I Reasoning for circular/elliptical orbits

[Chenkel]

TL;DR Summary

I'm wondering how objects achieve circular/elliptical orbit.

Hello everyone! So I have a fairly basic question about orbits, if I launch an object vertically from the surface of the earth, for most velocities it will come back to earth, if I launch at an angle it will have a velocity tangential to the vertical velocity, and this vertical velocity will change due to the acceleration of gravity, but I don't see why this tangential velocity is changing direction to achieve a circular or elliptical motion, obviously I'm missing something and I'm trying to fill that gap in my knowledge, if anyone can provide that insight I will be grateful, thank you!

 

[Dale]

This is not specific to gravity. Any time that you have any force which is not parallel to the velocity then that force will cause the tangential velocity to change direction. That is simply how forces behave. If you wish, I can step through the mathematical derivation, but I don’t know a non mathematical way.

 

[russ_watters]

...and if the object doesn't have escape velocity, it has to come back. If it has enough tangential velocity to miss...orbit!

 

[Chenkel]

This is not specific to gravity. Any time that you have any force which is not parallel to the velocity then that force will cause the tangential velocity to change direction. That is simply how forces behave. If you wish, I can step through the mathematical derivation, but I don’t know a non mathematical way.

I suppose I can see how the velocity will change direction when taking into account your analysis, and also taking into account Newton's first and second laws, so a derivation of the changing velocity vector is probably not necessary (but I do appreciate your willingness to guide me in this regard.) I feel I do not have a full understanding of how orbits are achieved. I might lack a little intuition, and mathematical knowledge, but hoping I can improve on these aspects.

 

[Chenkel]

...and if the object doesn't have escape velocity, it has to come back. If it has enough tangential velocity to miss...orbit!

I suppose with a little imagination and intuition that makes some sense, but for me the overall picture isn't completely coming in. What makes something achieve a mostly elliptical orbit, like Halley's comet for example?

One thing I'm curious about is the following.. The tangential velocity vector is corrected by a small change in velocity each second due to the gravitation of earth, let's call it dV/dt. I imagine for a circular orbit, dV/dt would trace out a circle, would it also trace out a circle for a very elliptical orbit?

 

[kuruman]

If the projectile is stays near the surface of the Earth which is approximated as flat, it will return to the same height from which it was launched. Give it more initial speed and it will land below the horizon. Give it even more speed and the Earth will curve away from it and the projectile will go into orbit (see picture).

 

[Drakkith]

if I launch at an angle it will have a velocity tangential to the vertical velocity, and this vertical velocity will change due to the acceleration of gravity, but I don't see why this tangential velocity is changing direction to achieve a circular or elliptical motion, obviously I'm missing something and I'm trying to fill that gap in my knowledge, if anyone can provide that insight I will be grateful, thank you!

What you're missing is that the tangential velocity IS changing. Let's set up an example:
Let's place a stationary point-mass (a mass with zero size) at the origin of a graph. Let's also place a point-particle with substantially smaller mass at (0,1), one unit above the mass, and let us also say that this particle is moving in the +X direction at 1 unit per second.

Our particle is attracted towards the mass via a force. This force itself is a vector that points from the particle towards the mass. Initially this force cannot act on the particle's X-velocity since it is acting solely in the Y-direction. But, if we advance by a second (ignoring the Y-acceleration that would happen in that second under the applied force) we find that the particle is at (1,1) on the graph and now the force vector CAN act on the X-velocity since the force vector now acts equally in the X and Y-directions.

But wait, we just skipped a whole second when advancing time. Perhaps things are different in a continuous world. As a test, let's see what happens when we only advance a half-second. Well, the particle is at (1/2,1) on the graph, the force vector still has an X-component, and the particle's x-velocity will be changed, just by a smaller amount than when we advanced straight to 1 second.

If we continue to make our time steps smaller, we find that ANY deviation in the X-direction away from zero will result in the force vector having an X-component, and thus affecting the particle's X-velocity, which was the original tangential velocity. So when you throw a ball on Earth, it IS being pulled back towards you by the Earth's gravity. Just not by very much in that short time and distance.

But, what about the Y-velocity? Well, it turns out that something similar happens to it as well, it just starts at zero and increases instead of starting at one and decreasing. In a perfectly circular orbit both the X and Y velocity vectors 'trade' magnitudes in such a way as to ensure that the overall velocity vector has the same magnitude at all times.

 

[Delta2]

If you want to set a mass in an elliptical or circular orbit, you must give to it the proper initial velocity both in magnitude and direction, and also the mass must be at the proper starting point. For example to set a mass $m$ in circular orbit of radius $R$ around the earth, one way is to put the mass at height R, and give it a tangential (perpendicular to the radius) velocity with magnitude $v$ such that $$m\frac{v^2}{R+R_e}=G\frac{M_em}{(R+R_e)^2}$$

To set it in an elliptical orbit I think the initial velocity is different in magnitude and in direction, I think it is not exactly perpendicular to the radius vector.

 

[Orodruin]

If the projectile is stays near the surface of the Earth which is approximated as flat, it will return to the same height from which it was launched. Give it more initial speed and it will land below the horizon. Give it even more speed and the Earth will curve away from it and the projectile will go into orbit (see picture).

View attachment 303632

That’s not really how it works though. Any bound orbit (ie, excluding escaping solutions) starting at the Earth’s surface will return to the Earth’s surface simply by virtue of the orbit being an ellipse. In order to achieve an actual orbit above the Earth, additional changes of velocity that are not due to gravity (eg, by rocket boosters) is necessary.

 

[Delta2]

@Orodruin if you put a mass m at random height above the surface of the Earth and give it a random velocity both in magnitude and direction, then the orbit , assuming no other bodies except the mass and earth, will be elliptical with the center of Earth at one foci of the ellipse and the other foci depending on the initial height, and initial velocity. Right?
PS. Also assume Earth as point mass in order to avoid collision issues.

 

[Orodruin]

@Orodruin if you put a mass m at random height above the surface of the Earth and give it a random velocity both in magnitude and direction, then the orbit , assuming no other bodies except the mass and earth, will be elliptical with the center of Earth at one foci of the ellipse and the other foci depending on the initial height, and initial velocity. Right?
PS. Also assume Earth as point mass in order to avoid collision issues.

If the orbit is bound, yes. The collision issue is handled simply by checking the perigee of the orbit. If that is closer than the radius of the Earth there will be a collision.

 

[Delta2]

If the orbit is bound, yes. The collision issue is handled simply by checking the perigee of the orbit. If that is closer than the radius of the Earth there will be a collision.

Sorry I have not read any good textbook on astrophysics, what do you mean by the term "orbit is bound", you mean provided that it does not escape to infinity(initial velocity greater or equal to escape velocity)?

 

[Orodruin]

Sorry I have not read any good textbook on astrophysics, what do you mean by the term "orbit is bound", you mean provided that it does not escape to infinity(initial velocity greater or equal to escape velocity)?

Yes.

 

[Chenkel]

In a Walter Lewin lecture I'm watching he says the escape velocity is$$v_e = \sqrt{\frac {2MG} R }$$Where M is the mass of Earth and R is the radius of earth. This is derived by taking the mechanical energy at infinite distance, and seeing what your velocity needs to be to escape the gravitational pull at that infinity, the potential energy at infinity is 0, so the additional speed the object needs to go is 0, that means the mechanical energy is more than or equal to 0 for any mass in the force field that has escape velocity, but I'm wondering if direction of velocity matters, my guess is that it doesn't and I would like to understand why, if the object is placed at infinity and the velocity is aiming toward decreasing potential energy then will the object escape the gravitational pull as it goes to the other side? Provided it doesn't hit the Earth on it's inward trajectory? My intuition is that the additional velocity as it moves to the other end of the orbit is more than enough velocity to escape the pull of earth, but it's a little hard for me to see this concretely for velocities not aiming in direction of increasing potential energy.

Also he says that the orbital velocity is$$v_o = \sqrt{\frac {MG} R }$$Does this mean that for any orbiting object it only has to go faster by a multiple of $\sqrt 2$ to escape the pull of earth?

If any of you can provide some insight that would be awesome, thank you.

 

[Orodruin]

but I'm wondering if direction of velocity matters, my guess is that it doesn't

It indeed does not. As long as the object has positive total energy (or zero total energy in the limit) it will escape to infinity unless it hits the Earth in between - but that is then due to the perigee of the trajectory being below the Earth surface.

Also he says that the orbital velocity isvo=MGRDoes this mean that for any orbiting object it only has to go faster by a multiple of 2 to escape the pull of earth?

This is the orbital velocity of a circular orbit. The velocity in an elliptic orbit changes throughout the orbit. But yes, if you would double the kinetic energy of an object in circular orbit, then it would reach escape velocity.

 

[Chenkel]

It indeed does not. As long as the object has positive total energy (or zero total energy in the limit) it will escape to infinity unless it hits the Earth in between - but that is then due to the perigee of the trajectory being below the Earth surface.This is the orbital velocity of a circular orbit. The velocity in an elliptic orbit changes throughout the orbit. But yes, if you would double the kinetic energy of an object in circular orbit, then it would reach escape velocity.

One thing which I have difficulty reconciling is that there is ever a point where the object 'escapes' the gravitational pull of earth, for example for any finite distance from Earth there will be a force pulling on the object, the escape velocity equation looks at the mechanical energy at infinity and any real object will never reach infinity, so maybe the question I'm asking is "why does energy more than zero imply that the object will have a non elliptical orbit?"

 

[Delta2]

@Orodruin if we give escape velocity to a mass, then its orbit will still be elliptical but with one foci at infinity?

 

[Chenkel]

@Orodruin if we give escape velocity to a mass, then its orbit will still be elliptical but with one foci at infinity?

So the object 'escapes,' but it is still influenced no matter how far it gets away. So really the escape velocity equation says what minimum velocity is needed to created an unbounded orbit, it's hard for me to see how this follows from it's derivation of 'mechanical energy for zero velocity at infinity.'

 

[jbriggs444]

If the projectile is stays near the surface of the Earth which is approximated as flat, it will return to the same height from which it was launched. Give it more initial speed and it will land below the horizon. Give it even more speed and the Earth will curve away from it and the projectile will go into orbit (see picture).

View attachment 303632

That picture looks wrong. It starts nearly tangent to the Earth's surface and ends nearly tangent to a larger radius. You can't get that behavior with an conservative inverse square central force.

It looks like a parabola with a vertical axis where an ellipse with a diagonal axis would have been correct. Cobbling up a better shape with mspaint, shifting to a horizontal axis and eyeballing to figure out where the foci live, I get a sub-orbital trajectory:

Note that since an orbit under an inverse square central force is a closed figure, it follows that an object projected from the Earth's surface at less than escape velocity will have an orbit that intersects with the surface of the Earth.

Edit: There are conservative central force laws that could produce the tangent behavior that was identified. For instance, one with a potential well at one radius surrounded by potential hills above and below.

 

[Delta2]

So the object 'escapes,' but it is still influenced no matter how far it gets away. So really the escape velocity equation says what minimum velocity is needed to created an unbounded orbit, it's hard for me to see how this follows from it's derivation of 'mechanical energy for zero velocity at infinity.'

Well by the way we apply conservation of mechanical energy for the escape velocity problem, we don't have any info on what is the exact trajectory that the object will follow on its way to infinity. I find it hard also to prove that given any "random" direction to initial escape velocity the object will reach infinity. My intuition also tells me that the initial direction matters as to whether the object escapes to infinity or not. All I can tell is that IF the object reaches infinity then its kinetic energy will be exactly zero.

 

[jbriggs444]

Well by the way we apply conservation of mechanical energy for the escape velocity problem, we don't have any info on what is the exact trajectory that the object will follow on its way to infinity. I find it hard also to prove that given any "random" direction to initial escape velocity the object will reach infinity. My intuition also tells me that the initial direction matters as to whether the object escapes to infinity or not. All I can tell is that IF the object reaches infinity then its kinetic energy will be exactly zero.

If you stipulate that the force law is central then that gives you conservation of angular momentum about the center. That is a tool that you can bring to bear on the proof.

If you stipulate that the force law is monotone decreasing with increasing radius, that let's you rule out potential barriers that are too high to cross. That gives you another tool.

You should be able to leverage these to demonstrate that an object whose trajectory has escape energy and ever has a strictly positive radial velocity will retain a strictly positive radial velocity forever after.

Edit: I think that one actually needs to tweak the "monotone decreasing" bit above so that the potential energy associated with the central force is monotone increasing even after accounting for the kinetic energy associated with the tangential velocity that is assured by conservation of angular momentum. An inverse square law qualifies. Other central force laws will qualify as well.

As long as the force law does not result in an in-spiral to the center, you should be able to argue that an object that has a inward or zero initial radial velocity will eventually turn around and go outward. [This is the assertion about which I feel least comfortable].

 

[Delta2]

ehm @jbriggs444 you say another condition except escape velocity, that is the strictly positive radial velocity, @Orodruin what is your take on this?

 

[jbriggs444]

ehm @jbriggs444 you say another condition except escape velocity, that is the strictly positive radial velocity, @Orodruin what is your take on this?

Be very careful in how you read that posting. I am talking about how one would attack a mathematical proof without directly invoking an inverse square law.

For a more general central force law, there is the possibility of an inward spiral. Let me see if I can Google something up. First hit is a homework exercise:

For an inverse square force law, a non-degenerate orbit is guaranteed to have a periapsis at a finite radius and will move outward afterward. So the launch angle is irrelevant as long as the trajectory does not intersect with the gravitating body.

 

[Filip Larsen]

I find it hard also to prove that given any "random" direction to initial escape velocity the object will reach infinity.

For the two-body problem (with no collision), the vis-viva equation gives a relationship between just the current speed and radial distance and the orbit shape with 1/a zero being equivalent to a parabolic trajectory. As can be seen, there is no angle between speed and position in this.

 

[Delta2]

For the two-body problem (with no collision), the vis-viva equation gives a relationship between just the current speed and radial distance and the orbit shape with 1/a zero being equivalent to a parabolic trajectory. As can be seen, there is no angle between speed and position in this.

Sorry can you make it more clear for me? I don't understand how the vis viva equation tell us that an object with escape velocity will reach infinity all that I can infer is that at infinity will have a certain speed which would be 0 if 1/a is zero.

 

[Delta2]

Be very careful in how you read that posting. I am talking about how one would attack a mathematical proof without directly invoking an inverse square law.

For a more general central force law, there is the possibility of an inward spiral. Let me see if I can Google something up. First hit is a homework exercise:

For an inverse square force law, a non-degenerate orbit is guaranteed to have a periapsis at a finite radius and will move outward afterward. So the launch angle is irrelevant as long as the trajectory does not intersect with the gravitating body.

Don't confuse me with what is happening with non central non inverse square force law, I am only interested in the case of central inverse square force law. You guys don't give me a clear answer ...

 

[jbriggs444]

Don't confuse me with what is happening with non central non inverse square force law, I am only interested in the case of central inverse square force law. You guys don't give me a clear answer ...

Clear answer: Launch angle is irrelevant.

 

[Delta2]

Clear answer: Launch angle is irrelevant.

and the clear why? Something about conservation of angular momentum and force being monotonically decreasing but I can't get the clear reasoning...

Launch angle is not totally irrelevant btw, what if I launch the object with escape velocity towards the center of the earth? e hehe

 

[jbriggs444]

and the clear why? Something about conservation of angular momentum and force being monotonically decreasing but I can't get the clear reasoning...

Without much math:

Case 1: Outward launch.

If you launch outward, the trajectory is increasingly more in a direct radial direction. Tangential velocity reduces based on conservation of angular momentum. Radial velocity reduces at a lower rate.

Case 2: Tangential launch.

If you launch tangentially, the trajectory bends outward based on centrifugal force. It's moving too fast to stay in circular orbit.

Case 3: Inward launch.

If you launch inward, the trajectory has a low point where angular momentum requires a tangential velocity that is big enough to include all of the object's available kinetic energy. So now you must be moving tangentially. You are back in case 2.

Case 4: Launch dead center at the point mass that is providing the central force.

You can't aim that accurately and you can't have a gravitating mass that small. We hope. Your predictions die in a singular confusion between quantum mechanics, general relativity and the mathematics of situations with measure zero.

Edit: Some thoughts about Newtonian Gravity and the continuum in this insights article by John Baez. That dead center shot is interesting to contemplate.

 

[russ_watters]

I find it hard also to prove that given any "random" direction to initial escape velocity the object will reach infinity. My intuition also tells me that the initial direction matters as to whether the object escapes to infinity or not. All I can tell is that IF the object reaches infinity then its kinetic energy will be exactly zero.

There is only one direction: directly away from the object. Even if you start at tangential to the surface(call that zero degrees), as the distance increases the angle increases toward 90 degrees. Try drawing a diagram of this.

 

[Ibix]

Something about conservation of angular momentum and force being monotonically decreasing but I can't get the clear reasoning...

In a conservative force field, work done around a closed loop is zero. Draw any path from some point at radius $r_1$ to some point at radius $r_2$, and another path connecting another two points at $r_1$ and $r_2$. Close the loop by adding paths of constant $r$ connecting the lower ends of the paths to each other and the upper ends to each other. Because the force is central, no work is done along the paths of constant $r$. So the work done along the other two sides must be equal. But those paths are arbitrary - so the work done moving from one radius to another cannot depend on anything except the radii.

 

[Orodruin]

and the clear why? Something about conservation of angular momentum and force being monotonically decreasing but I can't get the clear reasoning...

Launch angle is not totally irrelevant btw, what if I launch the object with escape velocity towards the center of the earth? e hehe

The clear why is that both the gravitational potential and the angular momentum barrier of the effective radial 1D problem both go to zero at infinity. Therefore, any solution with energy > 0 will approach infinite radius

 

[Delta2]

The clear why is that both the gravitational potential and the angular momentum barrier of the effective radial 1D problem both go to zero at infinity. Therefore, any solution with energy > 0 will approach infinite radius

Ehm , what do you mean by "effective radial 1D problem". On the second sentence you probably mean any solution with energy>escape energy not energy >0.

 

[Orodruin]

Ehm , what do you mean by "effective radial 1D problem".

Using conservation of angular momentum, the equations of motion In a central potential can be rewritten as
$$
E = \frac{m \dot r^2}{2} + V(r) + \frac{L^2}{2mr^2}
$$
This is a one-dimensional kinematic problem in potential $V(r) + L^2/2mr^2$ (the latter term is what is called the angular momentum barrier and prevents reaching $r=0$ unless $L=0$).

On the second sentence you probably mean any solution with energy>escape energy not energy >0.

No. $E=0$ is the escape energy.

 

[Orodruin]

To add on top of that, for any 1D kinematic problem in variable $x$ with potential $V(x)$, it holds that the energy
$$
E = \frac{m\dot x^2}{2} + V(x)
$$
is a constant of motion. With $\dot x$ changing continuously, the only way that it can change sign is for $\dot x = 0$ which is only possible when $V(x) = E$. These points are the classical turning points, ie, where the 1D motion changes direction.

For the case of the effective problem arising from the Kepler potential, the effective potential has the form $a/r^2 - b/r$. This means that as long as $E$ is finite there is a lower classical turning point. However, only if $E < 0$ does an upper turning point exist (this is fleshed out a bit with illustration in ch 10 of my book iirc). Therefore, if $E < 0$, the orbit will move back and forth between the radial turning points. If $E\geq 0$ there is no upper turning point, meaning that such a trajectory for which $\dot r >0$ will never turn around and instead grow without bound. Such a trajectory with $\dot r < 0$ will eventually encounter the lower turning point, turn around, and be a trajectory with $\dot r>0$.

 

[Delta2]

Using conservation of angular momentum, the equations of motion In a central potential can be rewritten as
$$
E = \frac{m \dot r^2}{2} + V(r) + \frac{L^2}{2mr^2}
$$
This is a one-dimensional kinematic problem in potential $V(r) + L^2/2mr^2$ (the latter term is what is called the angular momentum barrier and prevents reaching $r=0$ unless $L=0$).No. $E=0$ is the escape energy.

This generates many new questions but anyway since this is not my thread I will not question anymore, except how can it be that any E>0 the body escapes to infinity, sorry can't seem to grasp that.
Oh just saw you expanded on your previous post, thx.

 

[Delta2]

Oh now I understood it, by E you mean the total energy not the kinetic energy only, and taking into account that the potential energy is negative...

 

[Orodruin]

To add to the addendum regarding the effective problem in the Kepler potential:

To analyze the behaviour of the effective potential $a/r^2 - b/r$, consider $x = 1/r$ such that the effective potential is $x(ax -b)$ as expressed in $x > 0$. It is clear that this potential grows without bound as $x \to +\infty$ ($r\to 0$) and that it is zero when $x \to 0$ ($r\to \infty$). However, it is also clear that the minimum of the potential is smaller than zero and indeed lies in the $x>0$ region. Therefore, the existence of negative energy (bound) solutions is guaranteed. This holds regardless of the values of the positive* constants $a$ and $b$.

* The exception being the special case of no angular momentum, which is a bit pathological as already mentioned in this thread.

 

[Chenkel]

After reading all this discussion, I gave a humble effort to understand, but I feel it may be a little bit over my head with my current level in physics to see how mechanical energy more than or equal to 0 implies unbounded orbit for any velocity or orbital position, maybe I can understand a simple case that I've been having trouble reconciling, that question is the following.. What is the proof that for a one dimensional case of a mass being launched radially without any tangential velocity that if the mechanical energy is more than or equal to 0 that it will never reverse direction of it's velocity to come back to center. I tried with pen and paper to understand this simple case, but I am having trouble coming to terms with it. If anyone can shed some light on this it would be greatly appreciated, thank you!

 

[Orodruin]

What is the proof that for a one dimensional case of a mass being launched radially without any tangential velocity that if the mechanical energy is more than or equal to 0 that it will never reverse direction of it's velocity to come back to center.

This:

only if E<0 does an upper turning point exist

Fleshed out:
The mechanical energy in your case of purely radial motion is
$$
E = m\frac{\dot r^2}{2} - \frac{GmM}{r}
$$
If $E \geq 0$ then $\dot r^2 > 0$ for all $r$, meaning $\dot r \neq 0$, which it would need to be at a turning point.

 

[Chenkel]

This:

Fleshed out:
The mechanical energy in your case of purely radial motion is
$$
E = m\frac{\dot r^2}{2} - \frac{GmM}{r}
$$
If $E \geq 0$ then $\dot r^2 > 0$ for all $r$, meaning $\dot r \neq 0$, which it would need to be at a turning point.

Thank you, this works for one dimensional analysis and I understand it, but now I'm trying to gain some intuition for the non one dimensional case where velocity can never be less than 0.

I'm trying to analyze the rate of change of radial distance to see what's going on. These are the equations I have$$E = \frac 1 2 m v^2 - \frac {mMG} r$$$$\frac {dE}{dt} = mv\frac{dv}{dt}+\frac {mMG} {r^2} \frac {dr}{dt}$$$$\frac {dE}{dt} = 0$$$$\frac {dr}{dt}=-\frac {r^2 v \frac {dv}{dt}}{MG}$$I'm wondering if I made a mistake because dr/dt is less than 0 when dv/dt is more than 0. Is this equation correct?

 

[Orodruin]

Thank you, this works for one dimensional analysis and I understand it, but now I'm trying to gain some intuition for the non one dimensional case where velocity can never be less than 0.

I covered this above by looking at the effective 1D problem for the radial motion. The argument is exactly the same.

Split the velocity in the radial component $\dot r$ and the tangent component $v_\theta$ and use the fact that angular momentum $L = mrv_\theta$ is conserved and you will arrive at

$$
E = \frac{m \dot r^2}{2} + V(r) + \frac{L^2}{2mr^2}
$$
This is a one-dimensional kinematic problem in potential $V(r) + L^2/2mr^2$ (the latter term is what is called the angular momentum barrier and prevents reaching $r=0$ unless $L=0$).

 

[Orodruin]

I'm wondering if I made a mistake because dr/dt is less than 0 when dv/dt is more than 0.

That’s just telling you that speed increases when the radius decreases. You could have concluded that already from potential energy decreasing when you decrease $r$.

 

[Chenkel]

I covered this above by looking at the effective 1D problem for the radial motion. The argument is exactly the same.

Split the velocity in the radial component $\dot r$ and the tangent component $v_\theta$ and use the fact that angular momentum $L = mrv_\theta$ is conserved and you will arrive at

I think I have some intuition about angular momentum, for example an ice skater speeding up in angular velocity as the arms are brought closer to the body, but I haven't had any rigorous study of it yet. Is it a necessary prerequisite to understand angular momentum deeply? What is V(r)? Also I'm still trying to come to terms with the angular momentum barrier term $\frac {L^2} {2mr^2}$

 

[Orodruin]

When I wrote $V(r)$ I was talking about motion in a general central potential. In the particular case of gravity it is the gravitational potential energy.

Obviously understanding angular momentum never hurts. For the purposes here it is enough to know that it is constant in a central potential and equal to $mrv_\theta$.

 

[Delta2]

@Orodruin can't resist asking this question: How we talk about 1D radial problem if there is angular momentum, that is the component $v_{\theta}$ of velocity besides the radial component $v_r=\dot r \hat r$.

 

[Orodruin]

@Orodruin can't resist asking this question: How we talk about 1D radial problem if there is angular momentum, that is the component $v_{\theta}$ of velocity besides the radial component $v_r=\dot r \hat r$.

It is an effective 1D problem. The equation of motion (after using conservation of angular momentum) is the same as that of a particle moving in one dimension with the angular momentum barrier added to the potential. All information of the problem being higher-dimensional is now encoded in the value of the constant $L$.

 

[Delta2]

Ok I see now thx, for some reason I didn't give too much weight to the word effective.

 

[Vanadium 50]

This thread is full of circular reasoning!
(Sorry...couldn't help myself)

 

[russ_watters]

This thread is full of circular reasoning!
(Sorry...couldn't help myself)

You're just being hyperbolic.