Mechanics Kepler Problem, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Kepler Problem, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity! Mechanics Physics 151 Lecture 6 Kepler Problem (Chapter 3) What We Did Last Time Discussed energy conservation Defined energy function h Conserved if Conditions for h = E Started discussing Central Force Problems Reduced 2-body problem into central force problem Problem is reduced to one equation Used angular momentum conservation Energy conservation gives Now we must solve this 2 3 ( ) lmr f r mr = + 0L t∂ ∂ = 2 2 2 1 ( ) const 2 2 m lE r V r mr = + + = Inverse-Square Force Consider an attractive 1/r2 force Gravity or electrostatic force 1/r2 force dominates at large r Centrifugal force dominates at small r A dip forms in the middle r ( )V r′ 2( ) kf r r = − ( ) kV r r = − 2 2( ) 2 k lV r r mr ′ = − + k r − 2 22 l mr Unbounded Motion Take V’ similar to 1/r2 case Only general features are relevant E = E1 r > rmin Particle can go infinitely far r ( )V r′ 1E 2E 3E 21 2 mr Arrive from r = ∞ Turning point Go toward r = ∞ E V ′= 0r = A 1/r2 force would make a hyperbola 1 min( )E V r′= Mechanics Physics 151 Lecture 6 Kepler Problem (Chapter 3) What We Did Last Time Discussed energy conservation Defined energy function h Conserved if Conditions for h = E Started discussing Central Force Problems Reduced 2-body problem into central force problem Problem is reduced to one equation Used angular momentum conservation Energy conservation gives Now we must solve this 2 3 ( ) lmr f r mr = + 0L t∂ ∂ = 2 2 2 1 ( ) const 2 2 m lE r V r mr = + + = Inverse-Square Force Consider an attractive 1/r2 force Gravity or electrostatic force 1/r2 force dominates at large r Centrifugal force dominates at small r A dip forms in the middle r ( )V r′ 2( ) kf r r = − ( ) kV r r = − 2 2( ) 2 k lV r r mr ′ = − + k r − 2 22 l mr Unbounded Motion Take V’ similar to 1/r2 case Only general features are relevant E = E1 r > rmin Particle can go infinitely far r ( )V r′ 1E 2E 3E 21 2 mr Arrive from r = ∞ Turning point Go toward r = ∞ E V ′= 0r = A 1/r2 force would make a hyperbola 1 min( )E V r′= Bounded Motion E = E2 rmin < r < rmax Particle is confined between two circles r ( )V r′ 1E 2E 3E 21 2 mr Goes back and forth between two radii Orbit may or may not be closed. (This one isn’t) A 1/r2 force would make an ellipse Stable Circular Orbit Circular orbit occurs at the bottom of a dip of V’ Top of a bump works in theory, but it is unstable Initial condition must be exactly r stable r unstable E E 2 0 2 mr E V ′= − = 0 dVmr dr ′ = − = constr = 00 and r r r= = 0r Stable circular orbit requires 2 2 0 d V dr ′ > Orbit Equation We have been trying to solve r = r(t) and θ = θ(t) We are now interested in the shape of the orbit Switch from dt to dθ Switch from r to ( )r r θ= 2l mr θ= 2 d l d dt mr dθ = 2 3 0 l dVmr mr dr − + = 2 2 2 3 0 l d l dr l dV r d mr d mr drθ θ ⎛ ⎞ − + =⎜ ⎟ ⎝ ⎠ 2 1 1du d dr d d r r dθ θ θ ⎛ ⎞= = −⎜ ⎟ ⎝ ⎠ 1u r≡ 2d du dr du = − const Orbit Equation Solving this equation gives the shape of the orbit Not that it’s easy (How could it be?) Will do this for inverse-square force later One more useful knowledge can be extracted without solving the equation 2 1 2 2 ( ) 0udVd u mu d l duθ + + =Switching variables Solving Orbit Equation Integrating the diff eqn will give energy conservation One can use energy conservation to save effort Switch variables 2 1 2 2 ( ) 0udVd u mu d l duθ + + = 2 2 2 ( )2 2 mr lE V r mr = + + 2 2 2 ( ) 2 lr E V r m mr ⎛ ⎞ = − −⎜ ⎟ ⎝ ⎠ Integrate this… 1 2 2 2 2 ( )2 umVdu mE u d l lθ = − − − l dur m dθ = − Inverse Square Force Look it up in a math text book and find Just substitute α, β and γ Or… 2 2 2 2arccos 4 dx x x x β γ γα β γ β αγ ⎛ ⎞+⎜ ⎟= − ⎜ ⎟−+ + −⎝ ⎠ ∫ 2 kf r = − kV r = − 2 2 2 2 2du mE mku u d l lθ = − + − 2 2 22 2mE mku l l du d u θ= − + − ∫ ∫ Working It Out Yourself ( )2 22 2 2 4 2 2 2 2 4 2 2 2 2 4 2 22 2 2 22 2 1 1 sin sin mE mku mE m k mk l l l l l mE m k mkl l l mE m k l l du dud u u du u d θ ω ω ω ω = − = − + − + − − = − + ⎛ ⎞−⎜ ⎟− ⎜ ⎟+⎝ ⎠ = − = − ∫ ∫ ∫ ∫ ∫ Define as cosω 2 2 2 4 2 sinmE m k l l du dω ω= + 2 2 2 2 4 2 cos cos( ) mk l mE m k l l u ω θ θ − ′= − = + Solve this for u = 1/r e > 1 hyperbola θ’ is the turning point (perihelion) cos(θ – θ’) > –1/e limits θ e = 1 parabola Unbound Orbits ( )1 1 cos( )C e r θ θ ′= + − θ ′ θ θ ′−r θ ′ r Bound Orbits Ends of the major axis are Length of the major axis ( )1 1 cos( )C e r θ θ ′= + − θ ′ r ( )1/ 1r C e= ± 1 1 1 2 (1 ) (1 ) 2 ka C e C e E ⎛ ⎞ = + = −⎜ ⎟+ −⎝ ⎠ 2 2 21 Ele mk = + 2 mkC l = Major axis is given by the total energy E 2a 2b Minor axis is 2 21 2 lb a e mE = − = − Rotation Period We know that the areal velocity is constant Express τ in terms of a 2 ka E = − 2 2 lb mE = − 2 2 38 l kA ab mE π π= = − 21 2 2 dA lr dt m θ= = Area of the orbit Period of rotation 2 32 mkdAA dt E τ π= = − 3/ 22 ma k τ π= Period of rotation is proportional to the 3/2 power of the major axis Kepler’s Third Law of Planetary Motion