I'm attempting to numerically solve the 3-body problem. Using Newton's second law, I've derived a system of 6 second order differential equations, the first three being:

$$ m_1\frac{d^2x_1}{dt^2} = -G \Bigg[ \frac{m_1m_2(x_1-x_2)}{[(x_1-x_2)^2+(y_1-y_2)^2]^{3/2}} + \frac{m_1m_3(x_1-x_3)}{[(x_1-x_3)^2+(y_1-y_3)^2]^{3/2}} \Bigg]$$

$$ m_2\frac{d^2x_2}{dt^2} = -G \Bigg[ \frac{m_1m_2(x_2-x_1)}{[(x_2-x_1)^2+(y_2-y_1)^2]^{3/2}} + \frac{m_2m_3(x_2-x_3)}{[(x_2-x_3)^2+(y_2-y_3)^2]^{3/2}} \Bigg]$$

$$ m_3\frac{d^2x_3}{dt^2} = -G \Bigg[ \frac{m_1m_3(x_3-x_1)}{[(x_3-x_1)^2+(y_3-y_1)^2]^{3/2}} + \frac{m_2m_3(x_3-x_2)}{[(x_3-x_2)^2+(y_3-y_2)^2]^{3/2}} \Bigg]$$

Likewise, I've found equations for $\frac{d^2y_i}{dt^2}$ and have cancelled out the repeated mass terms. I would like to nondimensionalize all of these equations before linearizing and solving, but am not sure how. I'd like to use $M$ as the nondimensional unit for mass and $d$ as the nondimensional unit of distance.

So far I've tried letting $u = x/L$, $v=y/L$, and $\tau=t/T$. Any ideas on how I could nondimensionalize this system? Thank you!


2 Answers 2


You are going along the right track from what I presume to be the case when you attempt to nondimensionalize Newtonian gravitation. There is way to do it, that I know of, that will leave it without the scaling factors. Let us start with the most general form of gravitational attraction with,

$$ \vec{a}_{i} = \sum_{i = 0,j \neq i}^{N} - \frac{Gm_{j}}{r_{ji}^{3}} \vec{r}_{ji} = \sum_{i = 0,j \neq i}^{N} - \frac{Gm_{j}}{\big| \vec{r_{i}} -\vec{r_{j}} \big|^{3}} \left( \vec{r_{i}} -\vec{r_{j}} \right) .$$

You could give a scale test mass M and a scale test length L. Then with the use of the gravitational constant you could then define unit correct scaling factors for all other variables. For T it would be

$$ T = \sqrt{\frac{L^{3}}{GM}} .$$

You could continue on to acceleration, velocity, and so on as needed. I can then let $\vec{r}_{i} = L\vec{\chi}_{i}$ as the scaling constant is the same for every component of the vector then it can be pulled out as is the case with vector magnitudes, so $r_{i} = L \chi_{i}$ or $r_{ji} = L \chi_{ji}$. When you substitute these into the previous force law you'll end up with

$$ \frac{GM}{L^{2}} \vec{\alpha}_{i} = \sum_{i = 0,j \neq i}^{N} - \frac{G M\eta_{j}}{L^{3} \chi_{ji}^{3}} L\vec{\chi}_{ji} .$$

Dragging all the constants out and simplifying leads to

$$ \frac{GM}{L^{2}} \vec{\alpha}_{i} = \frac{GM}{L^{2}} \sum_{i = 0,j \neq i}^{N} - \frac{ \eta_{j}}{ \chi_{ji}^{3}} \vec{\chi}_{ji} .$$

Therefore the gravitational acceleration of particle $i$ due to all other particles can be put into completely non-dimensional position vectors $\vec{\chi}_{i}$ and masses $\eta_{j}$. This being

$$\vec{\alpha}_{i} = \sum_{i = 0,j \neq i}^{N} - \frac{ \eta_{j}}{ \chi_{ji}^{3}} \vec{\chi}_{ji} .$$

You can perform the same similar substitution in your example but for each position coordinate, mass, and acceleration then begin pulling them out given these scaling constants for each respective S.I. unit will be the same.

  • $\begingroup$ This is such a thorough explanation of what I needed to do. Thank you! $\endgroup$
    – J. Castor
    Sep 30, 2021 at 21:24
  • $\begingroup$ @J.Castor I'm glad I could be of assistance in this matter. $\endgroup$ Oct 2, 2021 at 18:59
  • $\begingroup$ @Thevictorioustruther Could you explain the steps you took to nondimensionalize time? I am able to get it but only after trying to multiple different combinations of L, G, and M. I was wondering if there is a more methodical way to get this? $\endgroup$ Jan 4 at 20:44

The equations of motion for the 3-bodies are:

\begin{align*} &m_1\,\frac{d^2}{dt^2}\mathbf{{R}_1}=\mathbf{F}_{12}+\mathbf{F}_{13}\\ &m_2\,\frac{d^2}{dt^2}\mathbf{{R}_2}=-\mathbf{F}_{12}+\mathbf{F}_{23}\\ &m_3\,\frac{d^2}{dt^2}\mathbf{{R}_3}=-\mathbf{F}_{13}-\mathbf{F}_{23}\\ &\text{where:}\\ &\mathbf{F}_{12}=\frac{m_1\,m_2\,G}{ \bigg(\mathbf{R}_{21}\cdot\mathbf{R}_{21}\bigg)^\frac{3}{2}}\,\mathbf{R}_{21} \quad,\mathbf{R}_{21}=\mathbf{R}_2-\mathbf{R}_1\\ &\mathbf{F}_{13}=\frac{m_1\,m_3\,G}{ \bigg(\mathbf{R}_{31}\cdot\mathbf{R}_{31}\bigg)^\frac{3}{2}}\,\mathbf{R}_{31} \quad,\mathbf{R}_{31}=\mathbf{R}_3-\mathbf{R}_1\\ &\mathbf{F}_{23}=\frac{m_2\,m_3\,G}{ \bigg(\mathbf{R}_{32}\cdot\mathbf{R}_{32}\bigg)^\frac{3}{2}}\,\mathbf{R}_{32} \quad,\mathbf{R}_{32}=\mathbf{R}_3-\mathbf{R}_2 \end{align*} \begin{align*} &\text{and}\\ &\mathbf{R}_i=\begin{bmatrix} x_i \\ y_i \\ \end{bmatrix}\quad, i=1..3\\ &\text{if you want to work with units $~[1]~$ thus}\\\\ &\mathbf{R}_i=L\,\begin{bmatrix} u_i \\ v_i \\ \end{bmatrix}\\ &\frac{d}{dt}=\frac{1}{T}\frac{d}{d\tau} \\ &\text{L and T are scaling factors with the units [m] and [s]} \\\\ &\text{the scale forces}\\ &\mathbf{F}_{kj}=\frac{m_k\,m_j\,G}{ L^3\,\bigg(\mathbf{R}_{jk}\cdot\mathbf{R}_{jk}\bigg)^\frac{3}{2}} \,L\,\mathbf{R}_{jk} \\ &\text{the EOM's}\\ &m_1\,\frac{L}{T}\,\frac{d^2}{d\tau^2}\mathbf{{R}_1}= \frac{1}{L^2}\left(\mathbf{F}_{12}+\mathbf{F}_{13}\right)\\\\& \Rightarrow\\& m_1\,\frac{L^3}{T}\frac{d^2}{d\tau^2}\mathbf{{R}_1}= \mathbf{F}_{12}+\mathbf{F}_{13}\\ &m_2\,\frac{L^3}{T}\,\frac{d^2}{d\tau^2}\mathbf{{R}_2}=-\mathbf{F}_{12}+\mathbf{F}_{23}\\ &m_3\,\frac{L^3}{T}\,\frac{d^2}{d\tau^2}\mathbf{{R}_3}=-\mathbf{F}_{13}-\mathbf{F}_{23}\\ \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.