# Missing Algebraic Step in Three-Body Newtonian Problem

• deuteron
deuteron
TL;DR Summary: I am missing an algebraic step in the below described three-body problem, any help is much appreciated

Consider the following setup

where the distance between the masses are ##d##, and they exert gravitational force on each other. We want to find the angular velocity of the rotation of such system, for which we first need to find the total force on the masses.

What I would've done is the following:

$$m_i\ddot r_i=F_i = G \frac {m_i m_j}{|\vec r_i -\vec r_j|^2} \frac {\vec r_i-\vec r_j}{|\vec r_i-\vec r_j|} +G\frac {m_i m_k}{|\vec r_i-\vec r-k|^2} \frac {\vec r_i-\vec r_k}{|\vec r_i-\vec r_k|} \\ = G\frac {m_i m_j}{d^3}(\vec r_i-\vec r_j)+G\frac {m_i m_k}{d^3}(\vec r_i-\vec r_k) \\ = \frac G {d^3}\ [ m_i m_j\vec r_i -m_im_j \vec r_j + m_im_k \vec r_i -m_im_k\vec r_k] \\ = \frac G {d^3}\ [(m_im_j +m_im_k)\vec r_i -m_im_j\vec r_j +m_im_k\vec r_k]$$

But I am stuck here. In the solution of the problem, the final equation to be reached was:

$$\frac {d^2r_i}{dt^2} =\frac G{d^3} (m_i+m_j+m_k)r_i$$

I feel like I am missing a very obvious algebraic step but can't figure it out.

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If the three masses are fixed on vertecies of regular trianlge of side d, I don't think gravity matters for the motion of the triangle.

anuttarasammyak said:
If the three masses are fixed on vertecies of regular trianlge of side d, I don't think gravity matters for the motion of the triangle.
That's not how I'm interpreting the question (but I could be wrong of course). I interpret the positions as the starting positions of the 3 masses and they have some initial rotational motion, and the question is how the positions evolve given those initial conditions.

@deuteron could you please clarify the question and initial conditions? Thanks.

anuttarasammyak said:
If the three masses are fixed on vertecies of regular trianlge of side d, I don't think gravity matters for the motion of the triangle.
As berkeman said, in the 3-body problem since the general solution has no closed form, one often look for special configurations or restrictions (e.g. circular restriction) that allow for a "simple" solution. And often the rotation complexity of the system is reduced by removing, say, the rotation of the two primary masses, as for example is apparent in the solution giving Lagrange points which is probably most relevant for this post.

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What is the net resultant gravitational force of masses 2 and 3 on mass 1, and what is the direction of the resultant force?

The acceleration of mass 1 is not ##r^"_1##.

Also, you have to assume that the 3 masses are equal if the system is undergoing purely rotation.

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berkeman said:
That's not how I'm interpreting the question (but I could be wrong of course). I interpret the positions as the starting positions of the 3 masses and they have some initial rotational motion, and the question is how the positions evolve given those initial conditions.
Lagrangian of the system
$$L=\frac{m_1}{2}(\dot{x_1}^2+\dot{y_1}^2+\dot{z_1}^2)+\frac{m_2}{2}(\dot{x_2}^2+\dot{y_2}^2+\dot{z_2}^2)+\frac{m_3}{2}(\dot{x_3}^2+\dot{y_3}^2+\dot{z_3}^2)$$
$$+G[\frac{m_1m_2}{r_{12}}+\frac{m_2m_3}{r_{23}}+\frac{m_3m_1}{r_{31}}]$$
where
$$r_{12}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$
$$r_{23}=\sqrt{(x_2-x_3)^2+(y_2-y_3)^2+(z_2-z_3)^2}$$
$$r_{31}=\sqrt{(x_3-x_1)^2+(y_3-y_1)^2+(z_3-z_1)^2}$$
with initial conditions, i.e. values of
$$x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3$$
$$\dot{x_1},\dot{x_2},\dot{x_3},\dot{y_1},\dot{y_2},\dot{y_3},\dot{z_1},\dot{z_2},\dot{z_3}$$
when t=0, should show us the motion at least by numerical solution by computer.

I assume that just observing initial condition is enough to discuss angular momentum of the system thanks to conservation of angular momentum.

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berkeman said:
That's not how I'm interpreting the question (but I could be wrong of course). I interpret the positions as the starting positions of the 3 masses and they have some initial rotational motion, and the question is how the positions evolve given those initial conditions.

@deuteron could you please clarify the question and initial conditions? Thanks.
The question doesn't explicitly state any initial rotation but I assumed that there should be, since otherwise the configuration should be a stable equilibrium.

FWIW I think the problem is about a ‘central configuration’ (e.g. see https://en.wikipedia.org/wiki/Central_configuration).

This particular central configuration consists of 3 (in general, unequal) masses, positioned at the vertices of an equilateral triangle.

When the system rotates with a particular angular velocity around its centre of mass (point C in the diagram) the configuration is stable.

deuteron said:
TL;DR Summary: I am missing an algebraic step in the below described three-body problem, any help is much appreciated

Consider the following setup

View attachment 331800

where the distance between the masses are ##d##, and they exert gravitational force on each other. We want to find the angular velocity of the rotation of such system, for which we first need to find the total force on the masses.

What I would've done is the following:

$$m_i\ddot r_i=F_i = G \frac {m_i m_j}{|\vec r_i -\vec r_j|^2} \frac {\vec r_i-\vec r_j}{|\vec r_i-\vec r_j|} +G\frac {m_i m_k}{|\vec r_i-\vec r-k|^2} \frac {\vec r_i-\vec r_k}{|\vec r_i-\vec r_k|} \\ = G\frac {m_i m_j}{d^3}(\vec r_i-\vec r_j)+G\frac {m_i m_k}{d^3}(\vec r_i-\vec r_k) \\ = \frac G {d^3}\ [ m_i m_j\vec r_i -m_im_j \vec r_j + m_im_k \vec r_i -m_im_k\vec r_k] \\ = \frac G {d^3}\ [(m_im_j +m_im_k)\vec r_i -m_im_j\vec r_j +m_im_k\vec r_k]$$

But I am stuck here. In the solution of the problem, the final equation to be reached was:

$$\frac {d^2r_i}{dt^2} =\frac G{d^3} (m_i+m_j+m_k)r_i$$

I feel like I am missing a very obvious algebraic step but can't figure it out.

The exact question is in German. For German speakers:

Wir wählen den Schwerpunkt ##C## dreier Massen ##m_1,\ m_2,\ m_3## als Koordinatesursprung und die Positionsvektoren ##r_1,\ r_2,\ r_3## zeigen zu diesen Massen. Die Massen wechselwirken untereinander über die Schwerkraft. Sind die Massen in einem gleichseitigen Dreieck angeordnet, so kann das System in seiner Ebene rotieren. Bestimmen Sie die Winkelgeschwindigkeit der Rotation, wenn die Kantenlänge des gleichseitigen Dreiecks mit ##d## angenommen wird.

Rough translation:
We choose the center of gravity ##C## of three masses ##m_1,\ m_2,\ m_3## as coordinate origin and the position vectors ##r_1,\ r_2,\ r_3## point to these masses. The masses interact with each other via gravity. If the masses are arranged in an equilateral triangle, the system can rotate in its plane. Determine the angular velocity of rotation if the edge length of the equilateral triangle is assumed to be ##d##.

deuteron said:
The exact question is in German. For German speakers:

Wir wählen den Schwerpunkt ##C## dreier Massen ##m_1,\ m_2,\ m_3## als Koordinatesursprung und die Positionsvektoren ##r_1,\ r_2,\ r_3## zeigen zu diesen Massen. Die Massen wechselwirken untereinander über die Schwerkraft. Sind die Massen in einem gleichseitigen Dreieck angeordnet, so kann das System in seiner Ebene rotieren. Bestimmen Sie die Winkelgeschwindigkeit der Rotation, wenn die Kantenlänge des gleichseitigen Dreiecks mit ##d## angenommen wird.

Rough translation:
We choose the center of gravity ##C## of three masses ##m_1,\ m_2,\ m_3## as coordinate origin and the position vectors ##r_1,\ r_2,\ r_3## point to these masses. The masses interact with each other via gravity. If the masses are arranged in an equilateral triangle, the system can rotate in its plane. Determine the angular velocity of rotation if the edge length of the equilateral triangle is assumed to be ##d##.
And the solution method is to first use Newton second on one of the masses to get the following:

$$m_i \frac {d^2r_i}{dt^2} = G\frac {m_im_j\ (r_i-r_j)}{|r_i-r_j|^3} + G\frac {m_im_k\ (r_i-r_k)}{|r_i-r_k|^3}$$

And then without any other steps, they "use the definition of senter of mass and set ##|r_i-r_j|=|r_i-r_k|=d##" to get:

$$\frac {d^2 r_i}{dt^2}=G\frac {m_i+m_j+m_k}{d^3} r_i$$

deuteron said:
The question doesn't explicitly state any initial rotation but I assumed that there should be, since otherwise the configuration should couldn't be in a stable equilibrium.
Fixed that for you.

deuteron said:
The exact question is in German. For German speakers:

Wir wählen den Schwerpunkt ##C## dreier Massen ##m_1,\ m_2,\ m_3## als Koordinatesursprung und die Positionsvektoren ##r_1,\ r_2,\ r_3## zeigen zu diesen Massen. Die Massen wechselwirken untereinander über die Schwerkraft. Sind die Massen in einem gleichseitigen Dreieck angeordnet, so kann das System in seiner Ebene rotieren. Bestimmen Sie die Winkelgeschwindigkeit der Rotation, wenn die Kantenlänge des gleichseitigen Dreiecks mit ##d## angenommen wird.

Rough translation:
We choose the center of gravity ##C## of three masses ##m_1,\ m_2,\ m_3## as coordinate origin and the position vectors ##r_1,\ r_2,\ r_3## point to these masses. The masses interact with each other via gravity. If the masses are arranged in an equilateral triangle, the system can rotate in its plane. Determine the angular velocity of rotation if the edge length of the equilateral triangle is assumed to be ##d##.
If the system is rotating about the fixed center of mass C, and all three masses are rotating at the same rate, such that the equilateral triangular shape remains fixed, then there is no component of force acting tangent to the rotation arc of each mass. Focusing on mass 1, this means that mass 2 must be equal to mass 3. This means that all three masses must be equal, and the center of mass must be equal to the center of the triangle for all time.

Chestermiller said:
This means that all three masses must be equal, and the center of mass must be equal to the center of the triangle for all time.
But assume one of the masses are insignificant relative to the two others then we have a classical L4/L5 configuration where masses are not equal, but they still maintain an equilateral triangle. The rate in this case is equal to the 2-body rate.

Filip Larsen said:
But assume one of the masses are insignificant relative to the two others then we have a classical L4/L5 configuration where masses are not equal, but they still maintain an equilateral triangle. The rate in this case is equal to the 2-body rate.
How can the center of mass be at the center of an equilateral triangle and the other two masses be at the vertices if one of the masses is zero?

Chestermiller said:
How can the center of mass be at the center of an equilateral triangle and the other two masses be at the vertices if one of the masses is zero?
If the system is rotating about the fixed center of mass C, and all three masses are rotating at the same rate, such that the equilateral triangular shape remains fixed, then there is no component of force acting tangent to the rotation arc of each mass.
I believe point C in the Post #1 diagram is the centre of mass and is not the geometrical centre of the equliateral triangle.

Imagine an equilateral triangle rotating about a perpendicular axis which does not pass through the triangle's centre.

Each vertex of the triangle rotates in a circle about the axis. The vertices are different distances from the axis - they have the equal angular velocities but different linear velocities.

With a mass at each vertex, the resultant force on a mass (due to the gravitational attraction from the other 2 masses) turns out to be directed through the centre of mass (C). I guess this is a consequence of the inverse square law.

Filip Larsen
Steve4Physics said:
I believe point C in the Post #1 diagram is the centre of mass and is not the geometrical centre of the equliateral triangle.

Imagine an equilateral triangle rotating about a perpendicular axis which does not pass through the triangle's centre.

Each vertex of the triangle rotates in a circle about the axis. The vertices are different distances from the axis - they have the equal angular velocities but different linear velocities.

With a mass at each vertex, the resultant force on a mass (due to the gravitational attraction from the other 2 masses) turns out to be directed through the centre of mass (C). I guess this is a consequence of the inverse square law.
Then the simplest setup of this problem is to put m2 and m3 on a line of constant y below the x axis, and to then quickly determine the x and y coordinates of the 3 masses, given that the center of mass is at the origin. The line joining m2 and m3 is below the x axis by a distance $$\frac{d\sqrt{3}}{2}\frac{m1}{m1+m2+m3}$$This way, the cartesian coordinates of the three masses can be established.

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At some point during the rotation, the altitude through m1 will be vertical. At that time, the coordinates of vertices 1, 2, and 3 will be $$\left(\frac{m_3-m_2}{m_1+m_3+m_3}\frac{d}{2},\frac{m_2+m_3}{m_1+m_2+m_3}\frac{d\sqrt{3}}{2}\right)$$ $$\left(\frac{m_1+2m_3}{m_1+m_3+m_3}\frac{d}{2},\frac{-m_1}{m_1+m_2+m_3}\frac{d\sqrt{3}}{2}\right)$$and$$\left(\frac{-(m_1+2m_2)}{m_1+m_3+m_3}\frac{d}{2},\frac{-m_1}{m_1+m_2+m_3}\frac{d\sqrt{3}}{2}\right)$$respectively.

Based on these values of the coordinates, in terms of the three masses and d, what is ##r_1##?

In terms of the unit vectors in the x- and y directions, the three masses, and d, what is the unit vector in the direction from the center of mass to vertex 1?

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## What is the three-body problem in Newtonian mechanics?

The three-body problem in Newtonian mechanics involves predicting the motion of three celestial bodies interacting with each other through gravitational forces. Unlike the two-body problem, which has a closed-form solution, the three-body problem is generally chaotic and does not have a general analytical solution.

## Why is the three-body problem considered difficult to solve?

The three-body problem is difficult to solve because the gravitational interactions between the three bodies create a complex system of differential equations that are highly sensitive to initial conditions. This sensitivity leads to chaotic behavior, making it impossible to find a general closed-form solution.

## What is meant by a "missing algebraic step" in the context of the three-body problem?

A "missing algebraic step" refers to a gap or an unclear part in the mathematical derivation or solution process when dealing with the three-body problem. This can occur due to the complexity of the equations involved or the omission of intermediate steps that are crucial for understanding the solution.

## How can one address the missing algebraic steps in solving the three-body problem?

To address missing algebraic steps, one can carefully review the mathematical derivations, check for simplification errors, and ensure that all assumptions and approximations are clearly stated. Additionally, numerical methods and simulations can be used to verify and supplement analytical approaches.

## Are there any special cases of the three-body problem that are solvable?

Yes, there are special cases of the three-body problem that are solvable. One example is the restricted three-body problem, where one of the bodies has a negligible mass compared to the other two, simplifying the equations. Another example is the Lagrangian points, where the three bodies form a stable configuration.

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