Closed Form Approximations for the Trajectories of the Three Body Problem AbuBakr Mehmood, Syed Umer Abbas Shah and Ghulam Shabbir Faculty of Engineering Sciences, GIK Institute of Engineering Sciences and Technology Topi, Swabi, NWFP, Pakistan Email:
[email protected]
Abstract: The problem of describing the free motions of three gravitating bodies under each other’s gravitational influence is one of the oldest of unsolved problems of classical mechanics. Henry Poincare proved in his dictum that due to the nature of the instabilities involved in the problem, it could not be solved. Yet closed form approximations for the problem can be found and it is on these lines that we will explore the problem. It will be shown that closed form approximations for the associated trajectories can be found.
Introduction: In this paper we will develop a procedure to find closed form approximations for the trajectories of three gravitating masses having spherical symmetry, when they perform free motion under each other’s gravitational influence. It will be assumed that the masses form an isolated system in free space, and that all the bodies have angular velocities less than 1 rad /s along with bounded position vectors. Other approaches can be found in [1-10]. To begin with, figure 1 shows a typical configuration of the three bodies. It can be shown that the centre of mass this system moves with zero acceleration for all time and hence we do not hesitate in attaching a frame of reference to the centre of mass. Reference frame oxy is therefore an inertial frame of reference.
1
eθ1
y
ex2 m2
ex1 m1
x2
eθ2
x1
Ө2 Ө1 Ө3
o
x
x3
m3
eθ3 ex3
Figure 1
Here xi is the position vector of a body of mass mi . and θi is the rotation angle of ∧
vector xi measured form the x axis. In addition e xi is a unit vector in the direction of ∧
∧
xi and eθi is a unit vector perpendicular to e xi where i = 1, 2, 3. For the position ∧
vectors we will use the notation xi = xi (t ) e xi for i = 1, 2, 3. It should be noted that the assumptions of angular velocities being smaller than 1 rad / s and position vectors being bounded are very critical to our derivations. Provided that the position vectors are less than infinity (that is bounded), we can always scale them in a manner such that they become much less than infinity. Stated .
explicitly, we aim to find approximations for xi (θi ) ∀ θi (t ) < 1 rad/s & xi (t ) ∞
where i = 1, 2, 3. We also assume that while performing three body motion, each body remains approximately in two body motion with a body having a mass equivalent to the sum of the masses of the other two bodies, placed at their centre of mass. Specifically for 2
each body, our assumption states that (a)
m1 approximately remains in two body motion with a body of mass m2 + m3
placed at the centre of mass of m2 and m3 given by x 23 = (b)
m2 x 2 + m3 x3 m2 + m3
∀t
m2 approximately remains in two body motion with a body of mass m1 + m3
placed at the centre of mass of m1 and m3 given by x13 = (c)
m1x1 + m3 x3 m1 + m3
∀t
m3 approximately remains in two body motion with a body of mass m1 + m2
placed at the centre of mass of m1 and m2 given by x12 =
m1x1 + m2 x 2 m1 + m2
∀t
Note that by the use of this assumption, say in (a), we attempt to replicate the individual effects of m2 and m3 on m1 by placing a single body of mass m2 + m3 at the centre of mass of m2 and m3 . However, use of this assumption would turn our solutions into mere approximations instead of exact solutions. Reason being that in order to claim that m1 remains in two body motion with m2 + m3 located at a point given by x 23 , we need to show that ∧
∧
e x1 . e x23 = −1 ∀ t
(1)
∧
where e x23 is a unit vector in the direction of x 23 . Since the above cannot be shown, it follows that the solution x1 (θ1 ) given by the use of this assumption would be an approximation instead of an exact solution. Stated another way, we are assuming that a body of mass m2 + m3 located at the centre of mass of m2 and m3 approximately replicates the effects of the two individual bodies m2 and m3 , on m1 . In accordance, ∧
∧
unit vectors e x1 and e x23 remain approximately collinear for all time. ∧
∧
e x1 . e x23 −1 ∀ t
Had it been possible to prove relation (1) , the three body problem would have been reducible to the two body problem. Similar arguments follow for cases (b) and (c).
3
(2)
For now we consider case (a) and figure 2 presents the configuration of our two body approximation of the three bodies m1 , m2 and m3 . eθ1
y
ex1
m1
x1 Ө1
Ө23
o x23
m2 + m3
ex23
x
eθ23
Figure 2 In accordance with our assumption, the vectors x1 and x 23 are approximately collinear for all time, and this fact is clearly illustrated in figure 2. This in turn implies .
∧
.
that θ1 (t ) θ 23 (t ) + π ⇒ θ1 (t ) θ 23 (t ) . Note that in figure 2, e x23 is a unit vector in the ∧
∧
direction of x 23 and eθ23 is a unit vector perpendicular to e x23 . Modeling this system using Newton’s law of Gravitation and Newton’s 2nd law, we get G (m + m ) ∧ .. 2 3 ex x1 = − x −x 2 1 23 1
(3)
Gm ∧ .. 1 ex x 23 = − x − x 2 23 23 1
(4)
Now resolving (3) and (4) in polar coordinates and comparing the coefficients of the unit vectors on both sides of each equation, we can show that the following relations hold true.
4
G (m + m ) .. . 2 2 3 x1 − x1 θ1 = − x −x 2 23 1 ..
.
(5)
.
x1 θ1 + 2 x1 θ1 = 0
(6)
Gm .. . 2 1 x 23 − x23 θ 23 = − x −x 2 23 1
(7)
..
.
.
x23 θ 23 + 2 x 23 θ 23 = 0
(8)
It should be noted that a simultaneous solution of equations (5) through (8) would yield x1 (t ) , θ1 (t ) , x23 (t ) and θ 23 (t ) . We could then make use of the expressions x1 (t ) and θ1 (t ) to find x1 (θ1 ) , hence describing the trajectory of m1 under the influence of m2 and m3 . However, the system of equations (5), (6), (7) and (8) does not seem solvable analytically. We therefore reject the idea of using this approach and adopt an alternate method. We recall that the problem is being solved for the case θ1 (t ) < 1 rad/s and . 2
x1 (t ) ∞ . We could find an approximation to (5) by noting that θ1 0 ∀ t ⇒ . 2
.
x1 θ1 0 ∀ t (since θ1 (t ) < 1 rad/s and x1 (t ) ∞ ). Hence the left hand side of (5) ..
. 2
..
can be approximated as x1 − x1 θ1 x1 , yielding the following approximation to (5).
G (m + m ) .. 2 3 x1 − x −x 2 23 1
(9a) .
It can be shown that if | x 2 | ∞ and | x3 | ∞ , then | x 23 | ∞ . Also, if θ1 (t ) < 1 .
.
.
rad /s then θ 23 (t ) < 1 rad/s (since θ1 (t ) θ 23 (t ) ∀ t ). Using these arguments, just
like we found (9a) to be an approximation to (5), we can find (9b) to be an approximation to (7). Gm .. 1 x 23 = − x −x 2 23 1
(9b)
x = x 23 − x1
(10)
We now define a vector
5
Since x is a vector along the direction of x 23 , it follows that a unit vector along the ∧
direction of x is the same as a unit vector along the direction of x 23 . We define e x to ∧
∧
be a unit vector along the direction of x . Hence, it follows that e x = e x23 . Note that
x1 , x 23 and x have the representation ∧
∧
∧
x = x (t ) e x , x1 = x1 (t ) e x1 , x 23 = x23 (t ) e x23
(11)
and ∧
∧
∧
e x = e x23 = − e x1
(12)
Making use of (10) and (11), we can write the sum of (9a) and (9b) as .. .. G (m1 + m2 + m3 ) x1 (t ) + x 23 (t ) = − x 2 (t )
(13)
Now use of (11), (12) and (13), along with the definition of x , enables us to show that .. G (m1 + m2 + m3 ) x (t ) = − x 2 (t )
(14)
We then integrate (9a ) , (9b) and (14) twice with respect to time and perform a few manipulations on the resulting equations to get . . (m2 + m3 ) x1 (t ) = x1o + x1o ∗ (t − to ) + x t − x [ ( ) o ∗ (t − to ) − xo ] (m1 + m2 + m3 ) .
.
.
(15)
.
where x10 = x1 (to ), x1o = x1 (to ), x0 = x(to ), x o = x (to ), and to is the starting time (normally taken to be zero). In order to find the analytic expression for the trajectory of m1 by use of (15) , we must find t (θ1 ) . Doing so would help us find x1 (θ1 ) since all variables in (15) would become expressible as functions of θ1 . Therefore we can rewrite (15) as (m2 + m3 ) (m2 + m3 ) x. o − x. 1o to x1 (θ1 ) = x1o − xo + (m1 + m2 + m3 ) (m1 + m2 + m3 ) . . (m2 + m3 ) (m2 + m3 ) + x1o − x o t (θ1 ) + x(θ1 ) (m1 + m2 + m3 ) (m1 + m2 + m3 )
All that remains to be done to find x1 (θ1 ) explicitly, is to find t (θ1 ) and x(θ1 ) and substitute these expressions into (16) . We now begin with the task of finding 6
(16)
t (θ1 ) and x(θ1 ) . Making use of relations (3), (4), (10), (11) and (12) we can show that the following results hold true. .. . G (m1 + m2 + m3 ) x − xθ = − x2 2
..
.
(17)
.
xθ + 2 x θ = 0
(18)
Here ‘ θ ’ is the rotation angle of vector x (which is the same as the rotation angle θ 23 of vector x 23 ). Solving equations (17) and (18) simultaneously, it can be shown that x(θ ) =
1
c cos θ + c 1
2
sin θ +
G (m1 + m2
+ m )
(19)
3
. 2
x θo 4 o
where c1 and c2 are constants of integration, determinable by incorporation of the initial conditions. Now since θ = θ 23 θ1 − π , we can write x(θ ) x(θ1 − π ) . Use of this relation enables us to derive the result x (θ1 ) =
where k1 = ± c12 + c22 , k2 =
1 k cos(θ1 − φ1 ) + k2
(20)
1
G (m1 + m2 + m3 ) . 2
xo4 θ o
c2 . In addition, t (θ ) 1 c 1
and φ1 = tan −1
can be shown to satisfy t (θ1 ) =
θ1
1 .
x θo 2 o
∫
θ1o
2
1 dθ + t 1 o k cos(θ1 − φ1 ) + k2
1
An evaluation of the integral in the above equation gives −1 2 2 k1 + k2 − 12 − k1 tan (0.5φ − 0.5θ1 ) + k1 + k 2 t (θ1 ) = [ (k + k )(k − k ) ] ∗ x 2 θ. k2 − k1 1 2 2 1 + k2 tan 2 (0.5φ −0.5θ1 ) o o
7
(21)
1
(k − k ) tan(0.5φ − 0.5θ1 ) 1 k ((k1 + k2 )(k2 − k1 )) 2 tan(0.5φ − 0.5θ1 ) + k1 tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 + k2 tan 2 (0.5φ −0.5θ1 )−k1k2 tan −1 (k2 −k1 )tan(0.5φ −0.15θ1 ) ((k1 +k2 )(k2 −k1 )) 2
− k22 tan 2 (0.5φ −0.5θ1 )tan −1 (k2 −k1 )tan(0.5φ −0.15θ1 ) ((k1 +k2 )(k2 −k1 )) 2
− k22 tan −1 (k2 −k1 )tan(0.5φ −0.15θ1 ) ((k1 + k2 )(k2 −k1 )) 2
−1 2 + k k 2 − 12 − k1 tan (0.5φ − 0.5θ1o ) + k1 + k 2 1 2 − ∗ [ (k + k )(k − k )] x 2 θ. k2 − k1 1 2 2 1 + k2 tan 2 (0.5φ −0.5θ1o ) o o
1
(k2 − k1 ) tan(0.5φ − 0.5θ1o ) k ((k1 + k2 )(k2 − k1 )) tan(0.5φ − 0.5θ1o ) + k1 tan 1 2 (( k k )( k k )) + − 1 2 2 1 1 2
−1
+ k2 tan (0.5φ −0.5θ1o )−k1k2 tan (k2 −k1 )tan(0.5φ −0.51 θ1o ) ((k1 + k2 )(k2 −k1 )) 2 −1
2
− k22 tan 2 (0.5φ −0.5θ1o )tan −1 (k2 −k1 )tan(0.5φ −0.51 θ1o ) ((k1 + k2 )(k2 −k1 )) 2
−k22 tan −1 (k2 −k1 )tan(0.5φ −0.51 θ1o ) ((k1 + k2 )(k2 −k1 )) 2
+to Substitution of (20) and (22) into equation (16) would then yield the expression for x1 (θ1 ) as . m2 + m3 m2 + m3 m2 + m3 . + − x1 (θ1 ) = x1o − x x x o o 1o to + * m1 + m2 + m3 m1 + m2 + m3 m1 + m2 + m3
m2 + m3 . . 1 xo ∗ + x1o − k1 cos(θ1 − φ1 ) + k2 m1 + m2 + m3
8
(22)
2 2 . xo θ o 1
2 1 k1 + k2 [ (k1 + k2 )(k2 − k1 )]− 2 − k1 tan (0.5φ − 0.5θ1 ) + k1 + k2 * 2 k −k + k2 tan (0.5φ −0.5θ1 ) 2 1
(k2 − k1 ) tan(0.5φ1 − 0.5θ1 ) k ((k1 + k2 )(k2 − k1 )) tan(0.5φ1 − 0.5θ1 ) + k1 tan 1 2 (( )( )) + − k k k k 1 2 2 1
2 − 2 . xo θ1o 1
1 2
−1
+ k2 tan (0.5φ1 −0.5θ1 )−k1k2 tan (k2 −k1 )tan(0.5φ1 −0.15θ1 ) ((k1 + k2 )(k2 −k1 )) 2 −1
2
− k22 tan 2 (0.5φ1 −0.5θ1 )tan −1 (k2 −k1 )tan(0.5φ1 −0.15θ1 ) ((k1 + k2 )(k2 −k1 )) 2
−k22 tan −1 (k2 −k1 )tan(0.5φ1 −0.15θ1 ) ((k1 + k2 )(k2 −k1 )) 2
2 1 k1 + k2 [ (k1 + k2 )(k2 − k1 ) ]− 2 − k1 tan (0.5φ − 0.5θ1o ) + k1 + k2 * 2 k −k + k2 tan (0.5φ −0.5θ1o ) 2 1
(k2 − k1 ) tan(0.5φ1 − 0.5θ1o ) k ((k1 + k2 )(k2 − k1 )) tan(0.5φ1 − 0.5θ1o ) + k1 tan 1 2 + − (( k k )( k k )) 1 2 2 1 1 2
−1
+ k2 tan (0.5φ1 −0.5θ1o )−k1k2 tan (k2 −k1 )tan(0.5φ1 −0.51 θ1o ) ((k1 + k2 )(k2 −k1 )) 2 −1
2
− k22 tan 2 (0.5φ1 −0.5θ1o )tan −1 (k2 −k1 )tan(0.5φ1 −0.51 θ1o ) ((k1 +k2 )(k2 −k1 )) 2
− k22 tan −1 (k2 −k1 )tan(0.5φ1 −0.51 θ1o ) ((k1 + k2 )(k2 −k1 )) 2
+to (23) Having found x1 (θ1 ) , we go on to find x2 (θ 2 ) followed by x3 (θ3 ) following a similar
sequence of steps. We do not hesitate in presenting the results directly, reason being that the procedure and all arguments are essentially the same for any pair of gravitating bodies, be it m1 and m2 + m3 , m2 and m1 + m3 or m3 and m1 + m2 , as stated explicitly in (a), (b) and (c). Figures 3 and 4 present the configurations in accordance with our assumptions (b) and (c).
9
ex2
y
m2
eθ2
x2 Ө2
o
x x13
Ө13
m1 + m3
eθ13
ex13
Figure 3
eθ12
y
ex12
m1 + m2 x12
Ө3
Ө12
o
x3
m3
ex3
eθ3
Figure 4
10
x
It follows that proceeding in exactly the same way as we did for (a) (while finding x1 (θ1 ) ), we can find x2 (θ 2 ) and x3 (θ3 ). It is worth noting that the constant k3 in x2 (θ 2 ) and k5 in x3 (θ3 ) are defined in a way analogous to k1 in x1 (θ1 ) . Similarly the procedure for defining k4 in case of x2 (θ 2 ) and k6 in case of x3 (θ3 ) is exactly analogous to the one for k2 in case of x1 (θ1 ). Same holds true when defining φ2 and
φ3 in x2 (θ 2 ) and x3 (θ3 ) respectively. Having presented the guidelines to be followed, we present equations (24) and (25). m1 + m3 m1 + m3 y. o − x. 2o to + m1 + m3 * x2 (θ 2 ) = x2o − yo + m1 + m2 + m3 m1 + m2 + m3 m1 + m2 + m3 . m1 + m3 . 1 + − x 2 o yo ∗ k3 cos(θ 2 − φ2 ) + k4 m1 + m2 + m3 2 1 2 k3 + k4 [ (k3 + k4 )(k4 − k3 ) ]− 2 − k3 tan (0.5φ2 − 0.5θ 2 ) + k3 + k4 * . 2 y 2 θ k 4 − k3 + k4 tan (0.5φ2 −0.5θ2 ) o 2o 3
(k4 − k3 ) tan(0.5φ2 − 0.5θ 2 ) k ((k3 + k4 )(k4 − k3 )) tan(0.5φ2 − 0.5θ 2 ) + k3 tan 1 2 + − (( )( )) k k k k 3 4 4 3
2 − y2 θ . o 2o 3
1 2
−1
+ k4 tan (0.5φ2 −0.5θ2 )−k3k4 tan (k4 −k3 )tan(0.5φ2 −0.15θ2 ) ((k3 + k4 )(k4 −k3 )) 2 −1
2
− k42 tan 2 (0.5φ2 −0.5θ2 )tan −1 (k4 −k3 )tan(0.5φ2 −0.15θ2 ) ((k3 + k4 )(k4 −k3 )) 2 − k42 tan −1 (k4 −k3 )tan(0.5φ2 −0.15θ2 ) ((k3 + k4 )(k4 −k3 )) 2
2 1 k3 + k4 [ (k3 + k4 )(k4 − k3 )]− 2 −k3 tan (0.5φ2 − 0.5θ 2o ) + k3 + k4 * 2 k 4 − k3 + k4 tan (0.5φ2 −0.5θ2 o )
(k4 − k3 ) tan(0.5φ2 − 0.5θ 2 o ) k ((k3 + k4 )(k4 − k3 )) tan(0.5φ2 − 0.5θ 2 o ) + k3 tan 1 2 + − k k k k (( )( )) 3 4 4 3 1 2
−1
+ k4 tan (0.5φ2 −0.5θ2o )−k3k4 tan (k4 −k3 )tan(0.5φ2 −0.51 θ2 o ) ((k3 + k4 )(k4 −k3 )) 2 −1
2
− k42 tan 2 (0.5φ2 −0.5θ2 o )tan −1 (k4 −k3 )tan(0.5φ2 −0.51 θ2 o ) ((k3 +k4 )(k4 −k3 )) 2 − k42 tan −1 (k4 −k3 )tan(0.5φ2 −0.51 θ2o ) ((k3 + k4 )(k4 −k3 )) 2
+to (24) 11
m1 + m2 m1 + m2 z.o − x. 3o to + m1 + m2 * x3 (θ3 ) = x3o − zo + m1 + m2 + m3 m1 + m2 + m3 m1 + m2 + m3 . m1 + m2 . 1 + x 3o − zo ∗ k5 cos(θ3 − φ3 ) + k6 m1 + m2 + m3
2 1 2 k5 + k6 [ (k5 + k6 )(k6 − k5 ) ]− 2 − k5 tan (0.5φ3 − 0.5θ3 ) + k5 + k6 * . z 2 θ k 6 − k5 + k6 tan 2 (0.5φ3 −0.5θ3 ) o 3o 5
(k6 − k5 ) tan(0.5φ3 − 0.5θ3 ) k ((k5 + k6 )(k6 − k5 )) tan(0.5φ3 − 0.5θ3 ) + k5 tan 1 2 ((k5 + k6 )(k6 − k5 )) 1 2
−1
+ k6 tan (0.5φ3 −0.5θ3 )−k5k6 tan (k6 −k5 )tan(0.5φ3 −0.15θ3 ) ((k5 + k6 )(k6 −k5 )) 2 −1
2
− k62 tan 2 (0.5φ3 −0.5θ3 )tan −1 (k6 −k5 )tan(0.5φ3 −0.15θ3 ) ((k5 + k6 )(k6 −k5 )) 2
(k6 −k5 )tan(0.5φ3 −0.5θ3 ) 1 ((k5 + k6 )(k6 −k5 )) 2
−k62 tan −1
2 2 k5 + k 6 − 12 − k5 tan (0.5φ3 − 0.5θ 3o ) + k5 + k6 − + − k k k k ( )( ) [ ] * z 2 θ . k 6 − k5 5 6 6 5 + k6 tan 2 (0.5φ3 −0.5θ3o ) o 3o 5
(k − k ) tan(0.5φ3 − 0.5θ3o ) 1 k ((k5 + k6 )(k6 − k5 )) 2 tan(0.5φ3 − 0.5θ3o ) + k5 tan −1 6 5 1 ((k5 + k6 )(k6 − k5 )) 2 + k6 tan 2 (0.5φ3 −0.5θ3o )−k5k6 tan −1 (k6 −k5 )tan(0.5φ3 −0.51 θ3o ) ((k5 + k6 )(k6 −k5 )) 2
− k62 tan 2 (0.5φ3 −0.5θ3o )tan −1 (k6 −k5 )tan(0.5φ3 −0.51 θ3o ) ((k5 + k6 )(k6 −k5 )) 2
− k62 tan −1 (k6 −k5 )tan(0.5φ3 −0.51 θ3o ) ((k5 + k6 )(k6 −k5 )) 2
+to (25) It should be noted that the following two definitions find use in our results (24) and (25). For derivation of (24), we defined y = x13 − x 2 , and similarly, in case of (25), we utilized z = x12 − x3 . Having derived all the required results, we now go on to summarize our accomplishments.
12
Summary: We aimed at deriving closed form approximations for the associated trajectories when three masses execute free motion under each other’s gravitational influence. The masses were defined to form an isolated system in free space. We assumed that each mass possesed a spherically symmetric mass distribution and that all the bodies had angular velocities less than 1 rad /s along with bounded position vectors. Also, we utilized a single body to replicate the effects of individual bodies in each body. We were able to derive expressions defining the trajectories of these masses in equations (23), (24) and (25). It is worth stating that our assumptions were both practical and feasible. Celestial bodies do posses mass distributions that are approximately spherically symmetric, and angular velocities encountered in celestial motion are doubtlessly less than 1 rad /s (excluding exceptions). Also, we assumed that position vectors remain bounded ( x k < ∞ ∀ k = 1, 2 and 3) . Using reasonable units of length, this condition can be satisfied at the scale of solar systems as well as galaxies. Furthermore, we can always scale the position vectors along their length so that the relation x k
