Use angular momentum to find the velocity (comet orbit)

In summary: But I don't know what the potential energy...The potential energy would be the same, but the radial component is larger because the comet has more energy at the radius.
  • #1
vbrasic
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3

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##.
 
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  • #2
vbrasic said:

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.
 
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  • #3
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}##.
 
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  • #4
tnich said:
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.
 
  • #5
vbrasic said:
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?
 
  • #6
tnich said:
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?
 
  • #7
vbrasic said:
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 vΦ in terms of v and the angle between r and v?
 
  • #8
tnich said:
Yes, angular momentum is conserved.
I'm not sure what you mean by ρ here.
How would you express vΦ 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.
 
  • #9
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.
 
  • #10
NFuller said:
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}##.
 
  • #11
vbrasic said:
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.
 
  • #12
NFuller said:
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?
 
  • #13
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}##.
 
  • #14
NFuller said:
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}##.
NFuller said:
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?
 
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  • #15
vbrasic said:
Does that seem like an okay way to do it?
Seems like a good idea. Try it and see if the answer is reasonable.
 

1. What is angular momentum?

Angular momentum is a physical quantity that measures the rotational motion of an object. It is a vector quantity and is calculated by multiplying an object's moment of inertia (a measure of its resistance to rotation) by its angular velocity (the rate at which it rotates).

2. How is angular momentum related to comet orbits?

Angular momentum plays a crucial role in determining the shape and size of a comet's orbit. As a comet moves through space, it maintains a constant angular momentum unless acted upon by an external force. This means that as the comet gets closer to the sun, its velocity increases to compensate for the shorter distance, maintaining a constant angular momentum.

3. What is the formula for calculating angular momentum?

The formula for angular momentum is L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity. In the case of a comet orbit, the angular velocity is equal to the speed of the comet divided by its distance from the sun, and the moment of inertia is determined by the mass and shape of the comet.

4. Can angular momentum be used to predict a comet's velocity?

Yes, by using the formula L = Iω and knowing the values for moment of inertia and angular velocity, we can calculate the angular momentum of a comet. This can then be used to determine the comet's velocity at different points along its orbit.

5. How does angular momentum affect the stability of a comet's orbit?

Angular momentum is a conserved quantity, meaning it remains constant unless acted upon by an external force. This means that a comet's orbit will be stable as long as there are no external forces, such as collisions with other objects or gravitational interactions with other celestial bodies, that can disrupt its angular momentum.

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