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Unformatted text preview: Homework 1 Mechanics 701, Fall 2011 Due: Friday, September 9 1 Problem 3-16 Prove that in a Kepler elliptic orbit with small eccentricity e the angular motion of a particle as viewed from the empty focus of the ellipse is uniform to first order in e . The empty focus is the focus that is not the center of attraction. It is this theorem that enables the Ptolemaic picture of planetary motion to be a reasonably accurate approximation. On this picture the Sun is assumed to move uniformly on a circle whose center is shifted from Earth by a distance called the equant . If the equant is taken as the distance between the two foci of the correct elliptical orbit, then the angular motion is thus described by the Ptolemaic picture accurately to first order in e . Solution: Let R ( t ) be the distance to the planet from the empty focus and ( t ) be the angle to the planet measured from the empty focus ( and the standard angle are measured relative to the same line in space). Since the planet moves along the ellipse, R + r = l = const. First we prove that if the planets elliptic trajectory satisfies 1 r = 1 + e cos r then R and satisfy 1 R = 1- e cos r From the triangle F 1 F 2 P , where F 1 , 2 are the foci and P is the position of the planet, we get R 2 + d 2- 2 Rd cos = r 2 = ( l- R ) 2 from where we immediately get the equation above using r = ( l 2- d 2 ) / 2 l . Next we take the identity r ( ) + R ( ) = const and differentiate with respect to time. This gives sin (1 + e cos ) 2 = sin (1- e cos ) 2 In addition, the sine-theorem for the triangle F 1 F 2 P states that r sin = R sin or sin 1 + e cos = sin 1- e cos Together they give = 1- e cos 1 + e cos = L z mr 2 1- e cos 1 + e cos = L z mr 2 (1- e cos )(1 + e cos ) 1 where we have used = L z /mr 2 (conservation of momentum). Thus = L z mr 2 (1 + e (cos - sin ) +...
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