Download Lecture Notes on Central Force - Classical Mechanics | PHYS 3210 and more Study notes Mechanics in PDF only on Docsity! PHYS 321 Lecture #8, 09/13/2002 Central Forces Conservation of Angular Momentum d r F dt = × r rl r (= 0 for central forces; ≠ 0 for others) This is basically Kepler’s 2nd Law (equal areas in equal times). Newton’s 2nd Law: 2 2 d r m F dt = r r Change to spherical polar coordinates sin cosx r= θ ϕ sin siny r= θ ϕ cosz r= θ Express Newton’s 2nd in spherical coordinates: all are time dependent. Example: sin cos cos cos sin sinx r r r= θ ϕ + θθ ϕ − θ ϕϕ& && & If we do this for y& and z& , and add all three (squared) components, we get 2 2 2 2 2 2 2 2 2 2sinv x y z r r r= + + = + θ + θϕr & && & & & We can do this another way: let us represent the change in the radius vector in spherical coordinates as ˆ ˆ ˆsinrdr dre rd e r d eθ ϕ= + θ + θ ϕ r This defines the orthogonal system ˆ ˆ ˆ, ,re e eθ ϕ where ˆ ˆ ˆre e eθ ϕ× = , etc. We find that ˆ ˆ ˆsinr dr v re r e r e dt θ ϕ = = + θ + θϕ rr & && , giving the same thing for 2v r . If we try to differentiate vr (in polar coordinates) to get the acceleration, we have to do so in two places: (1) , ,r θ ϕ for a moving object depend on time; (2) The unit vectors ˆ ˆ ˆ, ,re e eθ ϕ rotate as they move to different points in space, so they, too, depend on time. So we skip this hard problem and work instead with the (conserved) energy. For central forces, ( ) ( )r̂F e f r V r= = −∇ r ; the ability to express F r as a gradient depends on its form, since ( ) ˆ ˆ ˆ r̂ x y z dV dV V r x y z e r r r dr dr ∇ = + + ≡ . We can therefore integrate Newton’s 2nd Law with the integrating factor dr v dt = rr to get ( )21 2 const.E mv V r= + = Rewriting this ( )2 2 2 2 2 21 2 ( sin )m r r r V r E+ θ + θϕ + =& && ( )( )sin sin const.z m r r= θ θϕ =&l 2 2sin const.r θ ϕ =& 2 2 2 4 2 const.x y m r⇒ + = θ =&l l Suppose we let 0ϕ =& at some initial time; then the equation of motion implies 0 0, ,...ϕ = ϕ =&& &&& Therefore 0ϕ =& for all time. If we choose 0=ϕ& then 2 2 2sinr θϕ& = 0 in the energy equation, and we obtain 2 2 2 4m r θ = l& . Substituting back, we have ( ) ( ) 2 2 2 2 2 2 2 1 1 2 2 ( ) ( )m r r V r m r V r E m r + θ + = + + = l&& & . Which we can solve for r& : ( ) 2 2 2 2 E Vdr r dt m m r − = = ± − l& . The equation of an orbit is determined by the function ( )r θ . We can eliminate time by dividing r& by θ& , to get ( )2 2 2 2 2 ( ) E Vdr r mr F r d m m r − ≡ = ± − = θ θ & l & l . That is, in general the orbit equation can be found by quadrature:
