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Central Force Motion and Reduced Mass

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Central Force Motion and Reduced Mass “Central-Force-Motion.doc” Central Force Motion and Reduced Mass Consider an isolated system consisting of 2 particles m interacting under a central force f(r). Their masses are 2 Center of m1, m2 at vector positions r1, r2 respectively. Mass r = r1- r2 r = r1 - r2 r = | r | = | r1 - r2 | r2 r R m ^r = – 1 r r1 •• ^ The equations of motion are: m1 r 1 = f(r) r (a) •• ^ and m2 r 2 = -f(r) r (b) The equations are said to be "coupled", ie. they are interdependent and the knowledge of the solution of one is needed to solve the other (and vice versa) as they are both dependent on ^r . The problem is solvable if we create two new co-ordinates, (just like normal mode co-ordinates in vibrations and waves) which are superpositions of r1 and r2. It is convenient - because of the physical interpretation of the quantities - to use the co-ordinates: (m1r1 + m2r2) r = r1 - r2 and R = (m1 + m2) R is the position of the centre of mass of the system (by definition), and the magnitude of r is simply the distance between the two masses. The equation of motion for R is straight forward as there are no external forces. •• R = 0 which has the solution R = Ro + Vt. It depends on the choice of origin and initial conditions, let Ro and V = 0 (ie. velocity of centre of mass = 0). Rearranging (a) and (b): •• •• 1 1 ^ r 1 – r 2 = + f(r) r (m1 m2 ) m m 1 2 •• •• ^ or ( r 1 – r 2) = f(r) r (m1 + m2 ) m1m2 let us define µ = reduced mass = m1 + m2 •• •• ^ •• •• •• then µ( r 1 - r 2) = f(r) r and as r = r 1 - r 2 then µ••r = f(r) ^r is the equation of motion for r. (c) Equation (c) is the identical equation of motion for a single particle of mass µ being acted on by a force f(r) ^r . No trace of the 2 particle problem remains. (Unfortunately this method cannot be generalised. There is no way to reduce the equations of motion of 3 or more interacting particles to equivalent one body equations, and for this reason the exact solution of the 3 body problem is unknown.) Notice that | r | is the distance between the 2 bodies (not the distance to the centre of mass). Ze2 In our case f(r) = (although it can equally apply to gravity and hence planetary systems.) 2 4πε0r Now consider the orbital angular momentum: L = r x µv ( = µvr if mutually orthogonal) (d) ie.: by replacing m by µ in central force formula involving 2 bodies you reduce the problem to that of one hypothetical body with exactly the same physics as in single particle motion (equation (c)). Equation (d) can be proved by considering the rotation of 2 m bodies about a common centre of mass, connected by a 'thin' rod 2 of length R. Rotation Axis The moment of inertia of this system about an axis passing through the centre of mass, and perpendicular to the plane m1 containing the masses is (by definition): r 2 2 2 2 I = m1r1 + m2r2 = Σmiri (e) r 1 R But from moments about centre of mass: m1r1 = m2r2 (by definition) (Σmiri = 0) 2 2 ∴(m1 + m2) I = (m1 + m2) m1r1 + (m1 + m2) m2r2 2 2 2 2 2 = m1m2r1 + m1 r1 + m1m2r2 + m2r2 2 2 = m1m2r1 + m1m2r1r2 + m1m2r2 + m1m2r1r2 2 2 = m1m2 (r1 + 2r1r2 + r2 ) 2 = m1m2 (r1 + r2) m m 1 2 2 2 ∴I = (r1 + r2) = µR (f) m1 + m2 2 From the definition of angular momentum: L = Σmiviri = Σmi(riω)ri = Σmiri ω Substituting in equations (e) and (f): L = Iω = µR2ω for 2 body systems. However as v = Rω then L = µR2ω = µR x Rω = µRv (i.e.: equation (d)). Instead of 2 bodies rotating about a centre of mass, all the relevant physics is contained in the equation of motion of a hypothetical particle of mass µ. .
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