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PHY 2048

by: Garett Kovacek

PHY 2048 PHY 2048

Garett Kovacek
GPA 3.95


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This 8 page Class Notes was uploaded by Garett Kovacek on Thursday September 17, 2015. The Class Notes belongs to PHY 2048 at Florida State University taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/205540/phy-2048-florida-state-university in Physics 2 at Florida State University.


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Date Created: 09/17/15
Gravity Chapter 11 Kepler39s Laws Example of a successful phenomenology at the cradle of science astronomical observations by Brahe 1 All planets move on elliptical orbits with the sun at one focus Ellipse see mathworld 2 A linejoining any planet to the sun sweeps out equal areas in equal times figure 114 of Tipler Mosca 3 The square of the period of any planet is proportional to the cube of the semimajor axes of its orbit T2 Cr2 Astronomical unit AU The mean earthsun distance 1AU 150 gtlt1011 m 930 X 106 mi Newton s Law of Gravity Force of attraction between each pair of point particles figure 116 of TiplerMosca e Gm1 W2 A Gm1 m2 4 F12 2 T12 3 12 T12 12 Where G is the gravitational constant G 667 gtlt1011 N m2kg Freefall acceleration for objects near the surface of the earth GM 9 RZE 981m82 E Measurement of G Torsion balance first Cavendish 1798 Gravitational and Inertial Mass The mass which enters the force law is called inertial mass I F m a whereas the mass which enters the gravitational law is called gravitational mass In principle these two masses could disagree like the electric charge has nothing to do with the mass of inertia However all experimental evidence is that mI mG Therefore we omit the superscripts again Derivation of Kepler s Laws Kepler s first law With a bit more involved mathematics than we have presently at our disposal one can show that the only closed solutions to Newton39s two body force are elliptical orbits intermediate mechanics for physicists Kepler39s second law Figure 118 of TiplerMosca The area swept out by the radius vector 77 in the time dt is 1 1 L dAz lfx dtl l77gtltm27l dt dt 2 2m 2m where L lFgtlt l is the magnitude of the orbital angular momentum of the planet about the sun Since the force on the planet is along the line from the planet to the center of the ellipse there is no torque acting on the planet and L is conserved Therefore dA constant dt 2 m Kepler39s third law We will show this for the special orbit of a circle Then 77117712 112 mgam2 2 r FG r and 02 2W Gml T 7 follows which can be written as Kepler39s third law 2 47r 3 T2 Gmlr Gravitational Potential Energy Remember for conservative forces a potential energy is defined by dU Fd where d is the infinitesimal displacement of the particle For the gravitational force we have dU m1 m2 m1 m2 G T dr G T dr Integrating both sides gives UUm G 2 In astronomy a frequently used convention for the integration constant is U0 0 so that Uoo 0 holds Figure 119 of Tipler Mosca gives in this convention a plot of the potential energy of a point particle in the gravitational field of the earth Escape Speed Initial kinetic energy needed to escape from the earth39 surface to 7 oo where 216 is called escape speed Solving for v6 2GM E we V2 981 ms2 637 X 106 m 112 kms If E K7 l U7 lt 0 the system is bound If E K7 U7 2 0 the system is unbound Gravitational field of the earth


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