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 —