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16.346 Astrodynamics
Fall 2008
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Lecture 24 Basic Elements of the Three Body Problem
The Rotation Matrix # 2.1
Let i , i , i and i
ξ
, i
η
, i
ζ be two sets of orthogonal unit vectors. i
ξ
= l
1 i + m
1 i + n
1 i i
η
= l
2 i + m
2 i + n
2 i i
ζ
= l
3 i + m
3 i + n
3 i where l
1
, m
1
, . . . , n
3 are called direction cosines.
Any vector r can be expressed as r = x i x
+ y i y
+ z i z
⎡ i
R =
⎣ i i
· i
ξ
· i
ξ
· i
ξ i i i
· i
η
· i
η
· i
η i i i
· i
ζ
· i
ζ
· i
ζ
=
⎤ ⎡
⎦
= l
⎣ m
1
1 n
1
ξ l m n i
ξ
2
2
2
+ η i
η
+ ζ i
ζ
. Then l n
3 m
3
3
⎤
⎦ is the rotation matrix and
⎡ ⎤
⎣ x y
⎦
⎡
ξ
= R
⎣
η
⎤
⎦ z ζ or
⎡
⎣
ξ
⎤
η
⎦
= R T
⎡
⎣ x y
⎤
⎦
ζ z
Since
RR T = R T R = I so that R T = R
−
1 then R is an orthogonal matrix.
Kinematics in Rotating Coordinates #2.5
Use an asterisk to distinguish a vector resolved along fixed axes from the same vector resolved along the rotating axes: r
∗
= Rr v
∗
= Rv a
∗
= Ra where r , v , a are the position, velocity, and acceleration vectors whose components are understood to be projections along the moving axes.
� � � v
∗
= d r
∗ d r dt dt dt
�
= R d r
+ Ωr = R + ω × r = Rv where Ω = R T d R dt
⎡
=
⎣
ω
0 − ω
ζ
− ω
ζ
η
ω
0 − ω
ξ
ω
η
0
ξ
⎤
⎦
ω =
⎡ ⎤
⎣
ω
ω
ξ
η
⎦
ω
ζ
Ω T = − Ω
The angular velocity vector ω is identified as the angular velocity of the moving coordinate system with respect to the fixed system.
16.346 Astrodynamics Lecture 24

For the acceleration vector a
∗
= d 2 dt
= R r
∗
2
� d 2 dt
= R
� d r
2
2 dt 2
� r d r d Ω
+ 2 Ω + r + ΩΩr
+ 2 ω × d r d ω dt
+ dt dt dt
× r + ω × ( ω × r
�
) = Ra
The four terms which comprise the acceleration referred to rotating axes are called the
Observed , the Coriolis , the Euler , and the Centripetal accelerations, respectively.
(The observed velocity and acceleration vectors d r /dt and d 2 r /dt 2 will sometimes be denoted by v rel and a rel since they are quantities measured relative to the rotating axes.
The symbols v and a will be reserved for the total velocity and acceleration vectors which include the effects of the moving axes relative to the fixed axes.) v = d r
+ ω
× r dt a = d
2 dt r
2
+ 2 ω × d r d ω dt
+ dt
× r + ω × ( ω × r )
The Lagrange Solutions of the Three-Body Problem #8.1
Circular, coplanar orbits with constant angular velocity ω = ω i
ζ
≡
ω i z
1. Two bodies of equal mass m
1
= m
2
= m separated by the distance r mω
2 r
2
=
Gmm r 2
= ω
2
=
2 Gm r 3
2. Two bodies m
1 of rotation at a distance ρ and m
2 at a distance r
−
ρ from the center m
1
ω
2
ρ =
Gm
1 m
2 r 2 m
2
ω
2
( r − ρ ) =
Gm
1 r 2 m
2
= ω
2
=
G ( m
1
+ m
2
) r 3
3. Three bodies of equal mass m
1 lateral triangle with sides r mω
2 r sec 30
2
◦
=
�
= m
2
= m
Gmm
+ r 2
Gmm r 2
�
3
= m at the corners of an equi­ cos 30
◦
= ω
2
=
3 Gm r 3
4. Three bodies m
1
, m
2
, m
3 sides r at the corners of an equilateral triangle with m
1
ω
2
( r
1
− r cm
) + G m m
2
3
ω
ω
2
2
( r
2
( r
3
− r ) + G
− r cm
) + G m
1 m
2
( r
2
− r
1 r 3 m
2 m
1
( r
1
− r
2 m r 3
3 r m
1
3
( r
1
− r
3
) + G
) + G
) + G m
1 r m
2 m
3
3
( r
3 m
3 m
3 r 3
( r
3 r m
2
3
( r
2
− r
1
) = 0
−
− r r
2
3
) =
) =
0
0
16.346 Astrodynamics Lecture 24

Add the three equations to obtain r cm
= m
1 r
1
+ m
2 r
2
+ m
3 r
3 m
1
+ m
2
+ m
3
Let the origin of coordinates be at the center of mass. Then m
1 r
1
+ m
2 r
2
+ m
3 r
3
= 0 .
[ ω
2 r
3 − G ( m
2
+ m
3
)] r
1
+ Gm
2 r
2
+ Gm
3 r
3
= 0
[ ω
2 r
3 − G ( m
1
+ m
3
)] r
2
+ Gm
1 r
1
+ Gm
3 r
3
= 0
[ ω
2 r
3 − G ( m
1
+ m
2
)] r
3
+ Gm
1 r
1
+ Gm
2 r
2
= 0 or
[ ω
2 r
3 − G ( m
1
+ m
2
+ m
3
)] r
1
= 0
[ ω
2 r
3 − G ( m
1
+ m
2
+ m
3
)] r
2
= 0
[ ω
2 r
3 − G ( m
1
+ m
2
+ m
3
)] r
3
= 0
Hence
ω
2
=
G ( m
1
+ m
2
+ m
3
) r 3
5. Three collinear masses m
1
, m
2
, m
3 on ξ axis with r
1
= r i
ξ r
2
= ( r + ρ ) i
ξ r
3
= ( r + ρ + ρχ ) i
ξ
The force balance equations are m
1
ω
2 r +
Gm
1
ρ 2 m
2
+
Gm
1 m
3
ρ 2 (1 + χ ) 2
= 0 m
2
ω
2
( r + ρ ) −
Gm
2 m
1
ρ 2
+
Gm
2
ρ 2 m
χ 2
3
= 0 m
3
ω
2
( r + ρ + ρχ ) −
Gm
3 m
1
ρ 2 (1 + χ ) 2
−
Gm
3 m
2
ρ 2 χ 2
= 0
Replace last equation by sum of three equations m
1
Then m
1
ω
2 r +
Gm
1
ρ 2 m
2 r + m
2
(
+
ρ
Gm
1
2 m
(1 + χ
3
) 2 r + ρ )+
= 0 m
3
( r + ρ + ρχ ) = 0 .
2
( r + ρ )
−
Gm
2 m
1
ρ 2
+
Gm
2 m
ρ 2 χ 2
3
= 0 m
1 r + m
2
( r + ρ ) + m
3
[ r + ρ (1 + χ )] = 0
Third equation = r = − ρ m
2
+ (1 + χ ) m
3 m
1
+ m
2
+ m
3
First equation = ω
2
=
G ( m
1
+ m
2
+ m
3
) m
2
(1 + χ ) 2 + m
3
ρ 3 (1 + χ ) 2 m
2
+ (1 + ) m
3
Second equation = ⇒ ( m
1
+ m
2
) χ
5
+ (3 m
1
+ 2 m
2
−
( m
2
+ 3 m
3
) χ
2 −
(2 m
2
) χ
4
+ 3 m
+ (3
3 m
1
+ m
2
) χ
) χ
−
( m
2
+ m
3
3
) = 0
16.346 Astrodynamics Lecture 24