Equal masses and angular momentum, what is linear velocity ratio of two particles in uniform circular and parabolic orbitals? [closed]

Question

Question

There are two particles with same mass and angular momentum magnitude. Both of them has potential function propertional with $\frac{1}{r}$ under an attractive force. One is moving in a uniform circular orbit and other is moving in a parabolic orbit. What's the ratio of linear velocities $\frac{v_{\text{parabolic}}}{v_{\text{circular}}}$?

My Solution

I chose polar coordinates as coordinate system since motion of the both particles are limited in a plane (due to constant angular momentum).

The potential energy function of both particles are proportional with $\frac{1}{r}$. Also the force is attractive so, it's an attractive inverse-square law force which can also be determined using:

$$\frac{dV}{dr}=\frac{d}{dr}\left(\frac{\left|k\right|}{r}\right)=\left|k\right|\frac{d}{dr}\left(r^{-1}\right)=-\frac{\left|k\right|}{r^2}$$

I know that they have same amount of mass and angular momentum. So I can start with angular momentums.

$$J_{c}=J_{p}$$

$$mr_{c}v_{c}\sin(\vec{r}_{c}, \vec{p}_{c})=mr_{p}v_{p}\sin(\vec{r}_{p}, \vec{p}_{p})$$

We can cancel $m$ in both sides since they are equal. Motion of these two particles is limited in plane, thus sines would give 1.

$$r_{c}v_{c}=r_{p}v_{p}$$

$$\frac{v_{p}}{v_{c}}=\frac{r_{c}}{r_{p}}$$

Answer 1

For an object in circular motion, we have the standard circular motion equation: Under your above assumptions, if $k$ is the absolute value of your proportionality constant:

$$v_c = \sqrt{\frac{k}{r_c}}$$

$$v_c^2 r_c= k$$

An object on a parabolic trajectory, under the influence of an inverse-square centripetal force comes to rest at $r=\infty$. Under your provided assumptions, its constant specific orbital energy is:

$$\varepsilon_p = \frac{v^2}{2} - \frac{k}{r} = 0 $$

This allows you to calculate the velocity at any point in terms of the radial distance: $$v_p = \sqrt{\frac{2k}{r}}$$ $$v_p^2 r = 2k$$

We can now work out the velocity at any point on the parabola in terms of the velocity of the circular orbit $v_c$ and the radial distance $r$

$$\frac{v_p}{v_c}= \sqrt{\frac{2r_c}{r}}$$

Parabolic Trajectory and Circular Orbit with the same specific angular momentum

We know the angular momentum $J$ and the specific angular momentum $h = J/m$ of both the circular orbit, and the parabolic trajectory are equal:

$$h_c = r_c v_c\sin(\theta_c) = r_p v_p \sin \theta_p = h_p$$

Where $\theta_c$ is the angle from $\vec{r_c}$ to $\vec{v_c}$ on the circular orbit, and is constant at $\frac{\pi}{2}$, so $\sin(\theta_c) = 1$

$\theta_p$ is the angle from $\vec{r_p}$ to $\vec{v_p}$, which is not constant on the parabolic trajectory. However, at the periapis (which we'll call $r_q$, with its corresponding velocity $v_q$), $\theta_q$ is $\frac{\pi}{2}$, and $\sin(\theta_q) = 1$

And that gives us two equations, and two unknowns: $$v_q^2r_q = 2v_c^2r_c$$ $$v_q r_q = v_c r_c$$

Which means: $$v_q^2r_q = 2v_c v_q r_q$$

And as a result: $$\frac{v_q}{v_c} = 2 $$

Which means, for a parabolic orbit and a circular orbit around the same body, with the same specific angular momentum, the periapsis velocity of the parabola is twice the orbital velocity of the circle. Given that the angular momentum is equal:

$$\frac{r_q}{r_c} = 0.5 $$

And that gets us the final answer:

$$\frac{v_p}{v_c}= \sqrt{\frac{2r_c}{r}}, r\ge \frac{r_c}{2} $$