Robittybob's Law - orbital issue???

[Robittybob1]

@Janus - If you had to give a reason for that lesser change in alignment of the perigee what would you put that down to?

....

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

...

"Ratios depend on the resultant forces ....." It was these forces that were like the quantised components. Either the force vector adds to drag or GE or KE, and the whole shape of the orbit is changed with even the slightest change of force. OK it might be impossible to measure this change but logically it will change. Could the 2:1:1 ratios still be maintained even when the forces need resolving into the individual components? I know you say they aren't but how can I be convinced of that?

Edited February 28, 2016 by Robittybob1

[Janus]

@Janus - If you had to give a reason for that lesser change in alignment of the perigee what would you put that down to?

It's due to the angle between the orbital velocity vector and the radial line( the line draw between the craft and the orbit focus). At perigee and apogee, where a reduction or increase in velocity causes no shift in the semi-major axis, this angle is 90 degrees. At any other point, it will be some other angle. It increases to a maximum at the point where the radial line equals the semi-major axis distance and decreases back to zero at apogee and perigee.

Thus when you apply a velocity change, when it done at apogee or perigee, all of the velocity change is at a right angle to the radial line and none along the radial line (away from or towards the focus). At other point of the orbit, some of the velocity change will be along the radial line. The more of the velocity change that is along the radial line, the greater the effect it has on the direction on the orientation of the semi-major axis.

[Robittybob1]

@Janus - thanks again. The key word that I learned from that was the "focus". We can change the lines and angles but we never seem to change the focus. In your rocket orbiting the Earth the focus I presume was the gravitational center of the Earth and that never moves for the mass of the Earth is so large compared to the mass of the orbiting body.

 

But in the case of binary stars elliptically orbiting and one losing or gaining velocity from some external force the focus would probably move for the focus then would be the barycenter.

It is harder to imagine what could alter the elliptical orbit of a binary star without causing a mass change as well but can you think in terms of a change in position of the focus as well?

 

@ everyone - Have you ever dealt with a situation where the focus moves as well?

Edited February 28, 2016 by Robittybob1

[J.C.MacSwell]

@Janus - thanks again. The key word that I learned from that was the "focus". We can change the lines and angles but we never seem to change the focus. In your rocket orbiting the Earth the focus I presume was the gravitational center of the Earth and that never moves for the mass of the Earth is so large compared to the mass of the orbiting body.

 

But in the case of binary stars elliptically orbiting and one losing or gaining velocity from some external force the focus would probably move for the focus then would be the barycenter.

It is harder to imagine what could alter the elliptical orbit of a binary star without causing a mass change as well but can you think in terms of a change in position of the focus as well?

 

@ everyone - Have you ever dealt with a situation where the focus moves as well?

Like a satellite going around the Earth as the Earth accelerates toward the Sun?

[Robittybob1]

Like a satellite going around the Earth as the Earth accelerates toward the Sun?

Not so much for would any parameters of a rocket or satellite be affected by this movement of the focus? It is not particularly important to know this but it could ultimately be involved with orbiting binaries giving off gravitational radiation. Would each star or black hole lose equal amounts of orbital energy so that the "focus" (I'm assuming the focus is the barycenter) never shifted? I'll have to go back to my calculations on the Hulse-Taylor binary to see whether I considered gravitational radiation as a mass loss. If it was a mass loss would both stars lose the mass at the same proportional rate so the barycenter stays in the same place?

I was definitely using the RB law. I have a feeling if I took mass reduction into account my results could have been different.

Am I wrong to think of gravitational radiation as being associated with a mass loss?

Like a satellite going around the Earth as the Earth accelerates toward the Sun?

It could be more like Theia whacking into the Earth and the orbit of the Earth becoming an ellipse but because the mass of the Earth increased from that moment on the Sun Earth barycenter (focus) would have shifted toward the Earth a little.

Edited February 29, 2016 by Robittybob1

[J.C.MacSwell]

Not so much for would any parameters of a rocket or satellite be affected by this movement of the focus? It is not particularly important to know this but it could ultimately be involved with orbiting binaries giving off gravitational radiation. Would each star or black hole lose equal amounts of orbital energy so that the "focus" (I'm assuming the focus is the barycenter) never shifted? I'll have to go back to my calculations on the Hulse-Taylor binary to see whether I considered gravitational radiation as a mass loss. If it was a mass loss would both stars lose the mass at the same proportional rate so the barycenter stays in the same place?

I was definitely using the RB law. I have a feeling if I took mass reduction into account my results could have been different.

Am I wrong to think of gravitational radiation as being associated with a mass loss?

It could be more like Theia whacking into the Earth and the orbit of the Earth becoming an ellipse but because the mass of the Earth increased from that moment on the Sun Earth barycenter (focus) would have shifted toward the Earth a little.

I think that would have to be correct. If gravitational radiation is detectable as per recent news, then their has to be associated energy in the detection.

 

No matter what the source of the Earth's movement, "Theia whacking" or whatever, the satellite will be responding to the changes in Earth's position.

[Robittybob1]

I think that would have to be correct. If gravitational radiation is detectable as per recent news, then their has to be associated energy in the detection.

.....

So if there are two binary stars orbiting at a high rate giving off gravitational radiation and you say that is associated with mass loss, please hazard a guess or do you have any knowledge of what that really means back on the star itself? I have always likened PE as a mass gain, but references to that or other people mentioning such like are very very rare. Have you ever seen a reference to increased gravitational potential energy being associated with mass gain? I have only ever seen it written down once in 60 years. It seems such a difficult thing to measure, but it seems to follow on from Einstein's work of E=mc^2, if you have to add energy to lift an object its mass should increase.

But I have never seen this being discussed recently except for when these two BHs merged last year and 3 solar masses of pure energy were released as gravitational energy. Now I don't think of that as single atoms being destroyed in some unknown fashion, for what known process reduces mass to pure energy except like antimatter-matter interactions, so what is actually happening? For the same process is involved with the RB law where PE is split into energy loss and kinetic energy, so does that too involve a mass change?

 

This is such an unusual topic, I just hope we can resolve it.

Edited March 1, 2016 by Robittybob1

[Janus]

So if there are two binary stars orbiting at a high rate giving off gravitational radiation and you say that is associated with mass loss, please hazard a guess or do you have any knowledge of what that really means back on the star itself? I have always likened PE as a mass gain, but references to that or other people mentioning such like are very very rare. Have you ever seen a reference to increased gravitational potential energy being associated with mass gain? I have only ever seen it written down once in 60 years. It seems such a difficult thing to measure, but it seems to follow on from Einstein's work of E=mc^2, if you have to add energy to lift an object its mass should increase.

But I have never seen this being discussed recently except for when these two BHs merged last year and 3 solar masses of pure energy were released as gravitational energy. Now I don't think of that as single atoms being destroyed in some unknown fashion, for what known process reduces mass to pure energy except like antimatter-matter interactions, so what is actually happening? For the same process is involved with the RB law where PE is split into energy loss and kinetic energy, so does that too involve a mass change?

 

This is such an unusual topic, I just hope we can resolve it.

With the BH merger, the 3 solar mass conversion didn't occur until the moment of merger. Up until then the energy for the gravitational waves came from the loss of orbital energy due to the decreasing orbital radius.(To be more precise, the emission of gravitational waves drains energy from the orbiting pair which is compensated for by decreasing the radius of the orbit.

 

The power (joules/sec) emitted by a pair of orbiting masses is found by:

 

[math]\frac{32G^4}{c^5r^5}(M_1M_2)^2(M_1+M_2)[/math]

 

The rate at which the radius decreases(meters/sec) is:

[math]\frac{64G^3}{c^5r^3}(M_1M_2)(M_1+M_2)[/math]

 

TO find the rate of energy loss per decrease in radius(joules/meter) we divide the second equation into the first and get

[math]\frac{GM_1M_2}{2r^2}[/math]

 

The total orbital energy of those two masses is

 

[math]-\frac{GM_1M_2}{2r}[/math]

 

To find the rate at which the energy changes with a change of r we take the derivative with respect to r and get

[math]\frac{GM_1M_2}{2r^2}[/math]

 

Which is exactly the same as we got for the energy loss rate due to gravitational wave emission. Which means that the loss of energy due to gravitational wave emission equals that of orbital energy loss.

[Robittybob1]

With the BH merger, the 3 solar mass conversion didn't occur until the moment of merger. Up until then the energy for the gravitational waves came from the loss of orbital energy due to the decreasing orbital radius.(To be more precise, the emission of gravitational waves drains energy from the orbiting pair which is compensated for by decreasing the radius of the orbit.

 

The power (joules/sec) emitted by a pair of orbiting masses is found by:

 

[math]\frac{32G^4}{c^5r^5}(M_1M_2)^2(M_1+M_2)[/math]

 

The rate at which the radius decreases(meters/sec) is:

[math]\frac{64G^3}{c^5r^3}(M_1M_2)(M_1+M_2)[/math]

 

TO find the rate of energy loss per decrease in radius(joules/meter) we divide the second equation into the first and get

[math]\frac{GM_1M_2}{2r^2}[/math]

 

The total orbital energy of those two masses is

 

[math]-\frac{GM_1M_2}{2r}[/math]

 

To find the rate at which the energy changes with a change of r we take the derivative with respect to r and get

[math]\frac{GM_1M_2}{2r^2}[/math]

 

Which is exactly the same as we got for the energy loss rate due to gravitational wave emission. Which means that the loss of energy due to gravitational wave emission equals that of orbital energy loss.

Thank you Janus for those equations and explanations.

 

I'll will have to go back and see if the orbital parameters of the Hulse Taylor binary fit with those equations. It has been several years ago and a stroke (of a sort) intervened, but previously I was always told the reason I could not get the periods, the masses and the distances to work for that binary was due to relativity.

But I take it you are absolutely certain in all those equations M1 and M2 are not changing with the gradual loss of orbital energy.

 

In a documentary on the discovery of the merging BH they said they expected those BHs to have been orbiting each other for more than a billion years, so during that billion years they were losing gravitational energy but no loss of mass, yet there was 3 solar masses lost at the end due to ??? but only at the end, and in your words "the 3 solar mass conversion didn't occur until the moment of merger".

Is that "moment of merger" a time period or just an instant? And what was that mass loss due to in that time period/instant then?

Edited March 1, 2016 by Robittybob1

[J.C.MacSwell]

Thank you Janus for those equations and explanations.

 

I'll will have to go back and see if the orbital parameters of the Hulse Taylor binary fit with those equations. It has been several years ago and a stroke (of a sort) intervened, but previously I was always told the reason I could not get the periods, the masses and the distances to work for that binary was due to relativity.

But I take it you are absolutely certain in all those equations M1 and M2 are not changing with the gradual loss of orbital energy.

 

In a documentary on the discovery of the merging BH they said they expected those BHs to have been orbiting each other for more than a billion years, so during that billion years they were losing gravitational energy but no loss of mass, yet there was 3 solar masses lost at the end due to ??? but only at the end, and in your words "the 3 solar mass conversion didn't occur until the moment of merger".

Is that "moment of merger" a time period or just an instant? And what was that mass loss due to in that time period/instant then?

Keep in mind that the mass of the system is greater than the sum of the masses of it's constituent parts. The system has a different rest frame than M1, which is different again from M2.

[Robittybob1]

Keep in mind that the mass of the system is greater than the sum of the masses of it's constituent parts. The system has a different rest frame than M1, which is different again from M2.

Would you care to explain that a little more or give me a link to explore this idea (even a relevant YT will do, some lecture from a University would be good)? What is the search term for the situation you are describing please?

[Janus]

Thank you Janus for those equations and explanations.

 

I'll will have to go back and see if the orbital parameters of the Hulse Taylor binary fit with those equations. It has been several years ago and a stroke (of a sort) intervened, but previously I was always told the reason I could not get the periods, the masses and the distances to work for that binary was due to relativity.

But I take it you are absolutely certain in all those equations M1 and M2 are not changing with the gradual loss of orbital energy.

 

In a documentary on the discovery of the merging BH they said they expected those BHs to have been orbiting each other for more than a billion years, so during that billion years they were losing gravitational energy but no loss of mass, yet there was 3 solar masses lost at the end due to ??? but only at the end, and in your words "the 3 solar mass conversion didn't occur until the moment of merger".

Is that "moment of merger" a time period or just an instant? And what was that mass loss due to in that time period/instant then?

I'm not that versed enough in GR to give a definite answer as to what happens in those last instants, but to hazard a guess, I would say it has to do with what happens when the two black holes get within the photon sphere distance from each other. The photon sphere radius is that distance where something would have to be traveling at c in order to maintain an orbit around a black hole and is outside of the event horizon. There are no stable orbits inside of this.(even if you discount energy loss via gravitational waves, an object cannot orbit inside the photon sphere). Basically, in the region between the photon sphere and the event horizon, the laws of orbital mechanics that work outside of the photon sphere don't apply. So it just might be that in this region a mass loss is required to balance things out. But again, this is just a guess on my part, and not based on an actual analysis of the math, so take it with a grain of salt.

[Robittybob1]

I'm not that versed enough in GR to give a definite answer as to what happens in those last instants, but to hazard a guess, I would say it has to do with what happens when the two black holes get within the photon sphere distance from each other. The photon sphere radius is that distance where something would have to be traveling at c in order to maintain an orbit around a black hole and is outside of the event horizon. There are no stable orbits inside of this.(even if you discount energy loss via gravitational waves, an object cannot orbit inside the photon sphere). Basically, in the region between the photon sphere and the event horizon, the laws of orbital mechanics that work outside of the photon sphere don't apply. So it just might be that in this region a mass loss is required to balance things out. But again, this is just a guess on my part, and not based on an actual analysis of the math, so take it with a grain of salt.

Very interesting concept of orbiting inside the photon sphere for we often think of things orbiting the BH right up to the event horizon.

Well I did anyway.

Can we go back to the simplest of situations please. http://www.scienceforums.net/topic/93568-robittybobs-law-orbital-issue/page-3#entry908792

When we lift a mass on Earth are we increasing its mass according to Einstein's E= mc^2 equation?

Now if you work out how much additional mass that is it can be appreciated it isn't any significant mass change when a 1 kg mass is lifted 1 meter in Earth's gravitational field.

 

I'm not really asking whether it can be measured but whether it actually occurs.

 

What does the forum say?

On another forum site it said:

 

 

Gravity works like this: if I have two masses at infinite separation, and I bring them closer together, I can extract work from the process, purely due to the mutual gravity of the masses. The final mass of the bound system will be less than the sum of the two masses; the difference is the negative potential energy due to gravity.

Reference https://www.physicsforums.com/threads/potential-energy-and-mass-energy-equivalence.731885/

I think that is the exact same idea I'm exploring. Can anyone see the problem with that? Nothing like that is written up in Wikipedia on Gravitational PE.

Edited March 2, 2016 by Robittybob1

[J.C.MacSwell]

 

 

On another forum site it said:

I think that is the exact same idea I'm exploring. Can anyone see the problem with that, for nothing like that is written up in Wikipedia on Gravitational PE.

Possibly. I have always had a problem with what seemed to me an arbitrary (though convenient at times) choice of having gravitational PE as negative. It's something I would like to have a better grasp of.

[Robittybob1]

Possibly. I have always had a problem with what seemed to me an arbitrary (though convenient at times) choice of having gravitational PE as negative. It's something I would like to have a better grasp of.

But this is different than what you were talking about before when you said "keep in mind that the mass of the system is greater than the sum of the masses of its constituent parts."

[J.C.MacSwell]

But this is different than what you were talking about before when you said "keep in mind that the mass of the system is greater than the sum of the masses of its constituent parts."

That is the problem I would have with the example you gave. How can the mass of a system be less than the sum of it's parts?

[Robittybob1]

That is the problem I would have with the example you gave. How can the mass of a system be less than the sum of it's parts?

Is that just dependent on when it is weighed?

You can't weight them at any distance but you can calculate their mass by acceleration i.e. you are finding their inertial mass.

If the parts are weighed have their mass calculated when they are touching they have no GPE (they are at the least value of GPE in fact a negative value) that would less than when they are at an infinite distance apart (zero GPE but then also the maximal GPE value).

If you had their mass calculated they would have more inertial mass, therefore if the inertial masses were summed they would be greatest then when they are infinitely separated. But when they were together touching their combined inertial mass would be less than the sum of its parts (the original parts before they came together).

 

It was quite hard to build up to that position but I got there. Does that make any sense?

 

When they orbit "the final mass of the bound system will be less than the sum of the two masses; the difference is the negative potential energy due to gravity" but some of the energy will be in the form of KE.

Edited March 2, 2016 by Robittybob1

[swansont]

 

When we lift a mass on Earth are we increasing its mass according to Einstein's E= mc^2 equation?

 

 

 

Yes. You do work on the system and that energy is not translational kinetic energy (which is accounted for separately, as it's frame dependent).

[Robittybob1]

 

 

Yes. You do work on the system and that energy is not translational kinetic energy (which is accounted for separately, as it's frame dependent).

I was thinking that at the moment of the lift being finished it has finished moving within that frame, so it is like lifting a mass in a gravitational field 1 meter at the end of that meter the force gone into the lift has all be converted to potential energy and no kinetic energy remains.

On Earth that is like lifting an object onto a table, or stacking 2 cubes on top of each other.

To have the G field present in a frame, what sort of frame are we talking about then, is that an accelerating frame?

When you say "translational kinetic energy" what you do you mean?

Google definition of "translational kinetic energy" is "An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another)."

So it had motion temporarily but that slowed down as it reach the required height.

Does that make sense?

[swansont]

I was thinking that at the moment of the lift being finished it has finished moving within that frame, so it is like lifting a mass in a gravitational field 1 meter at the end of that meter the force gone into the lift has all be converted to potential energy and no kinetic energy remains.

On Earth that is like lifting an object onto a table, or stacking 2 cubes on top of each other.

To have the G field present in a frame, what sort of frame are we talking about then, is that an accelerating frame?

When you say "translational kinetic energy" what you do you mean?

Google definition of "translational kinetic energy" is "An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another)."

So it had motion temporarily but that slowed down as it reach the required height.

Does that make sense?

 

 

Mostly. We don't care much about what happens during the lift, only the state before and after.

 

The true equation for total energy of an object is E² = p²c² + m²c⁴

 

That reduces to the familiar E = mc² for an object at rest (p=0), but one must also note that it means that equation only applies at rest. Anyone who says a moving object has more mass has co-opted this equation and applied it in a new way, and has redefined what mass is.

 

So energy from any motion that isn't associated with this momentum, or any other energy, is reflected in the mass. Rotation, internal vibration, etc. Anything that isn't center-of-mass translation (as defined above). You heat up a cup of coffee and its mass increases. You spin a top and its mass increases. An atom or nucleus absorbs a photon and goes to an excited state, and its mass increases (that last example has been experimentally confirmed) (And this is not the so-called relativistic mass that has changed. None of the masses in these equations is that abomination)

[Robittybob1]

 

 

Mostly. We don't care much about what happens during the lift, only the state before and after.

 

The true equation for total energy of an object is E² = p²c² + m²c⁴

 

That reduces to the familiar E = mc² for an object at rest (p=0), but one must also note that it means that equation only applies at rest. Anyone who says a moving object has more mass has co-opted this equation and applied it in a new way, and has redefined what mass is.

 

So energy from any motion that isn't associated with this momentum, or any other energy, is reflected in the mass. Rotation, internal vibration, etc. Anything that isn't center-of-mass translation (as defined above). You heat up a cup of coffee and its mass increases. You spin a top and its mass increases. An atom or nucleus absorbs a photon and goes to an excited state, and its mass increases (that last example has been experimentally confirmed) (And this is not the so-called relativistic mass that has changed. None of the masses in these equations is that abomination)

OK I saw it being discussed on a YT where they asked does 2 * 1kg blocks weigh more if they are side by side on the scale or if they are stacked on top of each other?

They are not moving. So they are not confounded by relativistic mass. Did the one on top have energy added, so does it have more mass?

I would hazard a guess that since the top mass is at a greater distance from the Earth's center of gravity that it actually weighs less (as measured by scales) even though it has greater mass (inertial) or it could be balanced (the two effects cancel out) and hence weigh the same.

I regret not taking more notice of the answer they gave and I have not attempted to work it out mathematically, but just trying to work through the logic at the moment.

But what do you say is the answer please?

[swansont]

It was yes a few posts ago. Still is.

[Robittybob1]

It was yes a few posts ago. Still is.

I think you mean its mass has increased. OK, if I understood you correctly, when the Moon is tidally accelerated away from the Earth is its mass being increased for the same reason?

Edited March 2, 2016 by Robittybob1

[swansont]

I think you mean its mass has increased. OK, if I understood you correctly, when the Moon is tidally accelerated away from the Earth is its mass being increased for the same reason?

Yes.

[Robittybob1]

OK so when the reverse happens is it a mass loss? Is a decaying orbit is associated with a mass loss? Now it could be argued that any mass the Moon gained the Earth lost so the system is still the same, but in the case of decaying orbits of the binary stars radiating gravitational energy it is not an exchange anymore but a loss to the universe. Then would you say that the binary stars (as exemplified by the Hulse Taylor Binary system) are losing mass as their orbits decay?

Edited March 3, 2016 by Robittybob1