 3/4*R3 Density = mass/volume M/V = M/(3/4*R3) = 3 MR3 /4 . So M = V = R3/3), Now ve(r) = sqrt(2GM/r) becomes ve(r) = sqrt(2G(R3 /3r) = sqrt(8GR3 )/3r) If one escaping from the surface of the planet, then r = R, so, ve(R) = sqrt(8GR3 /3R) = sqrt(8GR2 /3) = Rsqrt(8G) Page 2 of 2 (3) (4)