First we start with the vis-visa equation

${ v }^{ 2 }=GM\left( \dfrac { 2 }{ r } -\dfrac { 1 }{ a } \right)$

where $v$ is the tangential velocity of the body on the Keplerian trajectory.

For a parabola where $a\rightarrow \infty$ , this becomes

${ v }^{ 2 }=GM\left( \dfrac { 2 }{ r } \right)$

At the perihelion, where the maximum tangential velocity is $v_0$

${ { v }_{ 0 } }^{ 2 }=GM\left( \dfrac { 2 }{ a(1-e) } \right)$

or

$a(1-e)=\dfrac { 2GM }{ { { v }_{ 0 } }^{ 2 } }$

Then we use the polar equation for Keplerian trajectories

$r=\dfrac { a(1-{ e }^{ 2 }) }{ 1+eCos(\theta ) }$

or

$r=a(1-e)\dfrac { 1+e }{ 1+eCos(\theta ) }$

After substitution, we have

$r=\dfrac { 2GM }{ { { v }_{ 0 } }^{ 2 } } \dfrac { 1+e }{ 1+eCos(\theta ) }$

Since

${ v }^{ 2 }=GM\left( \dfrac { 2 }{ r } \right)$

we have

${ v }^{ 2 }={ { v }_{ 0 } }^{ 2 }\dfrac { 1+eCos(\theta ) }{ 1+e }$

Finally we use the angular momentum expression for bodies in Keplerian trajectories, which is a constant

${ L }^{ 2 }={ r }^{ 2 }{ { v }_{ t } }^{ 2 }{ { m }^{ 2 } }=GM{ m }^{ 2 }a(1-{ e }^{ 2 })$

where $v_t$ is the transverse velocity of the body on the Keplerian trajectory, and $m$ is the mass of the body. With substitution and letting $e=1$ , this becomes

${ \left( \dfrac { 2GM }{ { { v }_{ 0 } }^{ 2 } } \dfrac { 1+e }{ 1+eCos(\theta ) } \right) }^{ 2 }{ { v }_{ t } }^{ 2 }={ \left( \dfrac { 2GM }{ { v }_{ 0 } } \right) }^{ 2 }$

or

${ { v }_{ t } }^{ 2 }={ { { v }_{ 0 } }^{ 2 }\left( \dfrac { 1+eCos(\theta ) }{ 1+e } \right) }^{ 2 }$

The radial velocity is thus

${ v }_{ r }=\sqrt { { { v } }^{ 2 }-{ { v }_{ t } }^{ 2 } } ={ v }_{ 0 }\sqrt { \left( \dfrac { 1+eCos(\theta ) }{ 1+e } \right) -{ \left( \dfrac { 1+eCos(\theta ) }{ 1+e } \right) }^{ 2 } }$

which, letting $e=1$ , simplifies to

${ v }_{ r }={ v }_{ 0 }\dfrac { Sin(\theta ) }{ 2 }$

so that the answer is

$\dfrac { \pi }{ 2 } =90°$

for maximum $v_r$

The difficulty is handling the special case where $e=1$ and $a\rightarrow \infty$ , but such that $a(1-e)$ is a finite real.

In the general case, the radial velocity for any Keplerian trajectory is

${ v }_{ r }=\dfrac { \sqrt { GM } eSin(\theta ) }{ \sqrt { a(1-{ e }^{ 2 }) } }$

which, after substituting $a(1-e)=\dfrac { 2GM }{ { { v }_{ 0 } }^{ 2 } }$ and letting $e=1$ will yield the same result as above. However, note that the maximum radial velocity occurs at $\dfrac { \pi }{ 2 } =90°$ for any $e$ .

The fact that $\ddot{\mathbf{r}} = -\tfrac{GM}{r^3}\mathbf{r}$ , where $M$ is the mass of the sun, means that $\mathbf{r} \wedge \ddot{\mathbf{r}} = \mathbf{0}$ , so that $\mathbf{h} = \mathbf{r} \wedge \dot{\mathbf{r}}$ is a constant vector such that $\mathbf{r} \cdot \mathbf{h} = 0$ . Thus the motion of the comet is planar, and $\mathbf{h}$ is conserved. If we define unit vectors $\hat{\mathbf{r}} = \cos\theta \mathbf{i} + \sin\theta\mathbf{j} \hspace{2cm} \hat{\mathbf{t}}\; = \; -\sin\theta\mathbf{i} + \cos\theta\mathbf{j}$ where $\mathbf{ i },\mathbf{j}, \mathbf{k}$ are mutually perpendicular unit vectors, with $\mathbf{i},\mathbf{j}$ in the plane of the comet's motion, with $\mathbf{i}$ pointing in the direction $\vec{OP}$ , then $\begin{aligned} \mathbf{r} & = \; r\hat{\mathbf{r}} \\ \dot{\mathbf{r}} & = \; \dot{r}\hat{\mathbf{r}} + r\dot\theta\hat{\mathbf{t}} \\ \ddot{\mathbf{r}} & = \; \big(\ddot{r} - r\dot\theta^2\big)\hat{\mathbf{r}} + (2\dot{r}\dot{\theta} + r\ddot{\theta}\big)\hat{\mathbf{t}} \end{aligned}$ and so the equation of motion of the comet is $\ddot{r} - r\dot{\theta}^2 \; = \; -GMr^{-2} \hspace{2cm} \tfrac{d}{dt}\big(r^2\dot\theta\big) \; = \; 0$ Thus $r^2\dot\theta = h$ is constant (note that $\mathbf{h} = h\mathbf{k}$ is the conserved vector, and $m\mathbf{h}$ is the angular momentum of the comet) and $\ddot{r} - h^2r^{-3} \; =\; -GMr^{-2}$ Putting $r = u^{-1}$ we have $\begin{aligned} \dot{r} & = \; -u^{-2} \times \tfrac{du}{d\theta} \times \dot\theta \; = \; -h\tfrac{du}{d\theta} \\ \ddot{r} & = \; -h\tfrac{d^2u}{d\theta^2} \times \dot\theta \; = \; - h^2u^2 \tfrac{d^2u}{d\theta^2} \end{aligned}$ and hence $\tfrac{d^2u}{d\theta^2} + u \; = \; \tfrac{GM}{h^2}$ leading to the general solution $u = \tfrac{GM}{h^2} + A\cos\theta + B\sin\theta$ . This has shown the conic nature of the paths.

For this problem we have $u = a^{-1}(1 + \cos\theta)$ and we want to maximize $\dot{r} = -h\tfrac{du}{d\theta} = \tfrac{h}{a}\sin\theta$ . The value of $\dot{r}$ is therefore maximised when $\sin\theta = 1$ , so when $\theta = \boxed{90^\circ}$ .

Cartesian to Polar Transformation

First, we know that a parabola has the property that any vertically incoming ray gets reflected towards its focal point. One can find such a point by noticing that at the point on the parabola where the tangent line is at $45^\circ$ , a vertically incoming ray will get reflected by $90^\circ$ . This means that the $y$ -coordinate of such a point is the $y$ -coordinate of the focal point.

If we use $y=ex^2-a$ as our general expression of our parabola, and the fact that a line at $45^\circ$ has a slope of $1$ , then

$y'=2ex=1$ $\Rightarrow x=\dfrac{1}{2e}$

Substituting back into our original equation, we get

$y=e\left(\dfrac{1}{2e}\right)^2-a$

$\Rightarrow a=\dfrac{1}{4e}$

$\Rightarrow y=\dfrac{1}{4a}x^2-a$

By using the transformation $x=-r\sin\theta, y=r\cos\theta$ , and doing some algebra to solve for $r$ , we get

$r(\theta)=\dfrac{2a}{1+\cos\theta}$

The Equations of Motion

First, we know that the radial velocity is equal to the rate of change of the radius with respect to time. Thus,

$v_r=\dfrac{dr}{dt}=\dfrac{dr}{d\theta}\dfrac{d\theta}{dt}$

$=2a\dfrac{\sin\theta}{(1+\cos\theta)^2}\dfrac{d\theta}{dt}$

Similarly, the rate of change of the radial velocity is the radial acceleration. In addition, since our origin is at the focus of the parabolic trajectory, and the sun is at the focus, then the radial acceleration is the only acceleration experienced by the comet; this acceleration is the acceleration due to gravity.

Since all we care about is the angle at which the maximum radial velocity happens, for simplicity I will regard the product $GM$ from Newton's equation for gravity as equal to $1$ . Thus, the acceleration due to gravity is $g=\dfrac{1}{r^2}$ . Therefore,

$\dfrac{dv_r}{dt}=\dfrac{dv_r}{d\theta}\dfrac{d\theta}{dt}$

$=\dfrac{1}{r^2}=\dfrac{(1+\cos\theta)^2}{4a^2}$

We can now combine the equation of velocity and the equation of acceleration, getting rid of the term $\dfrac{d\theta}{dt}$ , to get the differential equation

$\dfrac{dv_r}{d\theta}=\dfrac{\sin\theta}{2v_ra}$

Through separation of variables, we get

$\int_{v_r(0)}^{v_r(\theta)}vdv=\dfrac{1}{2a}\int_{0}^{\theta}\sin\theta' d\theta'$

Note that the prime on the $\theta$ does not mean derivative. It is just there to express the fact that the $\theta$ we are integrating over is different from the one in the bounds of the integral!

Now, notice that when the comet is at the bottom of the parabola, its direction of motion is perpendicular to its acceleration. This means that the only velocity the comet has at that point is tangential velocity. Thus, $v_r(0)=0$ .

Integrating and solving for $v_r$ we get

$v_r(\theta)=\sqrt{\dfrac{1}{a}(1-\cos\theta)}$

Finding the Maximum Radial Velocity

One can see that the term $1-\cos\theta$ reaches a maximum when $\cos\theta=0\Rightarrow \theta=90^\circ$ .

Since the maximum of our $v_r(\theta)$ function will happen when $1-\cos\theta$ is at its maximum, then $v_{r,max}$ happens when $\boxed{\theta=90^\circ}$ .