The Perturbed 2-Body Problem And The Role Of The Delaunay Variables In Solutions

 

                             Perturbation diagram to for Delaunay variable analysis

The perturbed two-body problem begins by mathematically defining the perturbation in terms of the quantities in the triangle above. This actually shows  how the presence of a third body (say m ₃  in the accompanying diagram) perturbs or disturbs the motion of m ₂. which is in orbit about a central mass m ₁. (And of course: m ₁  >>  m ₂  )  Then we can write for the perturbation (N.B.  m  is the reduced mass, m =   1/ (1/m ₂   + 1/ m ₃):

r'' =  -  m r  / r³   -   G m ₃  ((r -  r ₃  /  D ³  ) +   r ₃   /   r ₃  ³

=    -  m r  / r³   -  Ñ R

Where:   Ñ R  =   G m ₃  (r -  r₃  /  D ³  ) +   r ₃   /   r ₃  ³

And:  R  =   - G m ₃  (1/  D    +   r ₂ ·  r ₃   /   r ₃  ³)

Then, adding the perturbations we get:

r'' = -  m r  / r³   -  Ñ R  =  Ö( x'' ²  +  y'' ²  +  z'' ²  )

For which a set of coordinates: q _1,2,3   =  x, y, z are defined, and a set of momenta p _1,2,3   =  x', y', z'    associated with them.  Hence:

dq _i  / dt  = ¶ H  / ¶ p _i        and:     dp _i  / dt  = -¶ H  / ¶ q _i 

Leading to the Hamiltonian:  

H  =   ½ å ³ _i=1  P _i ²    -  m / r  -  R

If this form exists, then a, e, i  etc. must be functions of the time and also further computations may be needed, i.e.

But instead of functions with a, e, i one can make use of the Delaunay variables: L', G', H', respectively. Where:

L'  =  - ¶ H  / ¶ ℓ 

G' =  -   ¶ H  / ¶ g 

H’  =  -   ¶ H  / ¶ h

These Delaunay variables are defined by the equations:

ℓ  =   n(t  - T)

L =  Ö(m a)

G  =   L    Ö (1 -   e ²)

H =   G cos i

g   =  w

h  =   Ω

Where a is the orbit's semi-major axis,  Ω  is the longitude of the ascending node,  is the arguments of the perihelion, e is the eccentricity of the orbit and i is the inclination of orbit to the ecliptic plane.

In this case, the previous Hamiltonian: 

H  =   ½ å ³ _i=1  P _i ²    -  m / r  -  R

Transforms to:

H  (L,  ℓ  )  =   -  m ² / 2 L ²  -  k ²  m ₃    {1/  D (L,  ℓ  )  +   r ·  r ₃   /   r ₃  ² }

 A huge simplification arrives  if R = 0., which then reduces to the 2-body problem, which we've examined in previous blog posts.   Note that R in the second Hamiltonian is expressed:

R   =   k ²  m ₃    {1/  D (L,  ℓ  )  +   r ·  r ₃   /   r ₃  ² }

Now, for the case at hand (consult diagram), we may write:

D ²     =  r  ²   +   r ₃  ²    -   2 r ·  r ₃  cos S

 And:  r ·  r ₃  =   r   r ₃  cos S

Assume now that r ₃  is greater than r (perturbing an inner planet or body by an outer one):

1 /D      =  1/r [ 1 + (r / r ₃)  ²  -  2 (r / r ₃)  cos S] ^1/2

Then the above can be rewritten as the sum:

½ å ^¥ _{r= 0}  (r / r ₃) ⁱ  P _i 

=   1/  r ₃  [ P _o   +   (r / r ₃)  P ₁   + (r / r ₃)  ²  P ₂    +  ...]

The P _i   are functions of the angle S and are called Legendre polynomials.  In this case, the first three may be written:

P _o   =   1,      P ₁     =   cos S,    P ₂     =   ½ (3  cos ² S   -   1) 

Suggested Problem:

  Estimate the magnitude of the error in the modified Hamiltonian H  in terms of k and   m ₃.  (You can take  m ₃ =  mass of Jupiter,  m ₂ = mass  of Earth, and a ₃ =  semi-major axis of Jupiter).   Make sure your reference Hamiltonian conforms with these.