## anonymous one year ago d^2r/dt^2 = (4pi^2r)/|r|^2 Find 2 scalar differential equations for x(t), y(t)

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1. anonymous

$\frac{ d^{2}\vec{r} }{ dt^{2} } = \frac{ 4\pi^{2}\hat{r} }{ |\vec{r}|^{2} }$

2. IrishBoy123

i think you can get $$\ddot x(t) = 4 \pi^2 \frac{x}{\sqrt{x^2 + y^2}}$$ just by ploughing through mechanically, and you get the same for y, pattern matched, and you'd then need polar to take it further. but i have no idea if that contextually makes any sense. looks like a repulsive radial field, but clutching at straws maybe.

3. IrishBoy123

@Michele_Laino will have an interesting opinion

4. Michele_Laino

by definition we have: $\Large \hat r = \frac{{\vec r}}{r},\quad r = \left| {\vec r} \right|$ so, if we multiply both numerator and denominator of the right side, by r, we can write: $\large \ddot \vec r = \frac{{{d^2}\vec r}}{{d{t^2}}} = 4{\pi ^2}\frac{{\vec r}}{{{r^3}}},\quad r = \sqrt {{x^2} + {y^2}}$

5. Michele_Laino

oops.. $\Large \frac{{{d^2}\vec r}}{{d{t^2}}} = 4{\pi ^2}\frac{{\vec r}}{{{r^3}}},\quad r = \sqrt {{x^2} + {y^2}}$

6. Michele_Laino

now, using the cartesian coordinates, or the components of the vector r, we can write: $\Large \vec r = \left( {x,y} \right)$ so substituting into the differential equation, we get: $\Large \left( {\frac{{{d^2}x}}{{d{t^2}}},\frac{{{d^2}y}}{{d{t^2}}}} \right) = 4{\pi ^2}\frac{{\left( {x,y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}$

7. Michele_Laino

and finally we have the subsequent pair of differential equations: $\Large \left\{ \begin{gathered} \frac{{{d^2}x}}{{d{t^2}}} = 4{\pi ^2}\frac{x}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \hfill \\ \frac{{{d^2}y}}{{d{t^2}}} = 4{\pi ^2}\frac{y}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \end{gathered} \right.$