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I have the following problem that I don't know how to solve:

Let $\textbf{r}(t)$ be a curve and let $r(t):= |\textbf{r}(t)|$. Let the curve $\textbf{r}(t)$ be defined by:

$$\textbf{a}(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$

Where $\textbf{a}(t) = \textbf{r}''(t)$ is the acceleration of the particle and $G,M \in \mathbb{R}$. Suppose that $\textbf{r}(t)$ and $\textbf{r}'(t)$ are not parallel.

The objective is to determine the trajectory $\textbf{r}(t)$ in polar coordinates.

So basically we have the following equation:

$$\textbf{r}''(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$

The thing is that I don't know how I should proceed from now on. If we'd have just normal scalar function this would be a differential equation and I think I would be able to solve it, but like this I have no idea how to approach the problem. How can I solve this?


Edit: If $$\textbf{r}(t):=(x(t),y(t))$$ then the equation turns into the following system of equations:

$$x''=\frac{GM}{(\sqrt{x^2 + y^2})^3}x$$ $$y''=\frac{GM}{(\sqrt{x^2 + y^2})^3}y$$

I don't know if this helps but I think that this turns thing easier perhaps.

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    $\begingroup$ If you had a minus sign on the right side, this would be the Newton equations for planetary orbits and your solution would be the Kepler laws. Could you check this? $\endgroup$ Jun 16, 2020 at 18:22
  • $\begingroup$ I thought the same, but there is no minus sign $\endgroup$ Jun 16, 2020 at 18:41
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    $\begingroup$ I would have solve for : $$p''=x+iy=\dfrac{GMp}{(pq) ^{3/2} } $$ $$q''=x-iy=\dfrac{GMq}{(pq) ^{3/2} } $$ $\endgroup$
    – EDX
    Jun 16, 2020 at 18:41
  • $\begingroup$ Can you explain this pls? What are the $p$ and $q$? $\endgroup$ Jun 16, 2020 at 18:43

1 Answer 1

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Your equation is wrong (you forgot the minus sign). It should instead be $$\ddot{\mathbf{r}}=\frac{-GM}{r^3}\mathbf{r}$$ Or, $$\ddot{\mathbf{r}}=\frac{-GM}{r^2}\hat{\mathbf{r}} ~~~~~(1)$$ With that out of the way, let's get started. It is well known that in polar coordinates, $$\ddot{\mathbf{r}} =\left(\ddot{r} -r\dot{\theta }^{2}\right)\hat{\mathbf{r}} +( r\ddot{\theta } +2\dot{r}\dot{\theta })\widehat{\mathbf{e}_{\theta }}$$ However, since the gravitational force is entirely radial, in our case this is simply $$\ddot{\mathbf{r}} =\left(\ddot{r} -r\dot{\theta }^{2}\right)\hat{\mathbf{r}}$$ Then, returning to (1), $$\ddot{r} -r\dot{\theta }^{2}=\frac{-GM}{r^2}$$ Now here's the clever part: We'll make a change of variable $u \equiv \frac{1}{r}$. Therefore $$\dot{r}=\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{-1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{-1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t}=\frac{-1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\dot{\theta}$$ We now recall the fact that $ r\ddot{\theta } +2\dot{r}\dot{\theta }=0$ and also observe that $$r\ddot{\theta } +2\dot{r}\dot{\theta }=\frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}t}(r^2\dot{\theta})=0$$ Therefore $r^2\dot{\theta}$ is a conserved quantity, and we make the definition $h \equiv r^2\dot{\theta}=\frac{1}{u^2}\dot{\theta}$. Therefore, $$\dot{r}=\frac{-1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\dot{\theta}=\frac{-1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{h}{r^2}=-h\frac{\mathrm{d}u}{\mathrm{d}\theta}$$ So, $$\ddot{r}=\frac{\mathrm{d}}{\mathrm{d}t}\left(-h\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)=-h^2u^2\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}$$ Then returning to (1) again, we can substitute in our new variables to obtain $$-h^2u^2\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}-\frac{1}{u}\left(hu^2\right)^2=-GMu^2$$ $$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}+u=\frac{GM}{h^2}$$ This is a second order linear constant coefficient ODE for $u(\theta)$ which has the following solution: $$u(\theta)=A\sin(\theta)+B\cos(\theta)+\frac{GM}{h^2}=\frac{1}{r(\theta)}$$ And if we let $r(\theta)=r_{\text{max}}$ at $\theta=0$ and $r'(\theta)=0$ at $\theta=0$, then we find that $A=0$ and $B=\frac{1}{r_{\text{max}}}-\frac{GM}{h^2}$ Thus $$r(\theta)=\frac{1}{B\cos(\theta)+\frac{GM}{h^2}}$$ If we define $C \equiv \frac{h^2B}{GM} = \frac{h^2}{GMr_{\text{max}}}-1$, $$r(\theta)=\frac{h^2/GM}{1+C\cos(\theta)}$$ More work is needed to express this in terms of the semi minor axis $a$, the semi major axis $b$, and the eccentricity $\epsilon$, but I hope this is satisfactory enough. Finding $r(\theta)$ instead of $r(t)$ is considerably more useful, so I would recommend to stick to this sort of formulation. Please ask if anything is unclear!


THE PARAMETRIC FORM (if you're interested)

The parametric form of the ellipse is $$\mathbf{r}(t)=a\cos(t)\hat{\mathbf{i}}+b\sin(t)\hat{\mathbf{j}}$$ Which in polar coordinates is $$\mathbf{r}(t)=\sqrt{a^2\cos^2(t)+b^2\sin^2(t)}\hat{\mathbf{r}}+\operatorname{atan2}\left(b\sin(t),a\cos(t)\right)\widehat{\mathbf{e}_{\theta}}$$ Information on $\operatorname{atan2}$ can be found here

EDIT: A more general analysis:

Suppose our equation is $$\ddot{\mathbf{r}}=\frac{D}{r^3}\mathbf{r}$$ Where $D \in \mathbb{R}$ is some positive or negative constant. Then by similar analysis as before we reach $$r(\theta)=\frac{1}{B\cos(\theta)-\frac{D}{h^2}}$$ The solutions behave rather differently based on the sign of $D$. If it is negative, we get elliptical orbits as before. If $D$ is positive, however, the "orbits" (bit of an abuse of terminology, as they aren't periodic) are either parabolic or hyperbolic, depending on $h$, $r(0)$, and $r'(0)$.

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    $\begingroup$ Thank you but It is not wrong. There is no minus sign in the equation $\endgroup$ Jun 16, 2020 at 22:09
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    $\begingroup$ @Eduude It is you who is wrong. $\endgroup$
    – amWhy
    Jun 19, 2020 at 19:33
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    $\begingroup$ It seems kind of crazy that there would be no minus sign there. The $GM$ hints that this is a question about planetary motion... Which doesn't make much sense without the minus sign. The system is pretty boring without the minus sign anyway, as particles simply fly off in basically straight lines away from the origin. $\endgroup$
    – K.defaoite
    Jun 19, 2020 at 21:19
  • $\begingroup$ What I don't understand is why the minus sign matters. You specify that $G,M\in \mathbb R$, with no other restrictions. If you change the sign of $G$ or $M$, you can then add a minus sign. @Eduude $\endgroup$ Jun 19, 2020 at 21:37
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    $\begingroup$ Thank you @K.defaoite $\endgroup$ Jun 21, 2020 at 10:10

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