from a numerical integration of the solar...
TRANSCRIPT
From a numerical integration of the Solar System.
Perturbation Theory Purpose: calculate deviations from Keplerian 2-body
motion due to external influences. Examples: • Gravitational effects of a 3rd body (e.g., another planet or
satellite) • Non-spherical primary body (J2) • Radiation forces (e.g., light pressure on small particles) • EM forces (e.g., charged particles in a planet’s
magnetosphere) • Relativistic corrections (e.g., Mercury’s orbit)
Perturbation Theory (cont.): Approaches: at least 3 are commonly employed: 1. Calculate perturbing forces directly, resolve in radial,
azimuthal & normal direc?ons (R, B, N) and derive the effect of each on the orbital elements (a, e, etc.) • Simple, but inelegant & generally messy…
2. Write perturba?on in terms of a poten?al (the “Disturbing Func?on”, R) and derive the effect on the orbital elements.
• Conserva?ve forces only, but generally simpler and more direct. 3. Rewrite Kepler problem in terms of Hamiltonian (canonical
coordinates & momenta) and apply canonical perturba?on theory.
• Elegant & simple, but non-‐intui?ve.
Perturbation Theory: Lagrange’s Planetary Equations
(Danby pp 238-252)
Method 1: the perturbing force approach:
Method 2: the Disturbing Function approach:
Danby, p. 251
Note on Longitude Perturbations
The most troublesome orbital element is the 6th, the time of pericenter passage (T) or equivalently the longitude at t = 0. This is most often written as the “mean longitude at epoch”, ε:
! t( ) = " +# + M
$ !# + n t % T( ) = & + ntFrom Kepler's equation, E % esin E = M , we get & = !# + E % esin E % nt and when the orbit is perturbed
"& = !"# + 1% ecos E( ) "E % "esin E % n % "nt
Even if !"# , "E, "e, and "n remain small, the last term gets large as t '(.This is circumvented by introducing a different quantity, which depends on the history of the perturbation:
! t( ) $ &1 + n t( )0
t
) dt
so that "&1 satisfies the same equation, minus the "nt term. See Danby's text on Celestial Mechanics for more details.
Note that, since n2a3 = GM exactly for the unperturbed Kepler problem, we have
"nn $ % 32"aa( ).
Small e or i Perturbations
Note: See J. Burns, Amer. J. Phys. 44, 944 for a simplified derivation of the perturbation equations
Note that the 2nd and 3rd terms can often be Neglected if h1 and k1 are small.
Ditto for 2nd term here.
Satellite Orbits Around an Oblate Planet As an example of perturbation techniques, we calculate the effects of a planet’s oblateness on a
close satellite (natural or artificial). The planet’s gravitational potential can be expanded in a series of multipole terms; we keep only the first 2 terms in this expansion:
Where R is the planet’s equatorial radius, J2 is a dimensionless constant of order 10-3 to 10-2, and P2
is the 2nd Legendre polynomial: ( is the usual spherical polar co-ordinate.) The 2nd term in V is our perturbing potential, , which we must express in terms of a, e, i, … !
V r,!( ) = " µ
r 1" J2Rr( )2
P2 cos!( ){ } (17)
P2 cos!( ) = 1
23cos2! "1( ).
!
i.e.,! r,!( ) = "µJ2 R2
2r33cos2! "1( )
From geometry, we have cos! = sin isin # +$( ), so
!= "µJ2 R2
4r33sin2 i 1" cos2 # +$( )%& '( " 2{ } and
r "3 = a 1" e2( )%&
'("3
1+ ecos$( )3
Since the perturbation equations involve ) (i.e., mean longitude) rather than the true anomaly, $, it is necessary to expand the cosine
terms in terms of the mean anomaly, M = n t " T( ). We can further simplify the problem, however, by considering only the cumulative
perturbations over many orbits and neglecting the various short-period perturbations.
Accordingly, we average ! over one orbital period as follows:
! = "µJ2 R2
4i2*
3sin2 i 1" cos2 # +$( )%& '( " 2
r30
2*
+ dM
the integrals involved are
cos2$r3
dM0
2*
+ ,sin 2$
r3dM
0
2*
+ ,1r3
dM0
2*
+ ,
of which the 1st and 2nd are equal to zero (verification left as an exercise to the student). To evaluate the 3rd integral, we employ Kepler's 2nd law again:
1r3
dM0
2*
+ =1r3
d$dM
,-.
/01
"1
d$ = n 1r3"$( )"1
0
2*
+0
2*
+ d$ =nh
d$r0
2*
+ , since r 2 "$ = h
(Note that R is conventionally defined as the negative of the perturbing potential.)
!1r3
dM0
2"
# =n
ha 1$ e2( ) 1+ ecos%( )d% =2"
a3 1$ e2( )32
0
2"
#
!! = $µJ2 R2
2a3 1$ e2( )32
32sin2 i $1&' () (18)
Note the ! is a function of a,e, and i only, so Lagrange's Equations *
"a = "e = didt
= 0
i.e., there are no long-term perturbations in a, e or i. (There are short - term perturbations due to the parts of ! that averaged to zero over one orbit, however.) Applying Lagrange's Equation, we obtain
d+dt
= $ 32
n J2Ra( )2 cos i
1$ e2( ),-.
/.
01.
2. (19)
d #3dt
= 32
n J2Ra( )2
1$ 32 sin2 i $ 1
2sin2i tan i
2( )1$ e2( )2
,
-..
/..
0
1..
2..
(20)
and instead of d41
dt, we calculate
dMdt
=ddt
41 + n dt $ #3#( ) =
d41
dt$
d #3dt
+ n
= n 1+ 32
J2Ra( )2 1$ 3
2 sin2 i
1$ e2( )32
,
-.
/.
0
1.
2.
,
-.
/..
0
1.
2..
(21)
Note that all factors enclosed in { }
are ! 1 for small e and small i.
These results have considerable practical significance. Since J2 > 0 for any planet
flattened by rotation, we see that
(i) the line of nodes (i.e., !) regresses for prograde orbits, except for i = 90o , so the plane of the obit rotates backwards in inertial space, relative to a fixed direction.
(ii) for all i less than some critical value (46.o5), the line of apsides (i.e., !" ) advances, so the pericenter rotates forwards around the orbit, relative to a fixed direction.
(iii) for all i < sin#1 23= 54.o7, the mean anomaly increases at a rate greater than n,
the Keplerian mean motion; i.e., the particle's orbital period (pericenter $ pericenter) is less than that of a Keplerian orbit about a spherical planet of the same mass, M .
(i) and (ii) provide the means to determine J2 for planets with satellites in eccentric and/or
inclined orbits (E, Mars, J, S, U, N), as well as make possible the orbits of sun-synchronous Earth satellites such as LANDSAT. (iii) means that one must be careful in calculating planetary masses from observed values of a and "M for satellites.
In physical terms, (i) arises because an oblate planet exerts a torque on an inclined orbit, which causesits angular momentum vector to precess (see later lectures on planetary shapes). Items (ii) and (iii) are less obvious, but involve the breakdown of spherical symmetry in the Kepler problem.
{Nodal and apsidal precession
For i << !2
and e << 1, we have
!"" # 32
n J2Ra( )2
= # "$
and
"M # n 1+ 32
J2Ra( )2{ }
% the sidereal average mean motion is "& = n + "'1
= "M + !""
# n 1+ 3J2Ra( )2{ }
( nG , the observed mean motion,
where
n ( GMa3 is the osculating, Keplerian
mean motion.
The osculating semi-major axis, a, is determined by the actual orbital energy:
EK = 12
nG2 aG
2 ! GMaG
where aG is the observed mean radius.
" ! GM2a
# GM2a3 1+ 6 J2
Ra( )2{ }aG
2 ! GMaG! ! GM
2a
12
aG
a( )3
1+ 6 J2Ra( )2{ } + 1
2
aG
a = 1
writing aG
a = 1+ $ , we have 1+ 3$ + 6 J2Ra( )2
+1+ $ ! 2 ! 0
#$ ! ! 32
J2Ra( )2
i.e., aG ! a 1! 32
J2Ra( )2{ }
ASIDE: osculating vs. “geometric” elements
or
The osculating eccentricity is determined by noting that, for eG = 0, the particle must be at pericenter
(since nG > n) :
!aG = a 1" e( ) !e ! 3
2J2
Ra( )2
for eG = 0
Note that nG and aG do NOT satisfy Kepler's 3rd law:
nG2 aG
3 ! 1+ 6 J2Ra( )2
" 92
J2Ra( )2{ }
= GM 1+ 32
J2Ra( )2{ } # constant
This is due, of course, to the additional radial gravity provided by J2 :
V r( ) = " GMr 1+ 1
2J2
Ra( )2{ } @ $ = %
2 so
"&V&r
= " GMr2 " 3GM J2 R2
2 R4 = "nG2 r
!nG2 r3 = GM 1+ 3
2J2
Ra( )2( )… as above
If J2 were to be abruptly "turned off", the particle would be travelling faster than the circular
velocity at r = aG , and thus would be at the pericenter of an elliptical orbit with a > aG . This
is the true meaning of the osculating orbit.