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- Thread starter dwsmith
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- #2

- Jan 26, 2012

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Let $x_{1}=x$ and $x_{2}=x'$. Then you have the first-order system

\begin{align*}

x_{1}'&=x_{2}\\

x_{2}'&=-\frac{\mu}{ \left(x_{1}^{2}+y^{2}+z^{2} \right)^{3/2}}.

\end{align*}

What are $y$ and $z$ doing? Are they independent functions of time?

\begin{align*}

x_{1}'&=x_{2}\\

x_{2}'&=-\frac{\mu}{ \left(x_{1}^{2}+y^{2}+z^{2} \right)^{3/2}}.

\end{align*}

What are $y$ and $z$ doing? Are they independent functions of time?

Last edited:

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- #3

- Mar 5, 2012

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Can it be that you actually have: $r''(t)=- {\mu \over r^3}$?

If that is the case you can do the following:

$\ddot r=- {\mu \over r^3}$

Multiply left and right with $2 \dot r$.

$2\ddot r \dot r = -2 {\mu \over r^3} \dot r$

$\dot r^2 = {\mu \over r^2} + C$

- Jan 26, 2012

- 183

My guess - there are two more equations that go with this one!

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There but if it was straight forward in changing the first equation, I would have done the other two. Will 2 other coupled equations alter how they are re-written as first order equations?My guess - there are two more equations that go with this one!

$$

y''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^3}y

$$

$$

z''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^3}z

$$

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- #6

I had a typo by the wayLet $x_{1}=x$ and $x_{2}=x'$. Then you have the first-order system

\begin{align*}

x_{1}'&=x_{2}\\

x_{2}'&=-\frac{\mu}{ \left(x_{1}^{2}+y^{2}+z^{2} \right)^{3/2}}.

\end{align*}

What are $y$ and $z$ doing? Are they independent functions of time?

$$

x''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^2}x

$$

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- #7

- Jan 26, 2012

- 4,198

$$x''(t)=-\frac{\mu}{(x^{2}+y^{2}+z^{2})^{3/2}}\,x,$$

then you can let $x_{1}$ and $x_{2}$ be defined as above, and let $y_{1}=y$ and $y_{2}=y'$, as well as $z_{1}=z$ and $z_{2}=z'$. Then you get the following system:

\begin{align*}

x_{1}'&=x_{2}\\

y_{1}'&=y_{2}\\

z_{1}'&=z_{2}\\

x_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,x_{1}\\

y_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,y_{1}\\

z_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,z_{1}.

\end{align*}