# Dave's Solution to the Three-Body Problem

### Dave's Solution to the Three-Body Problem

"Simple seeks simplest (best) solution." -- Dave.

https://en.wikipedia.org/wiki/Three-body_problem.

In our closed system, the total and constant Energy (E) = total Kinetic Energy (K) + total Gravitational Energy (U) of our three-body system.

We recall our system equations with regards to acceleration (a) due to gravitational energy:

1a. $$\frac{d^{2}x_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}})$$;

2a. $$\frac{d^{2}y_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})$$;

3a. $$\frac{d^{2}z_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}})$$;

4a. $$\frac{d^{2}x_{2 }}{dt^{2}} = -G* (m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}})$$;

5a. $$\frac{d^{2}y_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}})$$;

6a. $$\frac{d^{2}z_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}})$$;

7a. $$\frac{d^{2}x_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}})$$;

8a. $$\frac{d^{2}y_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}})$$;

9a. $$\frac{d^{2}z_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}})$$

with $$r_{ij} = \sqrt{(x_{i }-x_{j })^{2} + (y_{i }-y_{j })^{2} + (z_{i }-z_{j })^{2} }$$ where G is the gravitational constant.

Now, we generate our final system equations corresponding to the energy conservation law for our closed system:

1. $$\frac{1}{2} * m_{1 }*(\frac{dx_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}})*\triangle r_{1 } =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*(\frac{dy_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})*\triangle r_{1 } =E_{y_{1 }}$$;

3. $$\frac{1}{2} * m_{1 }*(\frac{dz_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}})*\triangle r_{1 } =E_{z_{1 }}$$;

4. $$\frac{1}{2} * m_{2 }*(\frac{dx_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}})*\triangle r_{2 } =E_{x_{2 }}$$;

5. $$\frac{1}{2} * m_{2}*(\frac{dy_{2 }}{dt})^{2} - G* m_{2 }*( m_{2 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}})*\triangle r_{2 } =E_{y_{2 }}$$;

6. $$\frac{1}{2} * m_{2 }*(\frac{dz_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}})*\triangle r_{2 } =E_{z_{2 }}$$;

7. $$\frac{1}{2} * m_{3 }*(\frac{dx_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}})*\triangle r_{3 } =E_{x_{3 }}$$;

8. $$\frac{1}{2} * m_{3 }*(\frac{dy_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}})*\triangle r_{3 } =E_{y_{3 }}$$;

9. $$\frac{1}{2} * m_{3 }*(\frac{dz_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}})*\triangle r_{3 } =E_{z_{3 }}$$

where $$\triangle r_{k } = \sqrt{(x_{k}(t + \triangle t )-x_{k }(t))^{2} + (y_{k}(t + \triangle t)-y_{k }(t))^{2} + (z_{k}(t + \triangle t)-z_{k }(t))^{2} }$$.

Next, we list the important position functions corresponding to our spatial components:

10. $$x_{1 } = x_{1 }(t) = \frac{1}{2} * a_{11 }*t^{2} + b_{11 }*t + c_{11 }$$;

11. $$y_{1 } = y_{1 }(t) = \frac{1}{2} * a_{12 }*t^{2} + b_{12 }*t + c_{12 }$$;

12. $$z_{1 } = z_{1 }(t) = \frac{1}{2} * a_{13 }*t^{2} + b_{13 }*t + c_{13 }$$;

13. $$x_{2} = x_{2 }(t) = \frac{1}{2} * a_{21 }*t^{2} + b_{21 }*t + c_{21 }$$;

14. $$y_{2 } = y_{2 }(t) = \frac{1}{2} * a_{22 }*t^{2} + b_{22 }*t + c_{22 }$$;

15. $$z_{2 } = z_{2 }(t) = \frac{1}{2} * a_{23 }*t^{2} + b_{23 }*t + c_{23 }$$;

16. $$x_{3 } = x_{3 }(t) = \frac{1}{2} * a_{31 }*t^{2} + b_{31 }*t + c_{31 }$$;

17. $$y_{3 } = y_{3 }(t) = \frac{1}{2} * a_{32 }*t^{2} + b_{32 }*t + c_{32 }$$;

18. $$z_{3 } = z_{3 }(t) = \frac{1}{2} * a_{33 }*t^{2} + b_{33 }*t + c_{33 }$$

where $$a_{ij }$$ is the acceleration variable; $$b_{ij }$$ is the speed variable; $$c_{ij }$$ is the initial position variable.

Therefore, there are 27 (1a - 9a plus 1-18) system equations with 36 unknowns (independent variables) which helps to explain the rich diversity of possible orbits for the three-body problem... Of course, if we select the energy associated with each spatial component, then we have 27 system equations with 27 unknowns (independent variables).

And if the position functions, 10 - 18, are generally correct, then we have solved the three-body problem generally.

David Cole,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!
Guest

### Re: Dave's Solution to the Three-Body Problem

FYI:

Mass one, $$m_{1 }$$, is at point, $$p_{1 }$$ = $$(x_{1 }, y_{1 }, z_{1 }$$);

Mass two, $$m_{2 }$$, is at point, $$p_{2 }$$ = $$(x_{2 }, y_{2 }, z_{2 }$$);

Mass three, $$m_{3 }$$, is at point, $$p_{3 }$$ = $$(x_{3 }, y_{3 }, z_{3 }$$).
Guest

### Re: Dave's Solution to the Three-Body Problem

Some Computations:

$$\frac{d^{2}x_{1 }}{dt^{2}} = a_{11 }$$, $$\frac{d^{2}y_{1 }}{dt^{2}} = a_{12 }$$, $$\frac{d^{2}z_{1 }}{dt^{2}} = a_{13 }$$;

$$\frac{d^{2}x_{2 }}{dt^{2}} = a_{21 }$$, $$\frac{d^{2}y_{2 }}{dt^{2}} = a_{22 }$$, $$\frac{d^{2}z_{2 }}{dt^{2}} = a_{23 }$$;

$$\frac{d^{2}x_{3 }}{dt^{2}} = a_{31 }$$, $$\frac{d^{2}y_{3 }}{dt^{2}} = a_{32 }$$, $$\frac{d^{2}z_{3 }}{dt^{2}} = a_{33 }$$.

$$\frac{dx_{1 }}{dt} = a_{11 } * t + b_{11 }$$, $$\frac{dy_{1 }}{dt} = a_{12 } * t + b_{12 }$$, $$\frac{dz_{1 }}{dt} = a_{13 } * t + b_{13 }$$;

$$\frac{dx_{2 }}{dt} = a_{21 } * t + b_{21 }$$, $$\frac{dy_{2 }}{dt} = a_{22 } * t + b_{22 }$$, $$\frac{dz_{2 }}{dt} = a_{23 } * t + b_{23 }$$;

$$\frac{dx_{3 }}{dt} = a_{31 } * t + b_{31 }$$, $$\frac{dy_{3 }}{dt} = a_{32 } * t + b_{32 }$$, $$\frac{dz_{3 }}{dt} = a_{33 } * t + b_{33 }$$.
Guest

### Re: Dave's Solution to the Three-Body Problem

Guest wrote:Some Computations:

$$\frac{d^{2}x_{1 }}{dt^{2}} = a_{11 }$$, $$\frac{d^{2}y_{1 }}{dt^{2}} = a_{12 }$$, $$\frac{d^{2}z_{1 }}{dt^{2}} = a_{13 }$$;

$$\frac{d^{2}x_{2 }}{dt^{2}} = a_{21 }$$, $$\frac{d^{2}y_{2 }}{dt^{2}} = a_{22 }$$, $$\frac{d^{2}z_{2 }}{dt^{2}} = a_{23 }$$;

$$\frac{d^{2}x_{3 }}{dt^{2}} = a_{31 }$$, $$\frac{d^{2}y_{3 }}{dt^{2}} = a_{32 }$$, $$\frac{d^{2}z_{3 }}{dt^{2}} = a_{33 }$$.

$$\frac{dx_{1 }}{dt} = a_{11 } * t + b_{11 }$$, $$\frac{dy_{1 }}{dt} = a_{12 } * t + b_{12 }$$, $$\frac{dz_{1 }}{dt} = a_{13 } * t + b_{13 }$$;

$$\frac{dx_{2 }}{dt} = a_{21 } * t + b_{21 }$$, $$\frac{dy_{2 }}{dt} = a_{22 } * t + b_{22 }$$, $$\frac{dz_{2 }}{dt} = a_{23 } * t + b_{23 }$$;

$$\frac{dx_{3 }}{dt} = a_{31 } * t + b_{31 }$$, $$\frac{dy_{3 }}{dt} = a_{32 } * t + b_{32 }$$, $$\frac{dz_{3 }}{dt} = a_{33 } * t + b_{33 }$$.

1. $$\frac{1}{2} * m_{1 }*( a_{11 } * t + b_{11 })^{2} - m_{1 } * a_{11 } * \triangle r_{1 } =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*( a_{12 } * t + b_{12 })^{2} - m_{1 } * a_{12 } * \triangle r_{1 } =E_{y_{1 }}$$;

3. $$\frac{1}{2} * m_{1 }*( a_{13 } * t + b_{13 })^{2} - m_{1 } * a_{13 } * \triangle r_{1 } =E_{z_{1 }}$$;

4. $$\frac{1}{2} * m_{2 }*( a_{21 } * t + b_{21 })^{2} - m_{2 } * a_{21 } * \triangle r_{2 } =E_{x_{2 }}$$;

5. $$\frac{1}{2} * m_{2 }*( a_{22 } * t + b_{22 })^{2} - m_{2 } * a_{22 } * \triangle r_{2 } =E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*( a_{23 } * t + b_{23 })^{2} - m_{2 } * a_{23} * \triangle r_{2 } =E_{z_{2 }}$$;

7. $$\frac{1}{2} * m_{3 }*( a_{31 } * t + b_{31 })^{2} - m_{3 } * a_{31 } * \triangle r_{3 } =E_{x_{3 }}$$;

8. $$\frac{1}{2} * m_{3 }*( a_{32 } * t + b_{32 })^{2} - m_{3 } * a_{32 } * \triangle r_{3} =E_{y_{3 }}$$;

9. $$\frac{1}{2} * m_{3 }*( a_{33 } * t + b_{33 })^{2} - m_{3 } * a_{33 } * \triangle r_{3 } =E_{z_{3 }}$$;

where $$\triangle r_{k } = \sqrt{(x_{k}(t + \triangle t )-x_{k }(t))^{2} + (y_{k}(t + \triangle t)-y_{k }(t))^{2} + (z_{k}(t + \triangle t)-z_{k }(t))^{2} }$$.
Guest

### Re: Dave's Solution to the Three-Body Problem

FYI: E is the total energy for our three-body system, and it must remain constant while the energy associated with each spatial component may vary.

$$E = E_{x_{1 }} + E_{y_{1 }} + E_{z_{1 }} + E_{x_{2 }} + E_{y_{2 }} + E_{z_{2 }} + E_{x_{3 }} + E_{y_{3}} + E_{z_{1 }}$$.
Guest

### Re: Dave's Solution to the Three-Body Problem

Guest wrote:FYI: E is the total energy for our three-body system, and it must remain constant while the energy associated with each spatial component may vary.

$$E = E_{x_{1 }} + E_{y_{1 }} + E_{z_{1 }} + E_{x_{2 }} + E_{y_{2 }} + E_{z_{2 }} + E_{x_{3 }} + E_{y_{3}} + E_{z_{3 }}$$.
Guest

### Re: Dave's Solution to the Three-Body Problem

FYI: E is the total energy for our three-body system, and it must remain constant while the energy associated with each spatial component may vary.

$$E = E_{x_{1 }} + E_{y_{1 }} + E_{z_{1 }} + E_{x_{2 }} + E_{y_{2 }} + E_{z_{2 }} + E_{x_{3 }} + E_{y_{3}} + E_{z_{3 }}$$.

FYI: The Important Noether's Theorem is applicable here.

"Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law...",

https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:_Conservation_of_energy.

Therefore, we expect symmetrical orbits in our solutions to the three-body problem.
Guest

### Re: Dave's Solution to the Three-Body Problem

FYI: 'Weird Orbits - the three-body problem',

Guest

### Re26: Dave's Solution to the Three-Body Problem

FYI: '3-Body Problem - Periodic Solutions',

Guest

### Re: Dave's Solution to the Three-Body Problem

'On a General Solution of the Three-Body Problem',

https://www.math10.com/forum/viewtopic.php?f=63&t=8398.
Guest

### Re: Dave's Solution to the Three-Body Problem

Guest wrote:
Guest wrote:Some Computations:

1. $$\frac{1}{2} * m_{1 }*( a_{11 } * t + b_{11 })^{2} - m_{1 } * a_{11 } * \triangle r_{1 } =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*( a_{12 } * t + b_{12 })^{2} - m_{1 } * a_{12 } * \triangle r_{1 } =E_{y_{1 }}$$;

3. $$\frac{1}{2} * m_{1 }*( a_{13 } * t + b_{13 })^{2} - m_{1 } * a_{13 } * \triangle r_{1 } =E_{z_{1 }}$$;

4. $$\frac{1}{2} * m_{2 }*( a_{21 } * t + b_{21 })^{2} - m_{2 } * a_{21 } * \triangle r_{2 } =E_{x_{2 }}$$;

5. $$\frac{1}{2} * m_{2 }*( a_{22 } * t + b_{22 })^{2} - m_{2 } * a_{22 } * \triangle r_{2 } =E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*( a_{23 } * t + b_{23 })^{2} - m_{2 } * a_{23} * \triangle r_{2 } =E_{z_{2 }}$$;

7. $$\frac{1}{2} * m_{3 }*( a_{31 } * t + b_{31 })^{2} - m_{3 } * a_{31 } * \triangle r_{3 } =E_{x_{3 }}$$;

8. $$\frac{1}{2} * m_{3 }*( a_{32 } * t + b_{32 })^{2} - m_{3 } * a_{32 } * \triangle r_{3} =E_{y_{3 }}$$;

9. $$\frac{1}{2} * m_{3 }*( a_{33 } * t + b_{33 })^{2} - m_{3 } * a_{33 } * \triangle r_{3 } =E_{z_{3 }}$$;

where $$\triangle r_{k } = \sqrt{(x_{k}(t + \triangle t )-x_{k }(t))^{2} + (y_{k}(t + \triangle t)-y_{k }(t))^{2} + (z_{k}(t + \triangle t)-z_{k }(t))^{2} }$$.

Hmm. We are doubtful about equations, 1 - 9, above. The value, $$\triangle r_{k }$$, seems incorrect (too big or too small or inexact) for equations, 1-9. And we have more doubts too.
Guest

### Re: Dave's Solution to the Three-Body Problem

An Update:

Some Computations:

1. $$\frac{1}{2} * m_{1 }*( a_{11 } * t + b_{11 })^{2} - m_{1 } * a_{11 } * |x_{1}( t + \triangle t) - x_{1}(t)|= E_{x_{1}}$$;

2. $$\frac{1}{2} * m_{1 }*( a_{12 } * t + b_{12 })^{2} - m_{1 } * a_{12 } * |y_{1}( t + \triangle t) -y_{1}(t)|= E_{y_{1}}$$;

3. $$\frac{1}{2} * m_{1 }*( a_{13 } * t + b_{13 })^{2} - m_{1 } * a_{13 } * |z_{1}( t + \triangle t) - z_{1}(t)|= E_{z_{1}}$$;

4. $$\frac{1}{2} * m_{2 }*( a_{21 } * t + b_{21 })^{2} - m_{2 } * a_{21 } * |x_{2}( t + \triangle t) - x_{2}(t)|= E_{x_{2}}$$;

5. $$\frac{1}{2} * m_{2 }*( a_{22 } * t + b_{22 })^{2} - m_{2 } * a_{22 } * |y_{2}( t + \triangle t) -y_{2}(t)|= E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*( a_{23 } * t + b_{23 })^{2} - m_{2 } * a_{23} * |z_{2}( t + \triangle t) - z_{2}(t)|= E_{z_{2}}$$;

7. $$\frac{1}{2} * m_{3 }*( a_{31 } * t + b_{31 })^{2} - m_{3 } * a_{31 } * |x_{3}( t + \triangle t) - x_{3}(t)|= E_{x_{3}}$$;

8. $$\frac{1}{2} * m_{3 }*( a_{32 } * t + b_{32 })^{2} - m_{3 } * a_{32 } * |y_{3}( t + \triangle t) - y_{3}(t)|= E_{y_{3}}$$;

9. $$\frac{1}{2} * m_{3 }*( a_{33 } * t + b_{33 })^{2} - m_{3 } * a_{33 } * |z_{3}( t + \triangle t) - z_{3}(t)|= E_{z_{3}}$$.
Guest

### Re: Dave's Solution to the Three-Body Problem

An Update:

"Simple seeks simplest (best) solution." -- Dave.

In our closed system, the total and constant Energy (E) = total Kinetic Energy (K) + total Gravitational Energy (U) of our three-body system.

We recall our system equations with regards to acceleration (a) due to gravitational energy:

1a. $$\frac{d^{2}x_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}})$$;

2a. $$\frac{d^{2}y_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})$$;

3a. $$\frac{d^{2}z_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}})$$;

4a. $$\frac{d^{2}x_{2 }}{dt^{2}} = -G* (m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}})$$;

5a. $$\frac{d^{2}y_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}})$$;

6a. $$\frac{d^{2}z_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}})$$;

7a. $$\frac{d^{2}x_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}})$$;

8a. $$\frac{d^{2}y_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}})$$;

9a. $$\frac{d^{2}z_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}})$$

with $$r_{ij} = \sqrt{(x_{i }-x_{j })^{2} + (y_{i }-y_{j })^{2} + (z_{i }-z_{j })^{2} }$$ where G is the gravitational constant.

Now, we generate our final system equations corresponding to the energy conservation law for our closed system:

1. $$\frac{1}{2} * m_{1 }*(\frac{dx_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}}) * |x_{1}( t + \triangle t) - x_{1}(t)| =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*(\frac{dy_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})* |y_{1}( t + \triangle t) - y_{1}(t)|= E_{y_{1}}$$;

3. $$\frac{1}{2} * m_{1 }*(\frac{dz_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}} * |z_{1}( t + \triangle t) - z_{1}(t)|= E_{z_{1}}$$;

4. $$\frac{1}{2} * m_{2 }*(\frac{dx_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}}) * |x_{2}( t + \triangle t) - x_{2}(t)|= E_{x_{2}}$$;

5. $$\frac{1}{2} * m_{2}*(\frac{dy_{2 }}{dt})^{2} - G* m_{2 }*( m_{2 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}}) * |y_{2}( t + \triangle t) - y_{2}(t)|= E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*(\frac{dz_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}}) * |z_{2}( t + \triangle t) - z_{2}(t)|= E_{z_{2}}$$;

7. $$\frac{1}{2} * m_{3 }*(\frac{dx_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}}) * |x_{3}( t + \triangle t) - x_{3}(t)|= E_{x_{3}}$$;

8. $$\frac{1}{2} * m_{3 }*(\frac{dy_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}}) * y_{3}( t + \triangle t) - y_{3}(t)|= E_{y_{3}}$$;

9. $$\frac{1}{2} * m_{3 }*(\frac{dz_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}}) * |z_{3}( t + \triangle t) - z_{3}(t)|= E_{z_{3}}$$.

Next, we list the important position functions corresponding to our spatial components:

10. $$x_{1 } = x_{1 }(t) = \frac{1}{2} * a_{11 }*t^{2} + b_{11 }*t + c_{11 }$$;

11. $$y_{1 } = y_{1 }(t) = \frac{1}{2} * a_{12 }*t^{2} + b_{12 }*t + c_{12 }$$;

12. $$z_{1 } = z_{1 }(t) = \frac{1}{2} * a_{13 }*t^{2} + b_{13 }*t + c_{13 }$$;

13. $$x_{2} = x_{2 }(t) = \frac{1}{2} * a_{21 }*t^{2} + b_{21 }*t + c_{21 }$$;

14. $$y_{2 } = y_{2 }(t) = \frac{1}{2} * a_{22 }*t^{2} + b_{22 }*t + c_{22 }$$;

15. $$z_{2 } = z_{2 }(t) = \frac{1}{2} * a_{23 }*t^{2} + b_{23 }*t + c_{23 }$$;

16. $$x_{3 } = x_{3 }(t) = \frac{1}{2} * a_{31 }*t^{2} + b_{31 }*t + c_{31 }$$;

17. $$y_{3 } = y_{3 }(t) = \frac{1}{2} * a_{32 }*t^{2} + b_{32 }*t + c_{32 }$$;

18. $$z_{3 } = z_{3 }(t) = \frac{1}{2} * a_{33 }*t^{2} + b_{33 }*t + c_{33 }$$

where $$a_{ij }$$ is the acceleration variable; $$b_{ij }$$ is the speed variable; $$c_{ij }$$ is the initial position variable (coordinate).

Therefore, there are 27 (1a - 9a plus 1-18) system equations with 36 unknowns (independent variables) which helps to explain the rich diversity of possible orbits for the three-body problem... Of course, if we select the energy associated with each spatial component, then we have 27 system equations with 27 unknowns (independent variables). Yes! The time variable, t, is also independent. However, we do not count it as unknown...

And if the position functions, 10 - 18, are generally correct, then we have solved the three-body problem generally.

David Cole,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!

P.S. We apologize for any errors posted here.
Guest

### Re: Dave's Solution to the Three-Body Problem

Guest wrote:An Update:

"Simple seeks simplest (best) solution." -- Dave.

In our closed system, the total and constant Energy (E) = total Kinetic Energy (K) + total Gravitational Energy (U) of our three-body system.

We recall our system equations with regards to acceleration (a) due to gravitational energy:

1a. $$\frac{d^{2}x_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}})$$;

2a. $$\frac{d^{2}y_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})$$;

3a. $$\frac{d^{2}z_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}})$$;

4a. $$\frac{d^{2}x_{2 }}{dt^{2}} = -G* (m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}})$$;

5a. $$\frac{d^{2}y_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}})$$;

6a. $$\frac{d^{2}z_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}})$$;

7a. $$\frac{d^{2}x_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}})$$;

8a. $$\frac{d^{2}y_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}})$$;

9a. $$\frac{d^{2}z_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}})$$

with $$r_{ij} = \sqrt{(x_{i }-x_{j })^{2} + (y_{i }-y_{j })^{2} + (z_{i }-z_{j })^{2} }$$ where G is the gravitational constant.

Now, we generate our final system equations corresponding to the energy conservation law for our closed system:

1. $$\frac{1}{2} * m_{1 }*(\frac{dx_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}}) * |x_{1}( t + \triangle t) - x_{1}(t)| =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*(\frac{dy_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}}) * |y_{1}( t + \triangle t) - y_{1}(t)|= E_{y_{1}}$$;

3. $$\frac{1}{2} * m_{1 }*(\frac{dz_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}}) * |z_{1}( t + \triangle t) - z_{1}(t)|= E_{z_{1}}$$;

4. $$\frac{1}{2} * m_{2 }*(\frac{dx_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}}) * |x_{2}( t + \triangle t) - x_{2}(t)|= E_{x_{2}}$$;

5. $$\frac{1}{2} * m_{2}*(\frac{dy_{2 }}{dt})^{2} - G* m_{2 }*( m_{2 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}}) * |y_{2}( t + \triangle t) - y_{2}(t)|= E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*(\frac{dz_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}}) * |z_{2}( t + \triangle t) - z_{2}(t)|= E_{z_{2}}$$;

7. $$\frac{1}{2} * m_{3 }*(\frac{dx_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}}) * |x_{3}( t + \triangle t) - x_{3}(t)|= E_{x_{3}}$$;

8. $$\frac{1}{2} * m_{3 }*(\frac{dy_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}}) * y_{3}( t + \triangle t) - y_{3}(t)|= E_{y_{3}}$$;

9. $$\frac{1}{2} * m_{3 }*(\frac{dz_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}}) * |z_{3}( t + \triangle t) - z_{3}(t)|= E_{z_{3}}$$.

Next, we list the important position functions corresponding to our spatial components:

10. $$x_{1 } = x_{1 }(t) = \frac{1}{2} * a_{11 }*t^{2} + b_{11 }*t + c_{11 }$$;

11. $$y_{1 } = y_{1 }(t) = \frac{1}{2} * a_{12 }*t^{2} + b_{12 }*t + c_{12 }$$;

12. $$z_{1 } = z_{1 }(t) = \frac{1}{2} * a_{13 }*t^{2} + b_{13 }*t + c_{13 }$$;

13. $$x_{2} = x_{2 }(t) = \frac{1}{2} * a_{21 }*t^{2} + b_{21 }*t + c_{21 }$$;

14. $$y_{2 } = y_{2 }(t) = \frac{1}{2} * a_{22 }*t^{2} + b_{22 }*t + c_{22 }$$;

15. $$z_{2 } = z_{2 }(t) = \frac{1}{2} * a_{23 }*t^{2} + b_{23 }*t + c_{23 }$$;

16. $$x_{3 } = x_{3 }(t) = \frac{1}{2} * a_{31 }*t^{2} + b_{31 }*t + c_{31 }$$;

17. $$y_{3 } = y_{3 }(t) = \frac{1}{2} * a_{32 }*t^{2} + b_{32 }*t + c_{32 }$$;

18. $$z_{3 } = z_{3 }(t) = \frac{1}{2} * a_{33 }*t^{2} + b_{33 }*t + c_{33 }$$

where $$a_{ij }$$ is the acceleration variable; $$b_{ij }$$ is the speed variable; $$c_{ij }$$ is the initial position variable (coordinate).

Therefore, there are 27 (1a - 9a plus 1-18) system equations with 36 unknowns (independent variables) which helps to explain the rich diversity of possible orbits for the three-body problem... Of course, if we select the energy associated with each spatial component, then we have 27 system equations with 27 unknowns (independent variables). Yes! The time variable, t, is also independent. However, we do not count it as unknown...

And if the position functions, 10 - 18, are generally correct, then we have solved the three-body problem generally.

David Cole,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!

P.S. We apologize for any errors posted here.
Guest

### Re: Dave's Solution to the Three-Body Problem

An Update:

"Simple seeks simplest (best) solution." -- Dave.

In our closed system, the total and constant Energy (E) = total Kinetic Energy (K) + total Gravitational Energy (U) of our three-body system.

We recall our system equations with regards to acceleration (a) due to gravitational energy:

1a. $$\frac{d^{2}x_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}})$$;

2a. $$\frac{d^{2}y_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}})$$;

3a. $$\frac{d^{2}z_{1 }}{dt^{2}} = -G* ( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}})$$;

4a. $$\frac{d^{2}x_{2 }}{dt^{2}} = -G* (m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}})$$;

5a. $$\frac{d^{2}y_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}})$$;

6a. $$\frac{d^{2}z_{2 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}})$$;

7a. $$\frac{d^{2}x_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}})$$;

8a. $$\frac{d^{2}y_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}})$$;

9a. $$\frac{d^{2}z_{3 }}{dt^{2}} = -G* ( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}})$$

with $$r_{ij} = \sqrt{(x_{i }-x_{j })^{2} + (y_{i }-y_{j })^{2} + (z_{i }-z_{j })^{2} }$$ where G is the gravitational constant.

Now, we generate our final system equations corresponding to the energy conservation law for our closed system:

1. $$\frac{1}{2} * m_{1 }*(\frac{dx_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{x_{1 }-x_{2 }}{r_{12}^{3}} + m_{3 }* \frac{x_{1 }-x_{3 }}{r_{13}^{3}}) * |x_{1}( t + \triangle t) - x_{1}(t)| =E_{x_{1 }}$$;

2. $$\frac{1}{2} * m_{1 }*(\frac{dy_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{y_{1 }-y_{2 }}{r_{12}^{3}} + m_{3 }* \frac{y_{1 }-y_{3 }}{r_{13}^{3}}) * |y_{1}( t + \triangle t) - y_{1}(t)|= E_{y_{1}}$$;

3. $$\frac{1}{2} * m_{1 }*(\frac{dz_{1 }}{dt})^{2} - G* m_{1 }*( m_{2 }* \frac{z_{1 }-z_{2 }}{r_{12}^{3}} + m_{3 }* \frac{z_{1 }-z_{3 }}{r_{13}^{3}}) * |z_{1}( t + \triangle t) - z_{1}(t)|= E_{z_{1}}$$;

4. $$\frac{1}{2} * m_{2 }*(\frac{dx_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{x_{2 }-x_{1 }}{r_{21}^{3}} + m_{3 }* \frac{x_{2 }-x_{3 }}{r_{23}^{3}}) * |x_{2}( t + \triangle t) - x_{2}(t)|= E_{x_{2}}$$;

5. $$\frac{1}{2} * m_{2}*(\frac{dy_{2 }}{dt})^{2} - G* m_{2 }*( m_{2 }* \frac{y_{2 }-y_{1 }}{r_{21}^{3}} + m_{3 }* \frac{y_{2 }-y_{3 }}{r_{23}^{3}}) * |y_{2}( t + \triangle t) - y_{2}(t)|= E_{y_{2}}$$;

6. $$\frac{1}{2} * m_{2 }*(\frac{dz_{2 }}{dt})^{2} - G* m_{2 }*( m_{1 }* \frac{z_{2 }-z_{1 }}{r_{21}^{3}} + m_{3 }* \frac{z_{2 }-z_{3 }}{r_{23}^{3}}) * |z_{2}( t + \triangle t) - z_{2}(t)|= E_{z_{2}}$$;

7. $$\frac{1}{2} * m_{3 }*(\frac{dx_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{x_{3 }-x_{1 }}{r_{31}^{3}} + m_{2 }* \frac{x_{3 }-x_{2 }}{r_{32}^{3}}) * |x_{3}( t + \triangle t) - x_{3}(t)|= E_{x_{3}}$$;

8. $$\frac{1}{2} * m_{3 }*(\frac{dy_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{y_{3 }-y_{1 }}{r_{31}^{3}} + m_{2 }* \frac{y_{3 }-y_{2 }}{r_{32}^{3}}) * |y_{3}( t + \triangle t) - y_{3}(t)|= E_{y_{3}}$$;

9. $$\frac{1}{2} * m_{3 }*(\frac{dz_{3 }}{dt})^{2} - G* m_{3 }*( m_{1 }* \frac{z_{3 }-z_{1 }}{r_{31}^{3}} + m_{2 }* \frac{z_{3 }-z_{2 }}{r_{32}^{3}}) * |z_{3}( t + \triangle t) - z_{3}(t)|= E_{z_{3}}$$.

Next, we list the important position functions corresponding to our spatial components:

10. $$x_{1 } = x_{1 }(t) = \frac{1}{2} * a_{11 }*t^{2} + b_{11 }*t + c_{11 }$$;

11. $$y_{1 } = y_{1 }(t) = \frac{1}{2} * a_{12 }*t^{2} + b_{12 }*t + c_{12 }$$;

12. $$z_{1 } = z_{1 }(t) = \frac{1}{2} * a_{13 }*t^{2} + b_{13 }*t + c_{13 }$$;

13. $$x_{2} = x_{2 }(t) = \frac{1}{2} * a_{21 }*t^{2} + b_{21 }*t + c_{21 }$$;

14. $$y_{2 } = y_{2 }(t) = \frac{1}{2} * a_{22 }*t^{2} + b_{22 }*t + c_{22 }$$;

15. $$z_{2 } = z_{2 }(t) = \frac{1}{2} * a_{23 }*t^{2} + b_{23 }*t + c_{23 }$$;

16. $$x_{3 } = x_{3 }(t) = \frac{1}{2} * a_{31 }*t^{2} + b_{31 }*t + c_{31 }$$;

17. $$y_{3 } = y_{3 }(t) = \frac{1}{2} * a_{32 }*t^{2} + b_{32 }*t + c_{32 }$$;

18. $$z_{3 } = z_{3 }(t) = \frac{1}{2} * a_{33 }*t^{2} + b_{33 }*t + c_{33 }$$

where $$a_{ij }$$ is the acceleration variable; $$b_{ij }$$ is the speed variable; $$c_{ij }$$ is the initial position variable (coordinate).

Therefore, there are 27 (1a - 9a plus 1-18) system equations with 36 unknowns (independent variables) which helps to explain the rich diversity of possible orbits for the three-body problem... Of course, if we select the energy associated with each spatial component, then we have 27 system equations with 27 unknowns (independent variables). Yes! The time variable, t, is also independent. However, we do not count it as unknown...

And if the position functions, 10 - 18, are generally correct, then we have solved the three-body problem generally.

David Cole,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!

P.S. We apologize for any errors posted here, and for the 'never-ending' updates too.
Guest

### Re: Dave's Solution to the Three-Body Problem

Guest wrote:FYI: E is the total energy for our three-body system, and it must remain constant while the energy associated with each spatial component may vary.

$$E = E_{x_{1 }} + E_{y_{1 }} + E_{z_{1 }} + E_{x_{2 }} + E_{y_{2 }} + E_{z_{2 }} + E_{x_{3 }} + E_{y_{3}} + E_{z_{1 }}$$.

FYI: 'Modelling the Three-Body Problem in Classical Mechanics using Python:
An overview of the fundamentals of gravitation, the odeint solver in Scipy and 3D plotting in Matplotlib',