anonymous
  • anonymous
d^2r/dt^2 = (4pi^2r)/|r|^2 Find 2 scalar differential equations for x(t), y(t)
Physics
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ d^{2}\vec{r} }{ dt^{2} } = \frac{ 4\pi^{2}\hat{r} }{ |\vec{r}|^{2} }\]
IrishBoy123
  • 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.
IrishBoy123
  • IrishBoy123
@Michele_Laino will have an interesting opinion

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Michele_Laino
  • 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: \[\large \ddot \vec r = \frac{{{d^2}\vec r}}{{d{t^2}}} = 4{\pi ^2}\frac{{\vec r}}{{{r^3}}},\quad r = \sqrt {{x^2} + {y^2}} \]
Michele_Laino
  • Michele_Laino
oops.. \[\Large \frac{{{d^2}\vec r}}{{d{t^2}}} = 4{\pi ^2}\frac{{\vec r}}{{{r^3}}},\quad r = \sqrt {{x^2} + {y^2}} \]
Michele_Laino
  • 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: \[\Large \left( {\frac{{{d^2}x}}{{d{t^2}}},\frac{{{d^2}y}}{{d{t^2}}}} \right) = 4{\pi ^2}\frac{{\left( {x,y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}\]
Michele_Laino
  • Michele_Laino
and finally we have the subsequent pair of differential equations: \[\Large \left\{ \begin{gathered} \frac{{{d^2}x}}{{d{t^2}}} = 4{\pi ^2}\frac{x}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \hfill \\ \frac{{{d^2}y}}{{d{t^2}}} = 4{\pi ^2}\frac{y}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \end{gathered} \right.\]

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