# Three bodies, one exact orbit: a worked reference card Ergentics, LLC · Epoch 007 · 22 September 2026. This card explains one classical Euler orbit and gives exact values that can be checked by hand. It accompanies [paper v0.7](paper.md), especially Sections 3.6 and 5. It is a learning and reference resource, not a general three-body solver or an independently certified test suite. ## The physical setup Use consistent units in which the gravitational constant G is 1. Bodies 1 and 3 each have mass 1. Body 2 has mass 2. Initially, body 1 is one unit left of the center, body 2 is at the center, and body 3 is one unit right of the center. The outer bodies move counterclockwise around the center; their tangential speed is 3/2. Their opposite velocities keep the total momentum zero. Set theta = 3t/2. The equation is \[ \begin{aligned} \mathbf r_1(t)&=(-\cos\theta,-\sin\theta,0),& \mathbf v_1(t)&=(\tfrac32\sin\theta,-\tfrac32\cos\theta,0),\\ \mathbf r_2(t)&=(0,0,0),&\mathbf v_2(t)&=(0,0,0),\\ \mathbf r_3(t)&=(\cos\theta,\sin\theta,0),& \mathbf v_3(t)&=(-\tfrac32\sin\theta,\tfrac32\cos\theta,0). \end{aligned} \] In words: body 3 has x position cosine of three-halves t and y position sine of three-halves t. Body 1 has the opposite coordinates. Body 2 stays at the origin. All z positions and velocities are zero. The full period is 4pi/3. The initial velocities are part of the problem. Changing them while keeping the initial positions generally gives a different orbit. ## Exact values at three times Every table entry is an exact two-dimensional vector; append z = 0. Body 2's position and velocity are (0, 0) at all three times. The square root symbol means the positive root; sqrt(3) is another way to write it. | Time t | Angle theta | Body 1 position | Body 3 position | Body 1 velocity | Body 3 velocity | | --- | --- | --- | --- | --- | --- | | 0 | 0 | (−1, 0) | (1, 0) | (0, −3/2) | (0, 3/2) | | pi/9 | pi/6 | (−sqrt(3)/2, −1/2) | (sqrt(3)/2, 1/2) | (3/4, −3sqrt(3)/4) | (−3/4, 3sqrt(3)/4) | | pi/3 | pi/2 | (0, −1) | (0, 1) | (3/2, 0) | (−3/2, 0) | These values were derived from the displayed sine and cosine expressions; they were not obtained by numerical integration. For a software comparison, evaluate the expressions independently at the intended precision and account for rounding in both the reference and tested calculation. ## Why the equation works For either outer body, the middle body contributes an inward acceleration of 2. The other outer body, two units away, contributes an inward acceleration of 1/4. The total is 9/4. A unit-radius circle with angular speed 3/2 has inward acceleration (3/2)² = 9/4, exactly the required value. The attractions on body 2 cancel because the outer masses are equal and opposite. The adjacent distances remain 1, and the outer-to-outer distance remains 2. Therefore no collision occurs at any real time. This is a special solution with finite positive masses; no body is treated as massless. ## What the display represents ![Three collinear bodies at phase pi/6: body 2 at the origin, body 1 below-left, and body 3 above-right, with the outer bodies on opposite sides of a unit circle.](assets/euler-initial.svg) This website figure freshly renders the same t = pi/9 reference state in the epoch 008 style. Body numbers identify positions; they are not a color-only key. The circular path is a geometric locus, while the three numbered points show one instant. The image was generated directly from the exact state. It is not a screenshot or an integrated trajectory. [Enlargeable vector original](assets/euler-initial.svg). The equation is collision-free but **linearly unstable**. That means the linearized equations have a perturbation that grows exponentially. It does not mean the displayed exact solution stops satisfying Newton's equation. The paper derives the positive growth exponent sqrt(3sqrt(2) − 1/4). ## Five questions 1. Why does the far outer body contribute 1/4 rather than 1/2 to acceleration? 2. Why does body 2 remain at the origin even though gravity acts on it? 3. Does this formula solve the problem with the same positions and all three initial velocities zero? 4. A program reports the correct total energy and angular momentum. Has that proved it returned the trajectory for the requested initial data? 5. Is an exact periodic solution necessarily stable to small perturbations? ## Answer key 1. Newtonian acceleration from a point mass has magnitude Gm/r². The far mass is 1 and its distance is 2, giving 1/2² = 1/4. In the vector form, the separation vector in the numerator supplies the extra factor of r. 2. The two gravitational accelerations have equal magnitude and opposite direction. Their vector sum is zero. 3. No. Differentiating the formula fixes the nonzero outer velocities shown in the table. Zero velocities define a different initial-value problem. 4. No. A time-shifted circular orbit has the same energy and angular momentum but usually the wrong initial positions and velocities. A quarter-period shift moves either outer body by sqrt(2) in these units. Initial-data and trajectory comparisons distinguish that error. 5. No. This Euler orbit is an example: the formula is exact for all time, while the linearized equations contain a growing perturbation. ## A small software-checking exercise Use the formula and three table times as declared reference points. Check positions and velocities with fixed body labels, the specified units, and the specified initial time. Report numerical discrepancies and arithmetic precision; choose any acceptance tolerance before evaluating a candidate implementation. Include a deliberately time-shifted output to verify that the comparison can reject a wrong initial state. Passing these few points is useful diagnostic evidence but cannot prove an implementation correct for all times or all inputs. The card requires no headset, simulator, or live AI service to read. Its plain-language explanation and table supplement the equations and figure. Assistive-technology compatibility and learning effectiveness still need evaluation. No learner outcomes have been collected.