# From Gravity Equation-of-Motion to General Solution in Polar Coordinates

I'm having trouble getting the general solution of this differential equation.

The gravitational equation of motion is, for constants $M$ and $G$ and position vector $\vec{r}$,

$$\frac{d^2}{d t^2}\vec{r} = -\frac{MG}{r^2}\hat{r}$$

By using 2D polar coordinates (one angle $\theta$ and one "distance from origin" $r$), one can calculate $\frac{d^2}{dt^2}\vec{r}$ by taking two time derivatives of $\vec{r}=r\hat{r}$. The hat notiation $\hat{r}$ is the unit vector pointing from $\vec{r}$ in the direction of increasing $r$. Similarly, $\hat{\theta}$ is the unit vector pointing from $\vec{r}$ in the direction of increasing $\theta$. By appropriate application of the chain rule of derivatives (remembering to take derivatives of the unit vectors themselves as well), one can derive that (dot means time-dirivative)

$$\frac{d^2}{d t^2}\vec{r} = -\frac{MG}{r^2}\hat{r} = \hat{\theta}(2\dot{r}\dot{\theta}+r\ddot{\theta})+\hat{r}(\ddot{r}-r\dot{\theta}^2)$$

Which means

$$$$-\frac{MG}{r^2}=\ddot{r}-r\dot{\theta}^2 \tag{1}$$$$

and

$$$$0=2\dot{r}\dot{\theta}+r\ddot{\theta} \tag{2}$$$$ The right hand side of the last equation (2) turns out to be the time derivative of the angular-momentum-per-unit-mass $h$ (to a factor of $r$): $$\frac{d}{d t}h=\frac{d}{d t}(r^2\dot{\theta})=2r\dot{r}\dot{\theta}+r^2\ddot{\theta}=0$$ Showing that h is constant in time.

My question is, using what has been laid out, how does one combine the two differential equations (1) and (2) to get something that looks like it can be solved? I have looked at references, but certain steps in the derivations seem to lack explanation.

From what I've seen, it seems important to eliminate $t$ from the equations (1) and (2) to yield an equation with only $r$ and $\theta$.

Thank you for any help, I have been tearing my hair out over this.

---Update--- It appears the substitution $u=\frac{1}{r}$ leads to a $\ddot{r}$ in terms of $\frac{d^2 u}{d^2 \theta}$.

Use the substitution $u=\frac 1r$ to get a differential equation involving $u$ and $\theta$

$$\frac{dr}{dt}==-\frac{1}{u^2}\frac{du}{d\theta}\frac{d\theta}{dt}=-h\frac{du}{d\theta}$$

Where $h=r^2\dot\theta$. Now differentiate again:

$$\frac{d^2r}{dt^2}=\frac{d}{dt}(-h\frac{du}{d\theta})=-h\frac{d}{d\theta}(\frac{du}{d\theta})\frac{d\theta}{dt}=-h^2u^2\frac{d^2u}{d\theta^2}$$

• $\frac{\partial^2 r}{\partial^2 t} = \frac{\partial}{\partial t} (-\frac{1}{u^2}\frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial t})$ ... I'm not quite sure how to do this. – OrangeSherbet Aug 4 '15 at 17:05
• $=- h^2u^2\frac{d^2u}{ds^2}$ – David Quinn Aug 4 '15 at 17:08
• I'm struggling with figuring out how to take the derivative of that. It seems to me like $\frac{\partial}{\partial t} f(u,\frac{\partial u}{\partial \theta}, \frac{\partial \theta}{\partial t})$ – OrangeSherbet Aug 4 '15 at 17:17
• Chain rule. See my previous comment – David Quinn Aug 4 '15 at 17:18
• I know how to apply the chain rule with something like $u=\frac{1}{r}$, since $r=r(u)$,$\frac{\partial r}{\partial t}=\frac{\partial r}{\partial u}\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial t}$ but now we have $f = -\frac{1}{u^2}\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial t} = f(u,\frac{\partial u}{\partial \theta},\frac{\partial \theta}{\partial t})$ and I'm not sure how the chain rule extends to this many dependencies. Your previous comment appears to be the answer, not the process :( – OrangeSherbet Aug 4 '15 at 17:32

Ok here is the answer, thanks to the intermediate step provided by David Quinn.

Remember throughout that we have already proved by Equation (2) that the quantity $r^2\dot{\theta}=h$ does not change in time; it is constant.

It turns out that if you plot $1/r$ vs $\theta$ for something like an elliptical orbit, you see a graph of an offset sine wave. Using this as inspiration, let $r=\frac{1}{u}$. Since for a given orbit, $r$, $\theta$, and $u$ are all completely dependent on $t$ ($r$ is a function of only $u$, $u$ is a function of only $\theta$, $\theta$ is a function of only $t$), by the chain rule we can legally write

$$\dot{r} =\frac{dr}{dt} = \frac{dr}{du}\frac{du}{d\theta}\frac{d\theta}{dt} = -\frac{1}{u^2}\dot{\theta}\frac{du}{d\theta} = -h\frac{du}{d\theta}$$

Remembering that $h$ is proven to be constant, and all variables are completely dependent on time, and the order of derivatives can be changed, we can write

$$\ddot{r} = \frac{d^2r}{dt^2}=\frac{d}{dt}\frac{dr}{dt}=\frac{d}{dt}(-h\frac{du}{d\theta})=-h(\frac{d}{dt}\frac{du}{d\theta}) = -h\frac{d}{d\theta}\frac{du}{dt}=-h(\frac{d}{d\theta}\frac{du}{d\theta})\frac{d\theta}{dt}$$

Substituting $\dot{\theta}=hu^2$ yields

$$\ddot{r}=-h^2u^2\frac{d^2u}{d\theta^2}$$

Replacing $\ddot{r}$ and $\dot{\theta}$ and $r$ in Equation (1) yields an equation between only $\theta$ and $u$ $$-MGu^2 = - h^2u^2\frac{d^2u}{d\theta^2} - \frac{1}{u}h^2u^4$$

which simplifies to $$\frac{MG}{h^2} = \frac{d^2 u}{d\theta^2} + u$$ which is the harmonic oscillator equation

$$m\ddot{x} + \lambda\dot{x} + kx = 0$$ with $x=u$, $t=\theta$, $m=1$, $\lambda=0$, $k=1$, and an offset of $\frac{MG}{h^2}$. From here, solve the harmonic oscillator equation.