Use angular momentum to find the velocity (comet orbit)

[vbrasic]

Homework Statement

A comet orbits the sun. It's position in polar coordinates is given by, $$r(\phi)=\frac{1.8r_0}{1+0.8\cos{\phi}},$$ where $r_0$ is the position at closest approach. Its velocity at this point is given by $v_0$. Use the concept of angular momentum to find the following:

• The comet's speed at $\phi=180^{\circ}$.
• The components of velocity, $v_r$, and $v_{\phi}$ at $\phi=120^{\circ}$.

Homework Equations

$$L=r\times p$$
$$|L|=m|r||v|\sin{\theta}$$

The Attempt at a Solution

We have that angular momentum at closest approach is given by, $$L=r\times p=m|r||v|\sin{\theta}.$$ At closest approach, we then have that $$L=mr_0v_0\sin{\theta}.$$ If the orbit was a perfect circle, we would know for sure that the angle between $r_0$ and $v_0$ is $90^{\circ}$. However, in this case, the orbit is an ellipse, so I'm unsure how to find $\theta$.

 

[tnich]

Homework Statement

If the orbit was a perfect circle, we would know for sure that the angle between $r_0$ and $v_0$ is $90^{\circ}$. However, in this case, the orbit is an ellipse, so I'm unsure how to find $\theta$.

I suggest that you plot the orbit. That should give you an idea of what the angle is at closest approach.

 

[NFuller]

At the point of closest (and furthest) approach, the radial velocity is zero. This means that $v_{0}$ is directed only in the $\phi$ direction, so $\theta=90^{\circ}$.

 

[vbrasic]

I suggest that you plot the orbit. That should give you an idea of what the angle is at closest approach.

Okay, so we have that, $$r(\phi)=\frac{1.8}{1+0.8\cos{\phi}}.$$ We want to find $\phi$ at $r_0$. So, just plug in $r_0$ and solve for $\phi$. We have, $$r_0=\frac{1.8r_0}{1+0.8\cos{\phi}}\to1+0.8\cos{\phi}=1.8\to\cos{\phi}=1\to\phi=0,$$ which is very convenient, because as NFuller said, the tangential velocity is indeed perpendicular to position vector at perhelion and aphelion. So we have that $L_{0}=mr_0v_0$, as $\sin{90}=1$. Then, it remains to solve for $L_{180}$, which is aphelion. We have $L_{180}=m|r||v|,$ as again, $\sin{90}=1$. It remains to find $r(180)$ and plug in and isolate for $|v|$. We have $$r(180)=\frac{1.8r_0}{1-0.8}=9r_0.$$ So, plugging in and equating the two momentums by conservation, we have, $$mr_0v_0=9mr_0|v|,$$ such that $$|v|=\frac{v_0}{9}.$$ That seems okay to me. I have no idea how to proceed for the second part.

 

[tnich]

Okay, so we have that, $$r(\phi)=\frac{1.8}{1+0.8\cos{\phi}}.$$ I have no idea how to proceed for the second part.

You have the definition of angular momentum. What happens to the angular momentum as the comet goes around its orbit?

 

[vbrasic]

You have the definition of angular momentum. What happens to the angular momentum as the comet goes around its orbit?

It must be conserved? I mean. Should I calculate, $\rho(\phi)$ by differentiating expression for position?

 

[tnich]

It must be conserved? I mean. Should I calculate, $\rho(\phi)$ by differentiating expression for position?

Yes, angular momentum is conserved.
I'm not sure what you mean by ρ here.
How would you express v_Φ in terms of v and the angle between r and v?

 

[vbrasic]

Yes, angular momentum is conserved.
I'm not sure what you mean by ρ here.
How would you express v_Φ in terms of v and the angle between r and v?

I'm not exactly sure. I know that $v_{\phi}$ and $v_r$ are orthogonal? Or is this only for circular orbits? If they are orthogonal, then I know the direction of $v_{\phi}$ is orthogonal to the position.

 

[NFuller]

Since you know $L$, one component of the velocity can be found from
$$\mathbf{L}=\mathbf{r}\times\mathbf{p}=m\mathbf{r}\times\mathbf{v}$$
Just remember that only the component of $\mathbf{v}$ which is perpendicular to $\mathbf{r}$ will contribute to the cross product.

 

[vbrasic]

Since you know $L$, one component of the velocity can be found from
$$\mathbf{L}=\mathbf{r}\times\mathbf{p}=m\mathbf{r}\times\mathbf{v}$$
Just remember that only the component of $\mathbf{v}$ which is perpendicular to $\mathbf{r}$ will contribute to the cross product.

Okay. So we know that $v_{\phi}$ is orthogonal to $r$. So, based on conservation of angular momentum, we can compute $v_{120,\phi}$. By conservation of angular momentum, we know that, $$mr_0v_0=mr_{120}|v_{120,\phi}|\sin{90}.$$ We note thath $r_{120}=3.6r_0.$ Plugging in gives $mr_0v_0=3.6mr_0|v_{120,\phi}|\to|v_{120,\phi}|=\frac{v_0}{3.6}$. Then it remains to find $v_{120,r}$.

 

[NFuller]

Then it remains to find v120,rv120,rv_{120,r}.

I would use the fact that total energy is conserved to find the radial component of the velocity.

 

[vbrasic]

I would use the fact that total energy is conserved to find the radial component of the velocity.

But I don't know what the potential energy is?

 

[NFuller]

The potential energy is given by
$$PE=-G\frac{Mm}{r}$$
and the total energy by
$$E=\frac{1}{2}mv^{2}-G\frac{Mm}{r}$$
The problem is that you don't know the mass of the central body $M$. However, since you do know the velocity at apogee and perigee and that $E$ is constant, you can solve for $M$.
$$\frac{1}{2}mv_{0}^{2}-G\frac{Mm}{r_{0}}=\frac{1}{2}mv_{180}^{2}-G\frac{Mm}{r_{180}}$$
Once you have $M$ you can use conservation of total energy to find $v_{120}$.

 

[vbrasic]

The potential energy is given by
$$PE=-G\frac{Mm}{r}$$
and the total energy by
$$E=\frac{1}{2}mv^{2}-G\frac{Mm}{r}$$
The problem is that you don't know the mass of the central body $M$. However, since you do know the velocity at apogee and perigee and that $E$ is constant, you can solve for $M$.
$$\frac{1}{2}mv_{0}^{2}-G\frac{Mm}{r_{0}}=\frac{1}{2}mv_{180}^{2}-G\frac{Mm}{r_{180}}$$
Once you have $M$ you can use conservation of total energy to find $v_{120}$.

The potential energy is given by
$$PE=-G\frac{Mm}{r}$$
and the total energy by
$$E=\frac{1}{2}mv^{2}-G\frac{Mm}{r}$$
The problem is that you don't know the mass of the central body $M$. However, since you do know the velocity at apogee and perigee and that $E$ is constant, you can solve for $M$.
$$\frac{1}{2}mv_{0}^{2}-G\frac{Mm}{r_{0}}=\frac{1}{2}mv_{180}^{2}-G\frac{Mm}{r_{180}}$$
Once you have $M$ you can use conservation of total energy to find $v_{120}$.

I found a neater way to do it I think. We have that, $$v_r=\frac{dr}{dt}.$$ But, $$\frac{dr}{dt}=\frac{dr}{d\phi}\frac{d\phi}{dt},$$ by chain rule. We know that $$v_{\phi}=r\frac{d\phi}{dt}.$$ But we have a function to find $r(\phi)$. So we can just find $$\frac{d\phi}{dt}$$ and $$\frac{dr}{d\phi}$$ at $120$, and then solve for $v_r$ that way. Does that seem like an okay way to do it?

 

[NFuller]

Does that seem like an okay way to do it?

Seems like a good idea. Try it and see if the answer is reasonable.