<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt


{\bf Question}

Suppose that three particles of equal mass $m$ are placed at the
corners of an equilateral triangle of side length $d.$  Suppose
that these three particles subsequently move from rest under the
influence of their mutual gravitational forces.  Find their speed
when they have moved a distance $\frac{d}{2}$.

\vspace{.25in}

{\bf Answer}

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\put(2.9,.9){$\bullet$}

\put(0,1){\line(1,2){1.5}}

\put(3,1){\line(-1,2){1.5}}

\put(1.4,3.9){$\bullet$}

\put(1.5,4){\line(0,-1){.25}}  %makes triangle%

\put(1.5,3.5){\line(0,-1){.25}}

\put(1.5,3){\line(0,-1){.25}}

\put(1.5,2.5){\line(0,-1){.25}} %dotted line down centre%

\put(.65,2.4){\line(2,-1){.9}}   %line from D to O%

\put(0,0.8){\makebox(0,0){$B$}}

\put(3,0.8){\makebox(0,0){$C$}}

\put(1.5,4.2){\makebox(0,0){$A$}} %labels the dots%

\put(.6,2.6){\makebox(0,0){$D$}}

\put(1.7,2){\makebox(0,0){$O$}}

\put(1.4,3.5){\makebox(0,0){$\frac{\pi}{6}$}}

\put(.8,2.7){\line(2,-1){.3}}

\put(1.1,2.55){\line(-1,-2){.15}}

\put(0,0.5){\vector(1,0){3}}

\put(0,0.5){\vector(-1,0){0}}

\put(1.5,0.3){\makebox(0,0){$r$}} %makes and labels the bottom length%

\put(.8,3.3){\makebox(0,0){$ \frac{r}{2}$}}%the label r by 2%

\put(1.7,2.2){\vector(0,1){1.8}}

\put(1.7,2.2){\vector(0,-1){0}}

\put(1.85,3){\makebox(0,0){$x$}}

\end{picture}

From triangle $ADO$ $\ds x \cos \frac{\pi}{6} = \frac{r}{2}
\hspace{.2in} \Rightarrow x \sqrt 3 = r$

The force on $A$ is directed along $AO,$ and it has magnitude $\ds
\sqrt 3 \frac{Gm^2}{r^2}$

$\ds \Rightarrow m \ddot x = \sqrt 3 \frac{Gm^2}{r^2}
\hspace{.2in} \Rightarrow \ddot x = - \frac{\alpha}{x^2}
\hspace{.2in} \textrm{\ \ where\ \ } \alpha = \frac{Gm}{\sqrt 3}$

Now $\ds \frac{d^2x}{dt^2} = v \frac{dv}{dx} \hspace{.2in}
\Rightarrow \frac{dv}{dx} = - \frac{\alpha}{x^2} \hspace{.2in}
\Rightarrow \frac{1}{2}v^2 = \frac{\alpha}{x} + A \hspace{.1in}
(*)$

The initially conditions are $\ds v = 0, \hspace{.2in} r = d
\hspace{.2in} \Rightarrow x = \frac{d}{\sqrt 3}$

Putting this into equation * gives $\ds 0 =
\frac{\alpha}{\frac{d}{\sqrt 3}} + A \hspace{.2in} \Rightarrow A =
-\frac{\alpha \sqrt 3}{d}$

$$ {\rm therefore \ \ \ } \frac{1}{2} v^2 = \alpha \left(
\frac{1}{x} - \frac{\sqrt 3}{d} \right)$$

At $\ds x = \frac{d}{2}, \ \ \frac{1}{2} v^2 = \alpha \left(
\frac{2}{d} - \frac{\sqrt 3}{d} \right) \Rightarrow v =
\sqrt{\frac{2\alpha}{d}\left(2 - \sqrt 3\right)}$



\end{document}
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