Celestial Mechanics II
Orbital energy and angular momentumElliptic, parabolic and hyperbolic orbits
Position in the orbit versus time
Orbital Energy
KINETIC POTENTIAL
per unit mass
The orbital energy – the sum ofkinetic and potential energiesper unit mass – is a constant inthe two-body problem (integral)
The Vis-viva Law
Vis-viva law, continued
This is a form of theenergy integral, wherethe velocity is given asa function of distanceand orbital semi-majoraxis.
VERY USEFUL!
Geometry versus Energy
E<0 ⇒ 0 ≤ e < 1: Ellipse
E=0 ⇒ e = 1: Parabola
E>0 ⇒ e > 1: Hyperbola
Orbital shape determinedby orbital energy!
Special velocities
Circular orbit (r ≡ a):
Parabolic orbit (a → ∞):
!
vc
2 =µ
r
!
ve
2 =2µ
r
Circular velocity
Escape velocity
!
ve
= 2 " vc
Note: The parabolic orbit is a borderline case thatseparates entirely different motions: periodic (elliptic)and unperiodic (hyperbolic). The slightest change ofvelocity around ve may have an enormous effect (chaos)
Geometry vs Angular Momentum
⇒
The semilatus rectum p is positive for all kinds of orbits(note that a is negative for hyperbolic orbits). It is awell-defined geometric entity in all cases.!
h = µp
In terms of the perihelion distance q, we have:
!
p > 2q
!
p = 2q
!
p < 2q
ellipse parabola hyperbola
Orbital position versus time:The choice of units
Gravitational constant:
SI units ([m],[kg],[s]) G = 6.67259 10-11 m3kg-1s-2
Gaussian units ([AU],[M],[days]) k = 0.01720209895 AU3/2M
-1/2days-1
Kepler III:
m1 = 1; m2 = 1/354710 (Earth+Moon)P = 365.2563835 (sidereal year); a = 1
k2 replaces G
Orbital position versus time:Why we need it
• Ephemeris calculation – ephemeris = table of positions in the
sky (and, possibly, velocities) at giventimes, preparing for observations
• Starting of orbit integration – this means integrating the equation of
motion, which involves positions andvelocities
Orbital Elements
• These are geometric parameters that fullydescribe an orbit and an object’s position andvelocity in it
• Since the position and velocity vectors havethree components each, there has to be sixorbital elements
• Here we will only be concerned with three ofthem (for the case of an ellipse, the semi-major axis, the eccentricity, and the time ofperihelion passage)
“Anomalies”These are angles used to describe the location of anobject within its orbit
ν = true anomaly
E = eccentric anomaly
M = mean anomaly
M increases linearly with time and can thus easily becalculated, but how can we obtain the others?
Introducing the eccentric anomaly
Express this in terms ofE and dE/dt
Differential equation for E
Integration
Mean motion: n
Mean anomaly: M
Kepler’s Equation
Finding (position, velocity) for a given orbit at a given time:Solve for E with given M
Finding the orbit from (position, velocity) at a given time:Solve for M with given E
(much easier)
Position vs time in elliptic orbit
• Calculate mean motion n from a (via P)• Calculate mean anomaly M• Solve Kepler’s equation to obtain the
eccentric anomaly E• Use E to evaluate position and velocity
components
Solving Kepler’s equation
Iterative substitution:
(simple but with slow convergence)
Newton-Raphson method:
(more complicated but withquicker convergence)
Handling difficult cases, 1
Solving Kepler’s equation for nearly parabolic orbits isparticularly difficult, especially near perihelion, sinceE varies extremely quickly with M
Trick 1: the starting value of E
Under normal circumstances, we can take E0 = M
But for e ≈ 1, we develop sin E in a Taylor series:
!
M = E " esinE # E " sinE =1
6E3
+K and hence:
!
E0
= 6M( )1/ 3
Handling difficult cases, 2
Trick 2: higher-order Newton-Raphson
Taylor series development of M:
With
!
M0
= M E0( ) and
!
"E0
=M #M
0
dM /dE( )0
we get:
!
"E =M #M
0
dM
dE
$
% &
'
( ) 0
+1
2
d2M
dE2
$
% &
'
( ) 0
"E0
+1
6
d3M
dE3
$
% &
'
( ) 0
"E0( )2
+K
!
M = M0
+dM
dE
"
# $
%
& ' 0
(E +1
2
d2M
dE2
"
# $
%
& ' 0
(E( )2
+1
6
d3M
dE3
"
# $
%
& ' 0
(E( )3
+K
Two useful formulae
⇓ ⇓
Parabolic orbits
a→∞ and e=1, but p=a(1-e2)=h2/µ remains finite
!
p = 2q ; h = 2µq
Kepler III loses its meaningNo center, no eccentric anomalyNo mean anomaly or mean motion, but we will find substitutes
Parabolic eccentric anomaly NIntroduce:
!
N = tan"
2
From the angular momentum equation:
!
r2˙ " = h
we get:
!
1+ N 2( )2
˙ " =h
q2
=2µ
q3
But:
!
dN = 1+ N 2( )d"
2Thus:
!
1+ N 2( )dN =µ
2q3dt
and:
N plays a role similar to E in Kepler’s equation
Mean motion, mean anomalyIntroduce the parabolic mean motion:
!
n =µ
2q3
and the parabolic mean anomaly:
!
M = n(t "T)
We get:
!
M = N +1
3N3
Cubic equation for N(M), called Barker’s Equation
Solving Barker’s equationNumerically, it can be done by simple Newton-Raphsonusing N0 = M
Analytically, there is a method involving auxiliary variables:
Position vs time in parabolic orbit
• Calculate parabolic mean motion n from q• Calculate parabolic mean anomaly M• Solve Barker’s equation to obtain the
parabolic eccentric anomaly N = tan(ν/2)• Use N to evaluate position and velocity
components
The Hyperbola
Semi-major axisnegative for hyperbolas!!
q = a(1" e)
!
p = a(1" e2)
!
cos"# = $1
e
Hyperbolic deflectionImpact parameter: B (minimum distance along the straight line)
Velocity at infinity: V∞ Angle of deflection: ψ = π - 2φ
!
" tan#
2=
µ
V$h=
µ
V$2B
Gravitational focusing: q < B
Hyperbolic eccentric anomaly H
Differential equation for H
Hyperbolic Kepler Equation
Hyperbolic mean motion:
Hyperbolic mean anomaly:
Solving the hyperbolic Keplerequation
• The methods are the same as for theelliptic, usual Kepler equation
• The same tricks can be applied fornearly parabolic orbits
• The derivatives of trigonometricfunctions are replaced by those ofhyperbolic functions
Position vs time in hyperbolic orbit
• Calculate hyperbolic mean motion nfrom |a|
• Calculate hyperbolic mean anomaly M• Solve the hyperbolic Kepler equation to
obtain the eccentric anomaly H• Use H to evaluate position and velocity
components
Examples• q=2.08 AU
• Ellipse: a=2.48, e=0.16
• Hyperbola: a=-1.398 AU, e=2.5
ο positions 253 days afterperihelion
Ellipse: v=22.3 km/sParabola: v=29.3 km/sHyperbola: v=38.7 km/s
Velocity at perihelion —