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PHYS 705: Classical Mechanics Kepler Problem: Geometry of Kepler Orbits 2

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PHYS 705: Classical Mechanics Kepler Problem: Geometry of Kepler Orbits 2 1 PHYS 705: Classical Mechanics Kepler Problem: Geometry of Kepler Orbits 2 Focus-Directrix Formulation In the following, we will study the geometry of the Kepler orbits by considering the locus of points described by the Focus- Directrix formulation. directrix - Pick a focus P PD - Pick a directrix – a vertical line a PF d distance d away - Then, consider the set of points focus {P} that satisfy, PF PD PF = distance from P to focus is the eccentricity PD = distance from P to directrix 3 Focus-Directrix Formulation Now, express this in polar coordinate with the focus as the origin: directix PF = r P PD = d – r cos q r d r cosq O q Then, the condition PF PD gives, focus r dr cos q Solving for r, we have, Comparing with our previous Kepler orbit eq, r d r cos q r d 1 cos q r 1 cos q They are the same with d or d 4 Focus-Directrix Formulation: Hyperbola Recall the physical parameters for the orbit: l 2 2El 2 1 2 mk mk directrix P (Note: We have 0 for physical orbits.) r q Case 1: E 0 1 max The directrix is at d as indicated O d d d We have, r 1q cos 1 cos q Since 1, 1 1 , the denominator can be Note: ,d 0 zero for some angle qmax < p r can increase hyperbola without bound @ qmax This is a hyperbola ! line at focus 5 Focus-Directrix Formulation: Parabola 2El 2 Case 2: E 0 1 1 mk 2 The directix is at d as indicated directrix P d d Again, we have, r r 1q cos 1 cos q O Once again, the denominator can be zero d at qmax = p r can increase without bound at @ p This is a parabola ! 6 Focus-Directrix Formulation: Circular Orbit Case 3: E 0 There are three different sub-cases here. 2El 2 V ' recall 1 mk 2 mk 2 case 3a: E is imaginary r 2l 2 No orbits ! E mk 2 case 3b: E 0 2l 2 V ' l 2 r 0 r 1 cos q mk 0 r A circular orbit E 7 Focus-Directrix Formulation: Circular Orbit For circular orbits, we can combine the equations for E and r0 into one relation: mk 2 l 2 k E E 2 + r0 2l mk 2r0 We can also get this from the Virial Theorem: 1 1 k k T V (at circular orbit r r 0 ) 2 2r0 2 r 0 kk k ETV TV 2rr0 0 2 r 0 8 Focus-Directrix Formulation: Ellipse mk 2 2El 2 case 3c: E 0 0 1 recall 1 2 2l 2 mk directrix The directrix is at d as indicated P d r Again, we have, r 1 cos q O d Since 0 1, 1 1 , the denominator cannot be zero and r is bounded. An ellipse O is at one of the focii 9 Some Common Terminology - Pericenter: the point of closest approach of mass 2 wrt mass 1 m2 Perigee refers to orbits around the earth Perihelion refers to orbits around the sun m1 - Apocenter: the point of farthest excursion of apocenter pericenter mass 2 wrt mass 1 (with m1 fixed in space at Apogee refers to orbits around the earth one of the foci) aphelion refers to orbits around the sun 10 Perihelion and aphelion of Planets in Solar System Inner Planets Outer Planets 11 Focus-Directrix Formulation: Ellipse Now, we are going to look at the elliptic case 0 1 closer: (trying to relate geometry to physical parameters) a - The origin is with m1 fixed in space at one b rmin of the foci of the ellipse m1 - The extremal distances of the reduced mass O rmax (or m2) away from O (or m1) are given by: d d rmin perihelion vq largest 1q cosq 0 1 1 d d aphelion v smallest rmax q 1q cosq p 1 1 12 Focus-Directrix Formulation: Ellipse - The semi-major axis a: rmax r min 1 a 2 2 1 1 a 1 1 b rmin a 2 1 2 1 2 O rmax Now, putting in the physical constants, we have, l 2 1 a 2 2 1 mk 2El Thus, the semi-major axis 1 1 2 mk depends on E and k only ! l2 mk 2 k k 2 a mk2 El 2 E 2E 13 Focus-Directrix Formulation: Ellipse Goldstein does this differently: 1 l2 k Start with the energy equation: E mr 2 2 2mr2 r l2 k At the apsides, r 0 , so E 0 (@ apsides) 2mr2 r Write it as a quadratic equation in r: k l 2 r2 r 0 (this is an equation for the apsides) E2 mE Calling the two solutions for the apsides as: r 1 and r 2, we can write: 2 Recall E 0 r rrr1 2 rr 1 2 0 so a 0 Comparing the two eqs, one immediately gets: k k k r1 r 2 2 a a Ea r1 r 2 2 E 2E 14 Focus-Directrix Formulation: Ellipse Side Notes: The fact that a is a function of E only is an important assumption in the Bohr’s model for the H-atom. k k a agrees with r in the limit for circular orbits. 2E 0 2E Eccentricity can be expressed in terms of a: 2El 2 k Recall 1 , substitute E mk 2 2a 2l2 k l 2 1 2 1 mk 2 a mak (note: only true for ellipses and circles) or l2 mk a 1 2 15 Focus-Directrix Formulation: Ellipse Side Notes: The two focii are located at c a To show, start with: a r r c max min 2 rmax rmin Substituting r , r in, we have, max min c c rmax r min 1 c 2 2 2 1 1 1 From previous page, we have a 1 2 This gives our result: c a 1 2 16 Focus-Directrix Formulation: Ellipse Side Notes: The semi-minor axis b for an ellipse can be expressed as b a 1 2 Recall that if one draws a line from b a one focus to any point p on the p ellipse and back to the other focus, a a the length is fixed so that L 2 a (blue line) p But, we can also calculate it as, a b 2 2 L2 a 2 b a a a b a 1 2 17 Focus-Directrix Formulation: Summary Summary on conic sections for the Kepler orbits: 1E 0 hyperbola 1E 0 parabola 0 1E 0 ellipse k mk 2 E 0 E circle 2a 2l 2 r 1 cos q l 2 2El 2 1 mk mk 2 18 Focus-Directrix Formulation: Summary Kepler orbits with different energies E but with the same angular momentum l 19 Focus-Directrix Formulation: Summary In this picture, the directrix is fixed at one location so that d is the SAME for all orbits. But, and are different. Recall that d with l E So, d being the same for diff E A typical family of conic sections means l must also be different. 20 Kepler’s 3rd Law: Period of an Elliptical Orbit rdq 1 dA 1 Recall that dA r 2 d q and r 2q r dq 2 dt 2 Combining this with the angular momentum equation: l mr 2q dA r2 l l 2m dt dA 2 dt2 mr 2 m l Integrating this dt over one round trip around the ellipse gives the period t : t 2m 2 m 2 m t dt dA A p ab 0 l l l Now, plug in our previous results for the semi-major and semi-minor axes: 2m 2 m tpab p a a 1 2 l l 21 Kepler’s 3rd Law: Period of an Elliptical Orbit Square both sides, we have: 2 24m 2 4 2 t2 pa 1 l l2 l 2 Recalla 12 and a 1 2 mk mk 4ml2 2 4p 2 m 2 2 3 2 3 rd tp2 a t a This is Kepler’s 3 Law l mk k (Note: Recall that m is the reduced mass m here and not m1,2 so that Kepler original statement for orbits of planets around the sun is only an approximation which is valid for m mm Sun mm Sun m and m k mGmm1 Gm 2 2 Sun Sun so the proportionality 4 pm k 4 p Gm Sun is approximately independent of m . 22 Motion in Time: Eccentric Anomaly Overview on how to analytically solve for the position of an elliptic orbit as a function of time: (can also be done for hyperbolic/parabolic orbits – hw) (with the advent of computers, numerical methods replace this as the norm) Recall from our EOM derivations, we get the following from E conservation: 2 l 2 r EVr ( ) 2 m2 mr Inverting r and t, we get: r 2 k l 2 k t dr E 2 V( r ) m r2 mr r r0 23 Motion in Time: True & Eccentric Anomaly Our plan is to rewrite the equation in terms of eccentric anomaly defined by: r a 1 cos (basically, a change of variable r ) As we will see, this can simplify the (analytical) integration for the EOM. Before we continue onto the EOM, let get more familiar with this Recall that we have our Kepler orbit equation in polar coordinate in the reduced mass frame: r (choosing q ‘ = 0 at perihelion) 1 cos q Historically, q is called the true anomaly and it can be shown that q1 tan tan 2 1 2 24 Motion in Time: True & Eccentric Anomaly To derive this relation, we equate the two expressions for r: a 1 cos 1 cos q Recall for an ellipse, the semi-major axis can be written as: a 1 2 1 cos 1 2 1 cos q 1 2 1 cos q 1 cos 1 1 2 1 1 2 1 cos cosq 1 1 cos 1 cos cos cosq 1 cos 25 Motion in Time: True & Eccentric Anomaly Now, evaluate the following two quantities, cos 1(1cos )co s cos 1 1cos 1 cosq 1 1 cos1 1c cosos 1 co s cos1 (1cos )co s cos 1 1 cos 1 cosq 1 1 cos1 1c cosos 1 co s Then , we divide these two expressions: 1cosq 1cos 2 qq 2 sin 2 2 2sin 2 q 2 LHS: tan2 q 2 1 cosq 1 cos2 qq 2 sin 2 2 2cos 2 q 2 1 1 cos 1 2 RHS: tan 1 1 cos 1 2 26 Motion in Time: True & Eccentric Anomaly Putting LHS = RHS and taking the square root, we then have, 2q1 2 q 1 tan tanor tan tan 2 1 2 2 1 2 Going through one cycle around the orbit, we have these relations: Starting at perihelion r min a 1 , q0 0 Reaching aphelion r max a 1 , qp p Going back to perihelion r min a 1 , qp2 p 2 27 True & Eccentric Anomaly: Geometry The true and eccentric anomaly are related geometrically as follows: m line thrum (reduced mass) r q aphelion C O perihelion orbit ellipse bounding circle (O is at one of the focii of the orbit ellipse) (C is the center of bounding circle) 28 Motion in Time: Eccentric Anomaly Ok.
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