anonymous
  • anonymous
d^2r/dt^2 = (4pi^2r)/|r|^2 Find 2 scalar differential equations for x(t), y(t)
Physics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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: \[\Large \left\{ \begin{gathered} \frac{{{d^2}x}}{{d{t^2}}} = 4{\pi ^2}\frac{x}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \hfill \\ \frac{{{d^2}y}}{{d{t^2}}} = 4{\pi ^2}\frac{y}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \hfill \\ \end{gathered} \right.\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.