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In polar coordinates, we can write:

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

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

$$0=2\dot{r}\dot{\theta}+r\ddot{\theta}\tag{2}$$

We also know that $h=r^2\dot{\theta}$ is a constant. This can be shown by math or noted from the fact that gravity doesn't exert torque.

However, in polar coordinates we can write:

$\hat{r}= \cos\theta\space \hat{i}+\sin\theta\space\hat{j}$

$\hat{\theta}= -\sin\theta\space \hat{i}+\cos\theta\space\hat{j}$

Thus, taking derivatives with respect to $\theta$, we get:

$\frac{d\hat{r}}{d\theta}= -\sin\theta\space \hat{i}+\cos\theta\space\hat{j}=\hat{\theta}$

$\frac{d\hat{\theta}}{d\theta}= -\cos\theta\space \hat{i}-\sin\theta\space\hat{j}=-\hat{r}$

Then,

$\frac{d\hat{r}}{dt}=\frac{d\hat{r}}{d\theta}\frac{d\theta}{dt}= \dot{\theta}\space\hat{\theta}\tag{3}$

$\frac{d\hat{\theta}}{dt}=\frac{d\hat{\theta}}{d\theta}\frac{d\theta}{dt}= -\dot{\theta}\space\hat{r} \tag{4}$

So, reorganizing (2) as:

$\hat{r}=-\frac{\frac{d\hat{\theta}}{dt}}{\dot{\theta}}\tag{5}$

We substituting (4) in (1) we get:

$$\frac{d^2\vec{r}}{dt^2}=\frac{-GM}{r^2}\hat{r}=\frac{-GM}{r^2}\frac{\frac{-d\hat{\theta}}{dt}}{\dot{\theta}}=\frac{GM}{h}\frac{d\hat{\theta}}{dt}$$

Since h is constant we can integrate. Integrating $\frac{d^2\vec{r}}{dt^2}$ with respect to t, we can say:

$$\frac{d\vec{r}}{dt}=\frac{GM}{h}\hat{\theta}+\vec{u}$$

Now, to find the tangential component of $\frac{d\vec{r}}{dt}$, we can take the dot product of $\frac{d\vec{r}}{dt}$ and $\hat{\theta}$.

$$v_T=\frac{d\vec{r}}{dt}\cdotp\hat{\theta}=\hat{\theta}\cdotp(\frac{GM}{h}\hat{\theta}+\vec{u})$$

$$=\hat{\theta}\cdotp(\frac{GM}{h}\hat{\theta}+\vec{u})$$$$=\frac{GM}{h}+u\cos\theta$$

My confusion is because what I have shown is that there is no dependence on $r(t)$ in the tangential component of velocity. If the radius is changing but velocity remains constant, then h must be changing. So how can angular momentum be conserved? Is this a contradiction? Is there a flaw in my argument somewhere?

Also, one last note: $$\frac{dv_T}{dt}=\frac{d}{dt}(\frac{GM}{h}+u\cos\theta)$$$$=u\sin\theta\,\dot{\theta}\tag{6}$$

$$\frac{dv_T}{dt}=\frac{d}{dt}(\dot{r}\dot{\theta})$$$$=\ddot{r}\dot{\theta}+\dot{r}\ddot{\theta}$$

Since $0=2\dot{r}\dot{\theta}+r\ddot{\theta}$ in (2),

$$\frac{dv_T}{dt}=\dot{r}\dot{\theta}+r\ddot{\theta}$$$$=-\dot{r}\dot{\theta}\tag{7}$$

Finally, combining 6 and 7:

$$\frac{dv_T}{dt}= u\sin\theta\,\dot{\theta}=-\dot{r}\dot{\theta}$$ $$u\sin\theta=-\dot{r}$$

This can't be right! Integrating gives a non-conic: $$r=u\cos\theta+C$$

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  • $\begingroup$ One thing that is off is $\hat{\theta} \cdot \vec{u} = - u_x \sin \theta + u_y \cos \theta $. $\endgroup$ – secavara Jul 17 '20 at 18:31
  • $\begingroup$ @secavara True, except I don’t see how that could make the L/mr dependence that you’d expect for velocity $\endgroup$ – Thomas Clark Jul 17 '20 at 18:34
  • $\begingroup$ I don't see a mayor conflict between $\frac{d \vec{r}}{d t} \cdot \hat{\theta} = \frac{G M}{h} - u_x \sin \theta + u_y \cos \theta = \frac{L}{m r} = \frac{h}{r}$ and the general parameterization for an ellipse relative to focus in link. $\endgroup$ – secavara Jul 17 '20 at 18:58
  • $\begingroup$ Or a general conic section, to be more precise: link. $\endgroup$ – secavara Jul 17 '20 at 19:09
  • $\begingroup$ You're right. There isn't a serious conflict. I still don't see how $\frac{d\vec{r}}{dt}\cdot\hat{\theta}$ is impacted by $r$. I'll edit my question... $\endgroup$ – Thomas Clark Jul 17 '20 at 19:35
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Got it!

The dependence is there, although not obvious.

Referring to $v_T=\frac{MG}{h}+u\cos\theta$, we can note that the path will be some sort of conic section, or at least something that can be described in polar coordinates: $$r=f(\theta)$$ $$\theta=f^{-1}(r)$$ Thus, $$v_T=\frac{MG}{h}+u\cos\, (f^{-1}(r))$$

There's the dependence for what it's worth!

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