I Reasoning for circular/elliptical orbits

[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.

 

[Orodruin]

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

No, you are wrong. Since we are discussing unbound orbits, I’d rather say the discussion is full of hyperbolae.

Edit: And I remind myself again to read to the end of the thread before replying …

Edit 2: As it stands my joke became a bit … derivative …

 

[Drakkith]

Terrible puns. I'd tell you all to go suck on lemon slices, but all I have are these conic slices.

 

[jbriggs444]

Terrible puns. I'd tell you all to go suck on lemon slices, but all I have are these conic slices.

Who cares about the shape of a cross section? Fill it with ice cream already.

 

[malawi_glenn]

This thread is full of circular reasoning!

I hate when physics discussions go off on a tangent

 

[Orodruin]

I hate when physics discussions go off on a tangent

Me too, it gives me physical pain in both radius bones.

 

[Orodruin]

Me too, it gives me physical pain in both radius bones.

On the other hand, it can also be rather humerus …

 

[PeroK]

On the other hand, it can also be rather humerus …

You're being elliptical now.

 

[Filip Larsen]

Terrible puns.

I agree, all those puns are so mechanical. Someone ought to get a kick in latus rectum!

 

[Orodruin]

You're being elliptical now.

… and there the circle is complete