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Let $\vec{r}(t)=(x(t),y(t))$ , $\dot{\vec{r}}=(v_x(t),v_y(t))$ and put the sun at the origin. Using Newtons second law show that the following equations hold- $$\dot{x}=v_x\\ \dot{y}=v_y\\ \dot{v}_x=-k\frac{x}{(x^2+y^2)^{\frac{3}{2}}}\\ \dot{v}_y=-k\frac{y}{(x^2+y^2)^{\frac{3}{2}}}.$$

In and earlier question we were given the gravitational potential energy as $V(\vec{r})=-k\frac{m}{r}$ where $m$ and $k$ are constant. I am assuming that this is also relevant to the above question.

I am unsure in how to approach the problem. In the first equation for example do I assume that $x$ is the polar coordinate $r\cos\theta$? In which case how to I take the derivative with respect to time? Likewise for the 3rd and 4th equations, I am guessing that $|\vec{r}|=x^2+y^2$ but I don't understand how one arrives at the given derivative.

Thankful for all help.

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    Are you just suppose to derive these equation from Newtons law of gravitation and Newtons second law or solve the resulting equations? – 2017-02-15
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    Going to polar coordinates is something you do if you want to solve the equations. You don't need that to derive them. Start with Newton's law of gravitation which says that the force on a body from the sun is $\vec{F} = -\vec{\nabla} V(\vec{r})$. Now compute $\vec{\nabla} V$ and combind this with Newton's second law $\vec{F} = m\vec{a} = m \frac{d^2\vec{r}}{dt^2}$ to get the desired equations. – 2017-02-15
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    Thank you for your reply Winther. So if I am understanding you correctly then $$-\vec{\nabla} V(\vec{r})=m\vec{a}\\ \iff -k\frac{m}{r^3}\vec{r}=m\frac{\partial^2\vec{r}}{\partial t^2}\\ \iff -k\frac{\vec{r}}{r^3}=\frac{\partial^2\vec{r}}{\partial t^2}\\ \iff \vec{r}=-\frac{r^3}{k}\frac{\partial^2\vec{r}}{\partial t^2} .$$ Assuming my calculations are correct what is the next step? I am afraid I do not have a good grasp on the problem, I am not really sure what the terms $\dot{x}$ and $v_x$ represent? – 2017-02-15
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    Yes it's correct, but I would write it the other way around $\frac{d^2\vec{r}}{dt^2} = - \frac{k\vec{r}}{r^3}$. Now $\vec{r} = (x,y)$ and $r = (x^2+y^2)^{1/2}$ which can be used to write the equations for $\ddot{x}$ and $\ddot{y}$ (you get one equation for each component of $\vec{r}$). The final step is simply rewriting a second order ODE to a system of two first order ODEs by defining $V = \frac{dX}{dt}$ like e.g. $\frac{d^2 X}{dt^2} = B$ gives the system $\frac{dX}{dt} = V$ and $\frac{dV}{dt} = B$. This is true since $\frac{d^2X}{dt^2} = \frac{d}{dt} \frac{dX}{dt} = \frac{d}{dt}V = B$ – 2017-02-15

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