Robittybob's Law - orbital issue???

[Robittybob1]

After impact and forever in an ideal Newtonian case, once per elliptical orbit, until something changes the system. GR is different but that would be off topic.

I can see that OK. So did the impact cause a change of direction of motion plus a loss of kinetic energy?

Could you use "change of direction" in you answer so I understand what you think happens to the direction the impacted object takes please?

e.g "it doesn't change direction" or "there is a change of direction" it's entirely up to you. Thanks

An object at "r" will have more angular momentum on the parts beyond the center of gravity (COG) than the matter on the inner side of the COG. When the impact occurs there will be a rotation about the COG. Does this rotation cause the change of direction? Or is it just due to the fact that the object was slowed and no longer has the momentum to continue along the original tangential path and the force of gravity makes it change direction. As it falls it speeds up hence the kinetic energy increases. That scenario definitely is sequential.

[J.C.MacSwell]

I can see that OK. So did the impact cause a change of direction of motion plus a loss of kinetic energy?

Could you use "change of direction" in you answer so I understand what you think happens to the direction the impacted object takes please?

e.g "it doesn't change direction" or "there is a change of direction" it's entirely up to you. Thanks

I assumed an instantaneous slowing with no change in direction...an idealized impulse in the same direction as any would be drag. Once this takes place the body is free in the elliptical orbit, and will follow the path of that orbit.

[Robittybob1]

I assumed an instantaneous slowing with no change in direction...an idealized impulse in the same direction as any would be drag. Once this takes place the body is free in the elliptical orbit, and will follow the path of that orbit.

Understood completely thanks very much for your help, J.C MacSwell and Swansont in particular.

Edited February 23, 2016 by Robittybob1

[swansont]

... because you will have changed the eccentricity of the orbit. Is that the rest of your sentence?

"It won't be stepwise, because Work + Energy will be continually adding up properly, but as long as work is being done, and/or you aren't in a circular orbit, you can't say that the KE = 1/2|PE| relationship holds at any instant because you will have changed the eccentricity of the orbit."

 

 

No, because for any orbit with e ≠ 1, the KE = 1/2|PE| relationship does not generally hold at any instant. It does not require a change in eccentricity.

[Robittybob1]

 

 

No, because for any orbit with e ≠ 1, the KE = 1/2|PE| relationship does not generally hold at any instant. It does not require a change in eccentricity.

I was swayed by J.C.MacSwell's ideas, so you do disagree with him too? I think the change in eccentricity can be extremely small, and the falling part takes half the orbit to complete (I started off thinking that falling was instantaneous but no longer). The eccentricity stays there till more collisions even it out. So I agree now that "the KE = 1/2|PE| relationship does not generally hold at any instant". The eccentricity is the continuing reflection of past impacts.

I hope I understood his concept correctly.

[swansont]

I was swayed by J.C.MacSwell's ideas, so you do disagree with him too?

 

 

We weren't addressing the same issue, so I don't know why you would raise this question. KE = 1/2 |PE| fails to be generally true for any noncircular orbit, even one where there has been no impulse.

[Robittybob1]

 

 

We weren't addressing the same issue, so I don't know why you would raise this question.

I was trying to understand you. Maybe we don't fully understand each other. MacSwell's explanation seemed like the complete answer to me, so do you agree with him?

 

 

We weren't addressing the same issue, so I don't know why you would raise this question. KE = 1/2 |PE| fails to be generally true for any noncircular orbit, even one where there has been no impulse.

That impulse could have been from some event in the annals of history. How did the orbit become eccentric or non-circular as you say? It could be captured in orbit, but the whole event then would be the impulse. If it takes the whole orbit and only holds true on average over the whole orbit I'd still happy with that.

"KE = 1/2 |PE| fails to be generally true for any noncircular orbit" have you got an example please?

[swansont]

I was trying to understand you. Maybe we don't fully understand each other. MacSwell's explanation seemed like the complete answer to me, so do you agree with him?

 

 

I don't see anything wrong, but I wouldn't say it's a complete explanation, since it focuses on an impulse.

 

KE = 1/2 |PE| is ONLY true for a circular orbit. No work done on the object, so its KE is always the same. You deviate from that if there is an impulse, or if the orbit has another shape.

[Robittybob1]

 

 

I don't see anything wrong, but I wouldn't say it's a complete explanation, since it focuses on an impulse.

 

KE = 1/2 |PE| is ONLY true for a circular orbit. No work done on the object, so its KE is always the same. You deviate from that if there is an impulse, or if the orbit has another shape.

What do you mean by deviate? Is just that the rhythmic fluctuation of an eccentric orbit where KE is being changed into PE?

"At each point along the elliptical orbit the PE and KE adds up to the same quantity" Does that make sense if KE = 1/2 |PE|? Now energy is conserved in the case of the elliptical orbit we are not losing energy to drag (impulses)

"At each point along the elliptical orbit the PE and KE adds up to the same quantity" is more like throwing a ball up into the air. It will reach a high point where KE = 0 and PE is maximised. The RB law does not cover that type of situation.

Edited February 24, 2016 by Robittybob1

[Robittybob1]

Maybe J.C.MacSwell would care to comment as to whether the idea is complete.

 

It is realising that the falling takes half of the orbital circumference to complete that makes it easier to understand. In fact it falls too far and gains too much velocity and throws itself back out to the position it started at, when at the end of the impulse. The large amount of time between the loss of orbital energy and attaining the correct velocity for a circularised orbit is the thing I have come to appreciate. If one was to go back through the thread you would see how I struggled to understand as to whether this was instantaneous or delayed effect but J.C.MacSwell has convinced me it is delayed.

Edited February 24, 2016 by Robittybob1

[J.C.MacSwell]

 

 

I don't see anything wrong, but I wouldn't say it's a complete explanation, since it focuses on an impulse.

 

KE = 1/2 |PE| is ONLY true for a circular orbit. No work done on the object, so its KE is always the same. You deviate from that if there is an impulse, or if the orbit has another shape.

Except of course, like a broken clock being right twice per day...for elliptical orbits KE = 1/2 |PE| holds twice per orbit

[Robittybob1]

Except of course, like a broken clock being right twice per day...for elliptical orbits KE = 1/2 |PE| holds twice per orbit

And I suppose they would be at the points where the orbit could be treated as circular. Is that correct? Those positions are somewhere down the sides of the ellipse, definitely not at the perihelion (too slow) or the aphelion (too fast) positions.

Wiki on elliptical orbits

 

 

Conclusions:

The orbital period is equal to that for a circular orbit with the orbital radius equal to the semi-major axis (a),

Whereabouts on an ellipse is the point where the orbital radius is equal to the semi-major axis?

According to this site it is the point of maximum bulge where the semi minor axis meets the orbit path.

http://www.mrelativity.net/inertiagravity/Inertia&Gravity4_files/image004.gif

Edited February 24, 2016 by Robittybob1

[J.C.MacSwell]

And I suppose they would be at the points where the orbit could be treated as circular. Is that correct? Those positions are somewhere down the sides of the ellipse, definitely not at the perihelion (too slow) or the aphelion (too fast) positions.

That's right. Same speed but different direction.

[Robittybob1]

That's right. Same speed but different direction.

Ellipses are neat!

And I suppose they would be at the points where the orbit could be treated as circular. Is that correct? Those positions are somewhere down the sides of the ellipse, definitely not at the perihelion (too slow) or the aphelion (too fast) positions.

Wiki on elliptical orbits

Whereabouts on an ellipse is the point where the orbital radius is equal to the semi-major axis?

According to this site it is the point of maximum bulge where the semi minor axis meets the orbit path.

http://www.mrelativity.net/inertiagravity/Inertia&Gravity4_files/image004.gif

But at this point the velocity is not tangential. There are only two places where the velocity is tangential and that is at the perihelion and the aphelion but the speeds there are either too slow or too fast.

[Robittybob1]

What I have been thinking is we could test it out. Have a planet of mass m in a circular orbit r instantaneously impacting with meteor that takes away an amount of the KE (10%?) but does not add or subtract any mass. Calculate the circular orbit prior to the collision and the elliptical orbit after the collision and see if the RB law holds just for the amount of KE lost (represents drag). The the average PE based on the planet's parameters at the intersection of the orbit and the semi minor axis.

Does the change in KE = 1/2 * change in |PE|

Edited February 25, 2016 by Robittybob1

[swansont]

Except of course, like a broken clock being right twice per day...for elliptical orbits KE = 1/2 |PE| holds twice per orbit

 

Yes. That's why I have been saying generally true.

 

What do you mean by deviate? Is just that the rhythmic fluctuation of an eccentric orbit where KE is being changed into PE?

I mean deviate. You don't have those conditions. You will have more or less KE and PE at most places in the orbit, while the sum remains constant if there is no external work being done.

 

"At each point along the elliptical orbit the PE and KE adds up to the same quantity" Does that make sense if KE = 1/2 |PE|?

KE = 1/2 |PE| doesn't hold for the elliptical orbit, so this is moot. But for a circular orbit it holds, and if they are each constant then their sum will be, too.

 

Now energy is conserved in the case of the elliptical orbit we are not losing energy to drag (impulses)

"At each point along the elliptical orbit the PE and KE adds up to the same quantity" is more like throwing a ball up into the air. It will reach a high point where KE = 0 and PE is maximised. The RB law does not cover that type of situation.

The two situations are very similar. There's no external force in the system, so the mechanical energy is constant.

[Robittybob1]

 

 

Yes. That's why I have been saying generally true.

 

 

I mean deviate. You don't have those conditions. You will have more or less KE and PE at most places in the orbit, while the sum remains constant if there is no external work being done.

 

 

KE = 1/2 |PE| doesn't hold for the elliptical orbit, so this is moot. But for a circular orbit it holds, and if they are each constant then their sum will be, too.

 

 

The two situations are very similar. There's no external force in the system, so the mechanical energy is constant.

The origin of the RB relates to situations where there was an external force being applied. From the OP

 

I had said that the reduction in the potential energy is always split, half goes into additional kinetic energy and the other half is lost as drag, someone called that the "RB Law".

It is split when there is a loss of energy to an external influence. I had already understood how a planet moves in an elliptical orbit with the exchange or PE to KE and back. I was not intending to make any comment on the Kepler Laws.

When there is an 1:1 exchange there is an elliptical orbit

but when the exchange is split 50:50 there has been an external force.

RB Law implies there was nothing in between i.e you can't get a 75:25 split between PE and KE.

 

Do you think there are intermediary values? The complication we have discovered is that the exchange from one type to the other takes forever unless you circularise the orbit (for calculation purposes only) of the body.

 

That is why I was going to attempt the problem and investigate it further.

"Mechanical energy" is not a term I use that often but it maybe the preferred term to either PE or KE for those two are being continually exchanged (So ME is the sum of those two)

Edited February 25, 2016 by Robittybob1

[swansont]

The origin of the RB relates to situations where there was an external force being applied. From the OP

We've discussed several things since the OP.

 

It is split when there is a loss of energy to an external influence. I had already understood how a planet moves in an elliptical orbit with the exchange or PE to KE and back. I was not intending to make any comment on the Kepler Laws.

When there is an 1:1 exchange there is an elliptical orbit

but when the exchange is split 50:50 there has been an external force.

RB Law implies there was nothing in between i.e you can't get a 75:25 split between PE and KE.

 

Do you think there are intermediary values? The complication we have discovered is that the exchange from one type to the other takes forever unless you circularise the orbit (for calculation purposes only) of the body.

 

That is why I was going to attempt the problem and investigate it further.

"Mechanical energy" is not a term I use that often but it maybe the preferred term to either PE or KE for those two are being continually exchanged (So ME is the sum of those two)

If you are talking about a split between KE and PE then that sounds a lot like orbital energy, absent any external influence. You started with the exchange of KE and PE and back in an elliptical orbit, but also about the so-called RB law, so you need to be more precise regarding what you're talking about.

[J.C.MacSwell]

The origin of the RB relates to situations where there was an external force being applied. From the OP

It is split when there is a loss of energy to an external influence. I had already understood how a planet moves in an elliptical orbit with the exchange or PE to KE and back. I was not intending to make any comment on the Kepler Laws.

When there is an 1:1 exchange there is an elliptical orbit

but when the exchange is split 50:50 there has been an external force.

RB Law implies there was nothing in between i.e you can't get a 75:25 split between PE and KE.

 

Do you think there are intermediary values? The complication we have discovered is that the exchange from one type to the other takes forever unless you circularise the orbit (for calculation purposes only) of the body.

 

That is why I was going to attempt the problem and investigate it further.

"Mechanical energy" is not a term I use that often but it maybe the preferred term to either PE or KE for those two are being continually exchanged (So ME is the sum of those two)

In a slowly decaying almost circular orbit there is a approximate 50:50 split of the PE change(100%), between drag loss(50%) and KE gain(50%).

A change in the drag will, at least temporarily, change these ratios, and of course in elliptical orbits they are changing in (almost) any case.

Edited February 25, 2016 by J.C.MacSwell

[Robittybob1]

In a slowly decaying almost circular orbit there is a approximate 50:50 split of the PE change(100%), between drag loss(50%) and KE gain(50%).

A change in the drag will, at least temporarily, change these ratios, and of course in elliptical orbits they are changing in (almost) any case.

That is very similar to what I was saying but just one point is that in the case of an elliptical orbit I believe the change between the PE and KE is 1:1 (nothing was lost) but where there is loss due to drag it is 50:50.

On a quantum level it would read something like this "1 quanta of energy if loss to drag means there has to be a 2 quanta change in PE to get an increase of 1 quanta in KE".

You can't get other ratios for you can't split the quanta up into smaller components.

But I see the problem now is how to differentiate the effects of energy exchange during an elliptical orbit (where the quanta are exchanged 1:1) with the RB Law effect of 2:1:1 split.

We've discussed several things since the OP.

 

 

If you are talking about a split between KE and PE then that sounds a lot like orbital energy, absent any external influence. You started with the exchange of KE and PE and back in an elliptical orbit, but also about the so-called RB law, so you need to be more precise regarding what you're talking about.

I have found that with the new understanding of the time delay in the RB law mechanism it has become indistinguishable from an elliptical orbit. I am trying to develop the terminology to talk about it. I must try the example first and it might clarify my understanding.

An Object in circular orbit, encounters drag, it is slowed and falls (loss of PE) as if in an elliptical orbit. The elliptical orbit will have a shorter period. Which if thought of as a circular orbit (for calculation purposes) would have a reduced radius and an increased tangential velocity (hence KE increase).

That elliptical orbit will be maintained indefinitely so I am now left in a position of not being able to define when the RB Law completes the process.

So maybe we are forced to say the whole RB law is instantaneous for at the moment of impact energy is lost and the elliptical orbit is established instantaneously at that point too. The elliptical orbital characteristics accounts for the loss of PE and increase in KE.

So does that provide the solution to the paradoxical situation of something speeding up because it was slowed down?

Edited February 26, 2016 by Robittybob1

[J.C.MacSwell]

That is very similar to what I was saying but just one point is that in the case of an elliptical orbit I believe the change between the PE and KE is 1:1 (nothing was lost) but where there is loss due to drag it is 50:50.

On a quantum level it would read something like this "1 quanta of energy if loss to drag means there has to be a 2 quanta change in PE to get an increase of 1 quanta in KE".

You can't get other ratios for you can't split the quanta up into smaller components.

But I see the problem now is how to differentiate the effects of energy exchange during an elliptical orbit (where the quanta are exchanged 1:1) with the RB Law effect of 2:1:1 split.

Line 1: Two different circumstances

1)The 1:1 change between PE and KE is with no drag for an elliptical orbit...add a drag component and you change the ratio, generally not to any special 2:1:1 PE:KE:Drag split

2)The 50:50 ratio between KE and drag losses, which split the PE change, is generally approximate for a slowly decaying, substantially circular, orbit

Line 2: It does not have to be that way. The ratios depend on the resultant forces at any given point. Quantum considerations are insignificant here

Line 3: You can get other ratios

Line 4: A decaying long elliptical orbit will have varying ratios. This can be readily seen by comparing the resultant forces and velocities at various points. It's fairly simple Newtonian mechanics at any given point, even if more complex to track over time.

[Robittybob1]

Line 1: Two different circumstances

1)The 1:1 change between PE and KE is with no drag for an elliptical orbit...add a drag component and you change the ratio, generally not to any special 2:1:1 PE:KE:Drag split

2)The 50:50 ratio between KE and drag losses, which split the PE change, is generally approximate for a slowly decaying, substantially circular, orbit

Line 2: It does not have to be that way. The ratios depend on the resultant forces at any given point. Quantum considerations are insignificant here

Line 3: You can get other ratios

Line 4: A decaying long elliptical orbit will have varying ratios. This can be readily seen by comparing the resultant forces and velocities at various points. It's fairly simple Newtonian mechanics at any given point, even if more complex to track over time.

Thanking you again for such a detailed answer. Only thought that bugging me a bit is if one has established an elliptical orbit (from the circular orbit) and you hit the object again later with a second drag force has the orbit still got the characteristics of an elliptical orbit albeit a different set of parameters?

So the position of the perihelion could be shifted, that would be like a type of precession if it had some regularity to it, so the position of the aphelion and perihelion would change but the orbit would still be elliptical.

Precession may be the allowance for ratios other than 1:1 and 2:1:1, that is just a guess mind you for it is way beyond anything I have ever calculated.

[Janus]

Thanking you again for such a detailed answer. Only thought that bugging me a bit is if one has established an elliptical orbit (from the circular orbit) and you hit the object again later with a second drag force has the orbit still got the characteristics of an elliptical orbit albeit a different set of parameters?

So the position of the perihelion could be shifted, that would be like a type of precession if it had some regularity to it, so the position of the aphelion and perihelion would change but the orbit would still be elliptical.

Precession may be the allowance for ratios other than 1:1 and 2:1:1, that is just a guess mind you for it is way beyond anything I have ever calculated.

It depends on where the second drag is applied. If you wait until you are 180 degrees from where the first drag was applied, at first you would drive the orbit more circular. You will be at perihelion and will be reducing the aphelion distance. At a certain point you will reduce the aphelion distance to your distance and the orbit will be circular. After that, a continued reduction in your speed will cause you to be at aphelion and the point 180 degrees from you becomes the perihelion. At any other point of the orbit you will be changing the eccentricity, semi-major axis, and the position of the perihelion.

 

For example: Assume we have an orbit around the Earth with an eccentricity of 0.5 and a semi-major axis of 14,000 km. If we applied a 100m/s reduction of velocity at the moment our craft was at the semi-major axis distance from the orbital focus,( which in this orbits happens when the craft is 120 degrees from perigee)it will have the following effects:

The eccentricity will increase to 0.501

The semi-major axis will decrease to 13498.829 km

The perigee will decrease to 6735.9 km and shift 3.6 degrees further away from the current position of the craft.

[Robittybob1]

Thanks Janus. That shift in 3.6 degrees is a type of rotation of the semimajor axis of the new elliptical. Which way did the shift occur?

"Perigee the point in the orbit of the moon or a satellite at which it is nearest to the earth"

So we were slowing down, so you aren't going to climb so high, so you will turn earlier. If that logic is correct the angle shift is back against the direction of travel. Was that right?

 

The apogee : the point in outer space where an object traveling around the Earth (such as a satellite or the moon) is farthest away from the Earth.

 

So the apogee is reached earlier in the orbit.

Edited February 27, 2016 by Robittybob1

[Janus]

If you are orbiting clockwise, the perigee shift will be counter-clockwise. The greater the velocity change at your given position, the greater the shift (up to the point where you have reduced your velocity to zero and perigee is 180 degrees from your present position.

Also where you are in the orbit determines how much of a shift a given change of velocity will produce. For instance, the same 100 m/s reduction in velocity applied 41 degrees later in the orbit will cause a increase of eccentricity to 0.542, but a perigee shift of only 0.7 degrees.