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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. Now, coming back to the time evolution of the orbit: r 2 k l 2 t dr E 2 m r2 mr r0 Substituting the relations (on the right) into k l 2 E, a 1 2 the denominators of the above eq, we have: 2a mk
1/2 2 2kl 2 kkk 2 E2 2 a 1 m rmr2 m 2 ar 2 r
1/2 2 1 2k r 2 a 1 r r m2 a 2 29
Motion in Time: Eccentric Anomaly
Putting this back into the time integral, we have, r 2 m r 2 a1 t rdr r (this is Eq. 3.69 in Goldstein) 2k 2a 2 r0 Now, we substitute the eccentric anomaly into above:
By convention, we pick starting point r0 at perihelion 0 q 0 r Then, limit of integration becomes: dr d ro o Now, we need to do some more algebra to transform the integrant: First, the change of variable r a 1 cos gives, dr asin d and rdr a21 cos sin d 30
Motion in Time: Eccentric Anomaly
Now, we need to transform the square-root term: 1/2 2 2 2 a 1 cos2 a 1 a 1 cos r 2 a 1 r 2a 2 2a 2 1/2 2a21 cos a 2 1 cos 2 a2 1 2 2a 1/2 a2 1 cos 2 1 cos a2 a 2 2 2a 1/2 a21 cos 1 cos a 2 a 2 2 2a 31
Motion in Time: Eccentric Anomaly 1/2 a2 1 2cos 2 a 2 a2 2 2a 1/2 a 21 cos 2 2 Recall, a2 a sin sin 2 2 2 rdr a1 cos sin d Now, putting all the pieces back together, we have, r 2 m rdr m 2 a 1 cos ' sin 'd ' t 2k 2 k a sin ' r0 0 ma3 t1 cos ' d ' k 0 32
Motion in Time: Eccentric Anomaly
Carry through the integration, we have,
ma3 ma 3 (Note the dramatic t1 cos ' d ' sin simplification using ) k0 k
Integrating this over ONE period of the elliptic orbit, ma3 i.e., taking from 0 to 2p, we have: t 2 p k Again, we have Kepler’s 3rd law: 4p 2m t 2 a 3 k 33
Motion in Time: Eccentric Anomaly t Defining the mean anomaly M as: M 2 p (M is observable) t We arrive at the Kepler’s Equation:
2p k ma3 M t 2p sin t 4p 2 ma3 k
M sin
Standard (non-numeric) procedure in solving celestial orbit equation:
1. Solve Kepler’s equation (t) [transcendental] 2. Use the q transformation to get back to the true anomaly q t . 34
Hohmann Transfer
The most economical method (minimum total energy expenditure) of changing among circular orbits in a Kepler system such as the Solar system.
Simplifying assumptions: • Orbits are heliocentric (orbits with the Sun at the center of force) e.g. Orbit near Earth Orbit near Mars • Orbits are on the same orbital plane of the Sun
• Msun m satellite • Ignoring gravitational effects from other Planets • Thrusts (two) from rockets change only eccentricity and energy of an orbit BUT not the direction of angular momentum 35
Hohmann Transfer: Earth to Mars v m
1. Start from Earth’s “circular” orbit
2. Kick to go on transfer orbit (elliptic – rm
thick line) with new speed vnew 3. Kick to slow down when arrived at
Mars “circular” orbit with speed vm re (Both E’s and M’s orbits are close to
v new circular: Earth 0.0167, Mars 0.0934 )
Note: 1. Both kicks are done at points of tangency (@ perihelion and aphelion of the ellipse) so that transfer can be accomplished with speed-change only. 2. Both (speed) kicks won’t change the direction of L (stay on the same orbital plane). 3. Need to time operations so that Mars will actually be there at arrival. 36
Hohmann Transfer: Earth to Mars vm
Let calculate the required speed changes:
- For both circular and elliptic orbits, rm k E (a semi-major axis) 2a
- On Earth’s orbit, we have
k re E 2re 1 k vnew - Recall also, E mv 2 2 e r e k - Combining with above gives, v e This is the Earth’s orbital speed and mre the satellite will start with it as well. 37
Hohmann Transfer: Earth to Mars vm
Now, think of the position as the pericenter of the elliptic transfer orbit:
rm - The ellipse has its semi-major axis: r r a e m 2 - So, for the satellite to be on this orbit, it needs to have energy: re k1 k E mv2 new vnew rrem 2 r e
- Solve for v new ,
2 2k 1 1 2 k rm 2k rm vnew vnew mreem rr mrrr eem mre r e r m 38
Hohmann Transfer: Earth to Mars vm
- So, to get the satellite onto the elliptic transfer orbit, we need to give a “kick” to r speed it up (no change in direction) m
vv1 new v earth
2krm k v1 mrreem r mr e re - When satellite arrives at the aphelion of the transfer orbit (Mars’ circular orbit), thruster needs vnew to fire to slow down to get onto Mars’s circular orbit,
k1 2 k k E mvm (on Mars’s orbit) v v 2r 2 r new m m m mrm 39
Slingshot Effect (qualitative discussion)
Let consider a satellite swinging by a Planet (hyperbolic orbit) in a frame co- moving with Planet M.
v f Note: the hyperbolic orbit is symmetric with respect to the apside magnitude of v doesn’t change, (before and after) only its direction. M
v
vi vi v f 40
Slingshot Effect (qualitative discussion)
Now, if we consider the same situation in a “stationary” inertial reference frame (fixed wrt the Sun) in which the Planet M is moving to the right.
v ' is the velocity of the satellite v v 'i i f vi in the “stationary” frame.
vM M v ' f v ' f is the velocity of the satellite v f in the “stationary” frame.
vM vi
Note that v ' f has a larger magnitude than v 'i (with a boost in the direction of Planet M). 41
ISEE-3 Fly-by (1982-1985)
- ISEE-3 (International Sun-Earth Explorer 3) was originally parked in an orbit at the L1 Lagrange point to observe the Sun and the Earth. - 1982: NASA decided to reprogram it so that it will go to explore the Giacobini- Zinner comet scheduled to visit the inner Planets in September of 1985. - Design an orbit to get it from its parked orbit to the coming comet: cheaper than to design, build, and lunch a new satellite.
Some details: First burn v 10 mph A total of 37 burns Two close trips back to Earth and five flybys of the moon. One pass within 75 miles of the Luna surface Resulted in 20 min trip through the comet tail on 9/11/85. 42
ISEE-3 Fly-by (1982-1985)
(Lagrange Point)