How is Angular Momentum conserved in orbits?

[farolero]

So a light particle is orbiting a massive particle by gravity.

We take both particles as spot particles.

The light particle makes an eccentric orbit where maximum radius of the orbit equals 2 and minimum radius equals 1.

I suppose the mass of the massive particle such that the speed of the light particle at the farthest radius is 1 m/s

According Newton when you double the radius of an orbit its speed decreases by a factor of square root of two

So when the orbit is at a distance of 1 m its speed will be equal to square root of 2=1.41 m/s

So how is angular momentum conserved here for initial angular momentum is mvr=2 and final angular momentum=1.41?

 

[A.T.]

So a light particle is orbiting a massive particle by gravity...

According Newton...

This is General Relativity not Newton.

 

[Bandersnatch]

This is General Relativity not Newton.

He means a particle that is light, not a particle of light. I.e., a test particle.

The light particle makes an eccentric orbit where maximum radius of the orbit equals 2 and minimum radius equals 1.

I suppose the mass of the massive particle such that the speed of the light particle at the farthest radius is 1 m/s

According Newton when you double the radius of an orbit its speed decreases by a factor of square root of two

Have you actually changed the orbit here, though? What is the difference between the speed of a particle at the periapsis of an elliptical orbit vs the speed of a particle on a circular orbit with the radius equal to the apoapsis distance?

 

[Dale]

According Newton when you double the radius of an orbit its speed decreases by a factor of square root of two

Can you post the formula that you are referring to here. I don't recognize this immediately.

Please post a link or LaTeX formula, not just a jumble of characters, and any additional information that you have on this formula you referenced

 

[farolero]

i got it from here:

https://fisicaunopuradiversion.files.wordpress.com/2013/03/anscircle9.pdf

 

[Nugatory]

According Newton when you double the radius of an orbit its speed decreases by a factor of square root of t

That's for a circular orbit (and note that "THE radius" is not defined for an elliptical orbit).
Also, the magnitude of the angular momentum is only proportional to the product of the distance and the speed in circular orbits; for all others it's proportional to the product of the radius and the transverse component of the velocity. (It just so happens that in a circular orbit the transverse component of the velocity is the speed, so you can get by with using the speed in that special case).

So you're trying to use the circular orbit simplifications on an elliptical orbit, and that won't work. To take on elliptical orbits you're going to have to learn and use the general expression for angular momentum: ##\vec{L}=\vec{r}\times\vec{p}## where ##\vec{p}=m\vec{v}## is the linear momentum.

 

[farolero]

Thanks a lot that was very enlightening but I am not sure conservation of angular momentum would allow me to solve the problem because an eccentric orbit satellite is not an isolated system but interacts with earth.

For example the rotation of Earth is slowed down by tides and in exchange to keep angular momentum constant the moon goes higher.

How do i know that in an eccentric orbit angular moment is not interacting with Earth as it varies radius if i dot take Earth as a point mass but consider it has some moment of inertia?

Myself to solve this proble Id use potential energy and forget all about angular momentum.

 

[Nugatory]

Thanks a lot that was very enlightening but I am not sure conservation of angular momentum would allow me to solve the problem because an eccentric orbit satellite is not an isolated system but interacts with earth.

The mass of a satellite is enough less than the mass of the Earth that we can ignore the effects on the earth, and we still get extremely accurate results. Likewise, when we solve planetary motion problems, we treat the sun as a fixed point mass, and we still get good results.

Myself to solve this proble Id use potential energy and forget all about angular momentum.

You need both to solve the general problem of a small mass orbiting a larger fixed mass under the influence of gravity; if you ignore angular momentum you'll find yourself with more unknowns than you have equations. You'll likely encounter this solution towards the end of the first-semester of a calculus-based physics program; or you can google for "Kepler's laws derivation".

 

[farolero]

i found the formula to solve this problem::v=squareroot of(mu(2/r-1/a))
from:
https://en.wikipedia.org/wiki/Elliptic_orbit

and this is not the conservation of momentum formula

how can it be true this formula and the momentum conservation formula at the same time if they expect different results to solve this problem?

 

[Nugatory]

i found the formula to solve this problem::
v=squareroot of(mu(2/r-1/a))
from:
https://en.wikipedia.org/wiki/Elliptic_orbit

and this is not the conservation of momentum formula

how can it be true this formula and the momentum conservation formula at the same time if they expect different results to solve this problem?

That formula is derived from conservation of energy and conservation of momentum: Start with ##F=Gm_1m_2/r^2##, use conservation of energy and conservation of momentum to calculate the orbital speed of an object in an elliptical orbit, and you can calculate that result. In fact, if you go back to the wikipedia page and follow the link directly above the formula you will see how this is done. (Or, you can get hold of a suitable textbook, something like Kleppner and Kolenkow).

As for the resolution of the paradox that you think you've found: If you do a bit of algebra, you will find that there is a relationship between ##a## and ##r## that makes it possible for that formula to be true and for angular momentum to be conserved (the product of speed and distance is the same at the points of closest approach and greatest distance) if ##a## is chosen properly.

 

[farolero]

Yes my apologies youre right I calculated the a for speeds 1 and 2 and radius 2 and 1 and i obtained and a=0.33

I was wondering now if conservation of momentum couldn't be used to calculate potential energy instead of integrating to infinity.

The particle at 1 m radius is going at 2m/s and at 2 m radius at 1 m/s.

So the diference in cuadratic gravity potential energy between 2 and 1 m radius would be the difference between kinetic energy at both 1 and 2 radius

i just take that instead of shooting the projectile straight from Earth i shoot it in a slight curve that would keep angular momentum conservation true

 

[nasu]

Does not matter how you shoot it. Conservation of momentum in the reference frame centered on Earth is a consequence of the fact that the gravitational force is radial (central) not of the way you initiate the motion.

You could calculate the change in PE by knowing the change in KE, it's just conservation of energy.

 

[farolero]

Ok thanks so let's see if I analize the problem correctly:

By conservation of momentum when the radius of a satellite halves its speed doubles hence its kinetic energy cuadruples.

If the kinetic energy at radius 2 is x at radius 1 would be 4x so i substract both and obtain that when the radius doubles potential energy triples.

would this be right?

 

[farolero]

Also I have another question concerning orbits and momentum:

I throw a 1 kg ball horizontally at speed 0.01 m/s, I consider it starts and orbit interrupted by the ground.

In the initial instant there will be an angular momentum of 1*0.01*6000 km

but in the instant before it touches the ground the vector will be aiming almost down so its angular momentum would be something like 1*0.01*100m(the arm between the vector of the ball and the center of earth)

So how would you explain this?

 

[nasu]

PE is inverse proportional to distance between the bodies. You don't need all this.

Your description is not correct. Neither the conclusion. You calculate the change in KE, which is equal to the change in PE. You cannot draw conclusions about what the actual value does in the way you wrote above.

Note: This is about your post 13.

 

[Nugatory]

By conservation of momentum when the radius of a satellite halves its speed doubles hence its kinetic energy quadruples.

You mean "conservation of angular momentum", not "conservation of momentum". They're different things (and momentum is not conserved in planetary motion because the planet is not a closed system).

If the kinetic energy at radius 2 is x at radius 1 would be 4x so i substract both and obtain that when the radius doubles potential energy triples.

No. The potential energy of an object of mass ##m## at a distance ##r## from a massive object is ##-\mu{m}/r## (using the convention that the potential energy at infinity is zero, the sensible way of working this problem).

 

[Nugatory]

I throw a 1 kg ball horizontally at speed 0.01 m/s, I consider it starts and orbit interrupted by the ground.

That's not "throwing" the ball, it's "dropping" it... although you are correct that it will enter into a very eccentric elliptical orbit that will be interrupted by the ground.

but in the instant before it touches the ground the vector will be aiming almost down so its angular momentum would be something like 1*0.01*100m(the arm between the vector of the ball and the center of earth)

What is the transverse component of that velocity vector at that point? That's what goes into the angular momentum calculation... Not the speed.
And where did you get 100m from? The ball is still near as no never mind 6000 km from the center of the earth.

I said several posts above that you have to learn and use the definition of angular momentum to take on these problems. So we have ##\vec{L}=\vec{r}\times\vec{p}##. What are ##\vec{r}## and ##\vec{p}## at the moment of release and at the moment of impact with the ground? You'll have to write them in polar coordinates.

 

[Dale]

i got it from here:

https://fisicaunopuradiversion.files.wordpress.com/2013/03/anscircle9.pdf

Literally the first word on the page indicates that it is for circular orbits. The formula simply doesn't apply for your situation of an elliptical orbit.

Edit: never mind, I see you already covered that

 

[farolero]

Thanks a lot Nugatory and Dale I think I am beginning to make sense of it:

Angular momentum conservation just refers to the transversal component of the velocity of the particle.

Could I conclude then that to use conservation of angular momentum to obtain velocity is only valid at the perigee and apogee for those are the only places where net velocity equals the transversal component?

 

[Nugatory]

Could I conclude then that to use conservation of angular momentum to obtain velocity is only valid at the perigee and apogee for those are the only places where net velocity equals the transversal component?

No. You could conclude that it is especially easy to use conservation of angular momentum to obtain the velocity at one of those two points when you already know the velocity at the other.

But if you want to understand orbital motion, it makes a lot more sense to use conservation of angular momentum and conservation of energy to work out the equations of orbital motion; you've already posted a link to a wikipedia page that has links to how to do that. And once you've done that, then you know the velocity everywhere in the orbit, not just at perigee and apogee.

 

[Dale]

Could I conclude then that to use conservation of angular momentum to obtain velocity is only valid at the perigee and apogee for those are the only places where net velocity equals the transversal component?

Let me answer that question by stepping back a bit. In order to determine the behavior of a system you need a few pieces of information.

First, you need to know the laws of physics that govern the system. But in addition you also need so-called initial conditions. Because of the form of the laws of physics you need both initial positions and initial velocities. So, for two classical point particles that is 12 initial conditions.

You can fix 6 by choosing your coordinates so that the big mass is at rest in the center. You can fix two more by starting the small mass on the x axis, and one by rotating it so that the initial velocity is in the x-y plane.

So that leaves three initial conditions: the initial separation, the angular momentum, and the energy.

 

[David Neves]

So a light particle is orbiting a massive particle by gravity.The light particle makes an eccentric orbit where maximum radius of the orbit equals 2 and minimum radius equals 1.

I suppose the mass of the massive particle such that the speed of the light particle at the farthest radius is 1 m/s

You should not say "light particle" or "massive particle". The way you would phrase this in physics is to say "less massive" and "more massive". If you say "light", someone might think you are referring to electromagnetism. Also, in physics, if you say "massive", that just means with mass, as opposed to massless, without it necessarily being a "large" mass. Also, both particles are orbiting the center of mass of the system. You should say, "You have two particles, with mass m1 and m2, with m1 > m2, orbiting the center of mass, with m2 in an elliptical orbit." If you learn the correct terminology, it makes it easier for other people to understand what you are saying.

 

[Delburt Phend]

I think you caught it later; but back to the top. The force quads if the distance halves. But the velocity only doubles if the distance halves. If the radius doubles then the velocity halves. This is linear or arc velocity; and this change in linear velocity is required if angular momentum is to be conserved. But gravity does not double the velocity when the radius is halved in the lab. So how can angular momentum conservation work in the lab?

 

[jbriggs444]

I think you caught it later; but back to the top. The force quads if the distance halves. But the velocity only doubles if the distance halves. If the radius doubles then the velocity halves. This is linear or arc velocity; and this change in linear velocity is required if angular momentum is to be conserved. But gravity does not double the velocity when the radius is halved in the lab. So how can angular momentum conservation work in the lab?

The magnitude of the central force is irrelevant. All that matters is its direction. As long as it points toward or away from a fixed point, it does not alter angular momentum about that fixed point.