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Unformatted text preview: S. Widnall 16.07 Dynamics Fall 2008 Version 1.0 Lecture L18 Exploring the Neighborhood: the Restricted ThreeBody Problem The ThreeBody Problem In Lecture 1517, we presented the solution to the twobody problem for mutual gravitational attraction. We were able to obtain this solution in closed form. For bodies of comparable mass, the solution showed that the bodies orbited about their center of mass, with orbital periods that could be obtained analytically by transforming the twobody problem into a ”Kepler” problem with modified gravitational forces. For a body of negligible mass, such as a satellite orbiting the earth, we were able to place the earth at rest at the center of our coordinates, and obtain a simple expression for the orbit of the satellite. No such solution is available for the threebody problem. The general solution to the threebody problem must be obtained numerically, and even with these tools, the solution is chaotic: that is small changes in initial conditional lead to widely diverging behavior. For analysis of space missions, such a spacecraft moving between planets, or from earth to the moon, we may apply the model known as the restricted threebody problem, in which the mass of a small spacecraft is affected by the gravitational forces from two bodies, but the large bodies do not feel the inﬂuence of the spacecraft. The restricted threebody problem is most easily analyized/understood in a coordinate system rotating with the two primary bodies. Since the solution of the twobody problem is that of the two primary masses rotating at constant angular velocity Ω about their center of mass, in a coordinate system rotating with angular velocity Ω, these two bodies are stationary. The third body moves in their gravitational fields and experiences the additional acceleration terms associated with the rotation of the coordinate system. Consider a twobody system of mass M 1 and M 2 in an inertial references frame. We add a small mass m 3 which does not affect the large masses. 1 We consider planar x,y motion only. The equations of motion are x ¨ 1 = GM 2 ( x 1 − x 2 ) (1) − (( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 ) (3 / 2) x ¨ 2 = GM 1 ( x 2 − x 1 ) (2) − (( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ) (3 / 2) x ¨ 3 = GM 1 ( x 3 − x 1 ) GM 2 ( x 3 − x 2 ) (3) − (( x 3 − x 1 ) 2 + ( y 3 − y 1 ) 2 ) (3 / 2) − (( x 3 − x 2 ) 2 + ( y 3 − y 2 ) 2 ) ( 3 / 2) y ¨ 1 = GM 2 ( y 1 − y 2 ) (4) − (( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 ) (3 / 2) GM 1 ( y 2 − y 1 ) y ¨ 2 = − (( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ) (3 / 2) (5) ¨ = GM 1 ( y 3 − y 1 ) GM 2 ( y 3 − y 2 ) (6) y 3 − (( x 3 − x 1 ) 2 + ( y 3 − y 1 ) 2 ) (3 / 2) − (( x 3 − x 2 ) 2 + ( y 3 − y 2 ) 2 ) ( 3 / 2) The solution to the motion of M 1 and M 2 is a Kepler problem, where the two bodies move in circular orbits about their center of mass with a frequency G ( M 1 + M 2 ) Ω = R 3 , (7) where R is the distance between them. R...
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 Fall '09
 widnall
 Dynamics, Potential Energy, General Relativity, lagrange points

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