[Dave Lev]

No. Triton is not "broken".
Absolutely not. Triton is not a "broken" object.

Please look again on Triton shape:
https://en.wikipedia.org/wiki/Triton_(moon)#/media/File:Triton_moon_mosaic_Voyager_2_(large).jpg
Is it a regular shape???
Don't you see that it looks as a broken ball?
The shorter diameter in one side is less than 2/3 of the longer side.
Do you agree with that?

[chiralSPO]

No. Triton is not "broken".
Absolutely not. Triton is not a "broken" object.

Please look again on Triton shape:
https://en.wikipedia.org/wiki/Triton_(moon)#/media/File:Triton_moon_mosaic_Voyager_2_(large).jpg
Is it a regular shape???
Don't you see that it looks as a broken ball?
The shorter diameter in one side is less than 2/3 of the longer side.
Do you agree with that?

ummm... perhaps you are confused by the shadow on the moon (just like our own moon, triton is only illuminated by the sun on one side, so it would appear to have different "phases" depending on what angle it is observed at compared to where the sun is)

In the same wikipedia page that contains the image you linked to it reports Triton's radius as 1353.4±0.9 km... not exactly a huge deviation from spherical.

[Kryptid]

Please look again on Triton shape:
https://en.wikipedia.org/wiki/Triton_(moon)#/media/File:Triton_moon_mosaic_Voyager_2_(large).jpg
Is it a regular shape???
Don't you see that it looks as a broken ball?
The shorter diameter in one side is less than 2/3 of the longer side.
Do you agree with that?

Um, do you not know what a shadow is?

Here is a link showing other photographs of Triton. You can more clearly see its spherical shape: http://scienceblogs.com/startswithabang/2009/04/29/the-closest-kuiper-belt-object/

[Dave Lev]

Thanks Kryptid and chiralSPO

Yes, I was confused by the shadow on that moon.
However, as it is a real moon;
How do we know for sure that it drifts inwards?
It was quite difficult for us to find that our moon is drifting outwards.
So, how could we find the drifting direction while Triton is located so far away from us and it orbits around another planet (Neptune).
Please remember that we came briefly closer to that moon only in 1989 when Voyager 2 passed it in.
What kind of technique we have used in order to prove this assumption?
What is the expected inwards drifting rate per year? (Is it just few cm or more than few miles per year?)

[Kryptid]

Thanks Kryptid and chiralSPO

Yes, I was confused by the shadow on that moon.
However, as it is a real moon;
How do we know for sure that it drifts inwards?
It was quite difficult for us to find that our moon is drifting outwards.
So, how could we find the drifting direction while Triton is located so far away from us and it orbits around another planet (Neptune).
Please remember that we came briefly closer to that moon only in 1989 when Voyager 2 passed it in.
What kind of technique we have used in order to prove this assumption?
What is the expected inwards drifting rate per year? (Is it just few cm or more than few miles per year?)

I'm not sure if it actually has been measured or not. We know from physics that there is a source of energy loss from its orbit but do not know of any gains to counter it. So logically, its orbit must be decreasing.

On the other hand, I was able to find observational evidence for the orbital decay of the exoplanet WASP-43b. Basically, data about how long it took to go around its star was used to find that its orbital period is becoming very slightly smaller with time: http://iopscience.iop.org/article/10.3847/0004-6256/151/1/17/meta#aj521586s4

[Dave Lev]

I'm not sure if it actually has been measured or not. We know from physics that there is a source of energy loss from its orbit but do not know of any gains to counter it. So logically, its orbit must be decreasing.

Thanks
So, we don't have any sort of evidence which confirms that triton is drifting inwards.
It is just a logical outcome from an assumption of the source of energy loss.
Can we prove it?

If we can't, then why do we say so surly that it drifts inwards?
It is a mislead information
How can we use this unproved idea in order to prove some other unrealistic ideas?

On the other hand, I was able to find observational evidence for the orbital decay of the exoplanet WASP-43b. Basically, data about how long it took to go around its star was used to find that its orbital period is becoming very slightly smaller with time: http://iopscience.iop.org/article/10.3847/0004-6256/151/1/17/meta#aj521586s4

With regards to WASP-43b planet and its host WASP-43.
https://en.wikipedia.org/wiki/WASP-43b
It is located at 261 LY away.
"WASP-43 is a K-type star in the Sextans constellation that is about 80 parsecs (261 light years) away."
We have detected the WASP-43b planet due to "potential transiting event".
"WASP-43 was first flagged as host to a potential transiting event (when a body crosses in front of and dims its host star) by data collected by SuperWASP"
We have discovered that its orbital period, at the time of WASP-43b's discovery, was the second-shortest orbit yet detected"
"WASP-43b is a dense Hot Jupiter with a mass of 1.78 times the mass of Jupiter, but a radius of 0.93 times that of Jupiter's. The planet orbits its host star at a mean distance of 0.0142 AU every 0.813475 days (19.5234 hours);[5] this orbital period, at the time of WASP-43b's discovery, was the second-shortest orbit yet detected, surpassed only by WASP-19b"
However, how do we know that this one "second-shortest orbit" is not due to our current different point of view?
So, technically the orbital time could be identical, but due to different current location point, we might monitor different orbital time.
Do you agree with that?
Conclusions.
It is quite unrealistic to get into real conclusion from a system which is located so far any from us and based on different locations and different orbital time cycle.
Therefore, this WASP-43b verification can't contribute any real information for our discussion.
Hence, so far there is only one broken moon - Phobos, which our scientists claim that it drifts inwards.
Would you kindly advice how they have got this idea?

[evan_au]

how do we know that this one "second-shortest orbit" is not due to our current different point of view?

The Exoplanet Wasp-43b has an orbital period of 19.5 hours.

The shortest known orbital period for an Exoplanet (in September 2018) is SWIFT J1756.9-2508 b, with an orbital period of 4.3 hours around a millisecond pulsar.

Our point of view places the orbit of these planets on our line of sight to their parent star.

With an orbital decay period of <0.1 seconds per year, there is no way that the orbital period of Wasp-43b (261 light-years away) can be less than SWIFT J1756.9-2508 b.

It is quite unrealistic to get into real conclusion from a system which is located so far any from us and based on different locations and different orbital time cycle.

I'm afraid that I don't see the significance of the distance to the star when we are discussing planets which cross our line of sight?

Or why it may support or devalue the idea of tidal effects causing planets to migrate towards or away from their parent star.

In fact, SWIFT J1756.9-2508 b may be an interesting case: a millisecond pulsar is spinning hundreds of times per second. If tidal effects are significant, then the companion should be moving outwards? Of course, millisecond pulsars are almost perfect ellipsoids, of somewhat unknown stiffness, so how would we calculate the tidal energy loss or gain?

See: https://en.wikipedia.org/wiki/List_of_exoplanet_extremes#Orbital_characteristics

[Kryptid]

It is just a logical outcome from an assumption of the source of energy loss.
Can we prove it?

Yes. We know that tidal forces exist (we can detect them on Earth). We know that tidal forces transfer energy (obviously, since tides represent the movement of matter). We know that Newton's third law requires there to be equal and opposite reactions when energy is exchanged between two bodies. Since Triton is orbiting Neptune in a retrograde manner, it is creating tides that pull in the opposite direction against Neptune's rotation. Since Neptune is composed of matter that has friction, this inevitably leads to Neptune's rotation rate slowing down over time. However, due to conservation of angular momentum, we know that Neptune's spin can't disappear into nothingness. Newton's third law therefore requires that this loss of angular momentum also be transferred to Triton itself, slowing it down and thus pulling it into a closer orbit.'

Everything about tidal acceleration is based on things that we already know to be true from experiment.

So, technically the orbital time could be identical, but due to different current location point, we might monitor different orbital time.
Do you agree with that?

No, because our position relative to that system has nothing to do with how long we would perceive that it takes that planet to orbit that star. I don't know why you think it would.

[Dave Lev]

It is just a logical outcome from an assumption of the source of energy loss.
Can we prove it?

Yes. We know that tidal forces exist (we can detect them on Earth). We know that tidal forces transfer energy (obviously, since tides represent the movement of matter). We know that Newton's third law requires there to be equal and opposite reactions when energy is exchanged between two bodies. Since Triton is orbiting Neptune in a retrograde manner, it is creating tides that pull in the opposite direction against Neptune's rotation. Since Neptune is composed of matter that has friction, this inevitably leads to Neptune's rotation rate slowing down over time. However, due to conservation of angular momentum, we know that Neptune's spin can't disappear into nothingness. Newton's third law therefore requires that this loss of angular momentum also be transferred to Triton itself, slowing it down and thus pulling it into a closer orbit.'

Everything about tidal acceleration is based on things that we already know to be true from experiment.

Thanks for the explanation.
1. Tidal -  " We know that tidal forces exist (we can detect them on Earth)" - Yes. that is correct.
So, yes - there is a tidal between Earth and the Moon. However, the moon is drifting away from the Earth. This is clear for all of us.
In the same token, we know for sure that the Earth is also drifting away from the Sun (with or without tidal impact).
Those are pure evidences. Therefore, I don't understand why you are using the word "tidal" as a proof for drifting inwards, while there is no solid proof for that.
2. Orbiting in a retrograde manner
Thanks for the explanation. However, as I read it it seems to me again that it is a logical outcome.
It might be correct, but is also might be incorrect. There is no solid proof in that explanation.
Even if the following message is correct: "Newton's third law therefore requires that this loss of angular momentum also be transferred to Triton itself, slowing it down",
How do we know for sure that this "slowing down" must lead to: "pulling it into a closer orbit".
Actually I see a contradiction.
Based on Kepler law, if it drifts inwards, (so the radius is shorter), than the orbital velocity should be higher.
Therefore, the only way to keep Keler law while slowing down the orbital velocity is by drifting it outwards and increase its radius (assuming that there is no mass lose).
In any case, we didn't measure the real distance from Neptune to Triton and verify the real changes over time.
So far the only real measurements had been set between Moon-Earth and Earth-sun.
Therefore, do you agree that as long as we don't have a direct measurements of any moon or planet (in the solar system or outside it) which is drifting inwards to its parent Host, we can't say for sure that there is a solid evidence for drifting inwards movement due to tidal (or any other idea) in any orbital system.

[Kryptid]

Thanks for the explanation.
1. Tidal -  " We know that tidal forces exist (we can detect them on Earth)" - Yes. that is correct.
So, yes - there is a tidal between Earth and the Moon. However, the moon is drifting away from the Earth. This is clear for all of us.
In the same token, we know for sure that the Earth is also drifting away from the Sun (with or without tidal impact).
Those are pure evidences. Therefore, I don't understand why you are using the word "tidal" as a proof for drifting inwards, while there is no solid proof for that.

I already supplied you with an example of measurements of an extrasolar planet drifting inward, just as such a planet in such close orbit around its star is predicted to do via tidal braking. Our changing location relative to the system wouldn't change how long we observe it to take the planet to orbit its star.

2. Orbiting in a retrograde manner
Thanks for the explanation. However, as I read it it seems to me again that it is a logical outcome.
It might be correct, but is also might be incorrect. There is no solid proof in that explanation.
Even if the following message is correct: "Newton's third law therefore requires that this loss of angular momentum also be transferred to Triton itself, slowing it down",
How do we know for sure that this "slowing down" must lead to: "pulling it into a closer orbit".
Actually I see a contradiction.
Based on Kepler law, if it drifts inwards, (so the radius is shorter), than the orbital velocity should be higher.
Therefore, the only way to keep Keler law while slowing down the orbital velocity is by drifting it outwards and increase its radius (assuming that there is no mass lose).

Think about a spaceship in orbit around the Earth. It is travelling just fast enough to avoid losing altitude. Now we have the ship fire thrusters in the opposite direction of its movement to slow it down. The ship is travelling more slowly than before, but gravity is still pulling on it just as hard. For this reason, it will not travel as far before it falls the same amount of distance. This leads to a decrease in the altitude and therefore the diameter of the orbit. However (and this is the important part), the very act of falling into a tighter orbit will convert some of the ship's gravitational potential energy into kinetic energy, acting to speed it up again. So even though we slowed down the ship initially, its final velocity will be higher in its new orbit than the old one. The same thing happens with planets that fall into tighter orbits around their stars.

In any case, we didn't measure the real distance from Neptune to Triton and verify the real changes over time.
So far the only real measurements had been set between Moon-Earth and Earth-sun.

This is technically true. If there is another source of energy present that we don't know about, Triton could theoretically avoid falling into Neptune if it is gaining enough energy from that unknown source to offset tidal braking. However, we still know that tidal breaking must be a phenomenon that exists via logical deduction based on the laws of physics.

Therefore, do you agree that as long as we don't have a direct measurements of any moon or planet (in the solar system or outside it) which is drifting inwards to its parent Host, we can't say for sure that there is a solid evidence for drifting inwards movement due to tidal (or any other idea) in any orbital system.

Not any more than we can say that we don't have solid evidence that the Empire State Building would fall if we dropped it out of a giant aircraft from 80,000 feet. Such a thing has obviously never been observed before, but we certainly know enough about physics to deduce that it must fall. Same thing with tidal acceleration.

[Professor Mega-Mind (OP)]

...............Orbital Pictogram
 Before the laws , before the formulas , and before the equations , there was the system , and it was good !
 I can see that you have a good understanding of the Earth/Moon system .  You know that instead of being directly under the Moon , as it " should " be , the tidal bulge is displaced forward by Earth's 24 hr. rotation .  The bulge " wants " to stay directly under Luna , BUT the Earth's crust resists deformation .  It takes time for the Moon's pull to lift that bulge , it also takes energy.  By the time the bulge is raised , it has rotated out " ahead " of Luna .  It now exerts a slight forward pull on the Moon , even as the Moon exerts a slight backwards pull on the bulge .  These two pulls cause the Moon to speed up , and the Earth to slow down .  The caveat is that forcing the Earth's crust thru that massive distortion slows it's rotation down , and generates huge amounts of heat in the planet's crust .  If you took the Moon , and magically reversed it's revolution , all of the above would remain true EXCEPT ; the bulge's pull would now slow Luna's revolution , instead of accelerating it .  It would also slow Earth's rotation even more , as Luna spiralled down onto it .
  The same dynamics apply to both Phobos , and Triton .  The difference being that they are tiny compared to their planets .   
In conclusion , these systems effectively trade off rotational inertia .  In the process , they convert some of it into heat , thus the apparently disappearing momentum of the systems . 
 Alright man , pinball time ! ...P.M.

[Dave Lev]

extrasolar planet
 

I already supplied you with an example of measurements of an extrasolar planet drifting inward, just as such a planet in such close orbit around its star is predicted to do via tidal braking. Our changing location relative to the system wouldn't change how long we observe it to take the planet to orbit its star.

I can agree with that statement if we are moving in full synchronization and do not change our distance and relative location over time.
In this case, we will see the same relative orbital cycle of the extrasolar planet around its parent star.
However, what is the chance for that?
1. Bobbing around the galactic disc.
We know that the Sun is bobing up and down from the galactic disc in some sort of a sine wave.
Currently we are moving upwards from the galactic disc.
In the same way, any star in Orion arm (and at any other arm) is bobbing up or down.
So, what is the chance that Wasp-43 or SWIFT J1756.9-2508 stars are located exactly at the same distance from the galactic disc as we do and ride on the same sine wave as we do at the same velocity as we do?
I would say that the chance for that is less than 1 to one Billion.
There is good chance that as we go up they might go down.
So, there is good chance that we see the extrasolar planet orbital cycle from different location over time.
As an example -
A. Lets assume that we are located directly at the orbital disc of the extrasolar planet (but at a distance of 280 Ly away) - In this case, the planet crosses its parent star exactly at the center. Hence, we actually should see that the planet is moving in one striate line, left and right.
B. Let's assume that we are located few degrees above or below the extrasolar planet orbital disc - In this case, we should start to see the elliptical shape of the orbital cycle.
C. As we move higher, the elliptical cycle will be wider.
D. At some point, if the elliptical cycle will be wide enough, we might even see that the planet cycle doesn't cross any more its parent star.
So, let's assume that four years ago, we have been located directly on the extrasolar planet orbital disc. However, after four years we have moved a little bit higher than that  extrasolar planet orbital disc.
The outcome is that we see a shorter orbital cycle, even if the real orbital cycle is the same.
There is another issue - Accuracy.
Did we measure the time based on atomic clock?
If not, less than one sec per four years might be in the range of the measured accuracy clock.
So, I have proved that as our relative location is changing over time, it effects the orbital cycle that we see.
As we can't prove that there is full synchronization between the Sun orbital cycle around the center of the galaxy to those extrasolar systems, and as we don't know the accuracy of the measured time, do you agree that there is no meaning for that time decrease?

However, let's focus again on Triton and Phobos.
As all of us are located in the same solar system, we all move in full synchronization.
So, if we can set a compensation based on relative orbital location, why don't we measure their orbital cycle time frame and see if there is any change over time?
If that idea works Ok for extrasolar planet, why it can't work for solar planets and moons?

[Kryptid]

I can agree with that statement if we are moving in full synchronization and do not change our distance and relative location over time.
In this case, we will see the same relative orbital cycle of the extrasolar planet around its parent star.
However, what is the chance for that?
1. Bobbing around the galactic disc.
We know that the Sun is bobing up and down from the galactic disc in some sort of a sine wave.
Currently we are moving upwards from the galactic disc.
In the same way, any star in Orion arm (and at any other arm) is bobbing up or down.
So, what is the chance that Wasp-43 or SWIFT J1756.9-2508 stars are located exactly at the same distance from the galactic disc as we do and ride on the same sine wave as we do at the same velocity as we do?
I would say that the chance for that is less than 1 to one Billion.
There is good chance that as we go up they might go down.
So, there is good chance that we see the extrasolar planet orbital cycle from different location over time.
As an example -
A. Lets assume that we are located directly at the orbital disc of the extrasolar planet (but at a distance of 280 Ly away) - In this case, the planet crosses its parent star exactly at the center. Hence, we actually should see that the planet is moving in one striate line, left and right.
B. Let's assume that we are located few degrees above or below the extrasolar planet orbital disc - In this case, we should start to see the elliptical shape of the orbital cycle.
C. As we move higher, the elliptical cycle will be wider.
D. At some point, if the elliptical cycle will be wide enough, we might even see that the planet cycle doesn't cross any more its parent star.
So, let's assume that four years ago, we have been located directly on the extrasolar planet orbital disc. However, after four years we have moved a little bit higher than that  extrasolar planet orbital disc.
The outcome is that we see a shorter orbital cycle, even if the real orbital cycle is the same.

Why would I measure a merry-go-round as spinning faster from a mountain top than I would if I were standing right beside it? That makes no sense.

There is another issue - Accuracy.
Did we measure the time based on atomic clock?
If not, less than one sec per four years might be in the range of the measured accuracy clock.

They gave the accuracy level in the article: −0.02890795 ± 0.00772547 seconds per year. The number ± 0.00772547 seconds represents the uncertainty in the measurement.

So, I have proved that as our relative location is changing over time, it effects the orbital cycle that we see.

Except that you did not. Walking towards or away from a merry-go-round will not affect how fast its rotation looks to me. My height relative to it won't affect it either. Nor will walking to the left or the right.

As we can't prove that there is full synchronization between the Sun orbital cycle around the center of the galaxy to those extrasolar systems, and as we don't know the accuracy of the measured time, do you agree that there is no meaning for that time decrease?

No, I do not agree. The accuracy level was stated in the article. You must think that astrophysicists are morons if you don't think they take all of these things into consideration when making measurements like this.

However, let's focus again on Triton and Phobos.
As we located in the same solar system, we all move in full synchronization.
So, if we can set a compensation based on relative orbital location, why don't we measure their orbital cycle time frame and see if there is any change over time?
If that idea works Ok for extrasolar planet, why it can't work for solar planets and moons?

It might work, but unlike a star with a planet going around it, a distant planet with a moon is not self-luminous and the masses involved are much, much smaller. I don't know if our current technology can measure such a thing or not.

[Dave Lev]

They gave the accuracy level in the article: −0.02890795 ± 0.00772547 seconds per year. The number ± 0.00772547 seconds represents the uncertainty in the measurement.

Thanks

Are you sure that those numbers represents accuracy?
They don't say even one word about time accuracy in this article. It is just stated that:
http://iopscience.iop.org/article/10.3847/0004-6256/151/1/17/meta#aj521586s4
"On the other hand, for the solid curve in Figure 2, the overall orbital decay rate is dP/dt = δP/Pq = −0.02890795± 0.00772547 s year−1, which is one order smaller than the values in previous work. Therefore, with our newly observed transits, we obtain a very different orbital decay rate. These results indicate that if there is any orbital decay, the decay rate shall be much smaller than those values proposed in previous works. "
So it represents the overall orbital decay rate.
Hence, what is the real clock accuracy?

[Kryptid]

Are you sure that those numbers represents accuracy?

The number ± 0.00772547 seconds per year represents the uncertainty in the measurement. This means that the true value for the decay rate could be anywhere between -0.03663342 and -0.02118248 seconds per year. Both of those extremes are still negative numbers. So yes, the data points to a decaying orbit.

Again, astrophysicists are not morons. They wouldn't publish data that had a useless level of accuracy (at least not without noting that the accuracy was untrustworthy in their articles).

[Dave Lev]

Are you sure that those numbers represents accuracy?

The number ± 0.00772547 seconds per year represents the uncertainty in the measurement. This means that the true value for the decay rate could be anywhere between -0.03663342 and -0.02118248 seconds per year. Both of those extremes are still negative numbers. So yes, the data points to a decaying orbit.

Again, astrophysicists are not morons. They wouldn't publish data that had a useless level of accuracy (at least not without noting that the accuracy was untrustworthy in their articles).

Thanks

So, "The number ± 0.00772547 seconds per year represents the uncertainty in the measurement."
But they don't say what kind of clock they have used.
If the clock is Atomic, with almost infinite accuracy, than this ± 0.00772547 can represents the real accuracy of the measurements.
However, if for example they have used a clock with accuracy of ± 0.03 seconds per year, then the total real uncertainty in the measurement must be:
± 0.00772547 ± 0.03.
Do you agree that in this case it could change dramatically the meaning of the measurement?
Unless, the ± 0.00772547 represents the uncertainty in the measurement plus the uncertainty in the clock.

In any case, if we focus in the solar system:
I understand why our scientists assume that Triton is drifting inwards.
However, would you kindly explain why they also assume that Phobos is drifting inwards?
Is it based on the same idea as Triton?

[Kryptid]

So, "The number ± 0.00772547 seconds per year represents the uncertainty in the measurement."
But they don't say what kind of clock they have used.
If the clock is Atomic, with almost infinite accuracy, than this ± 0.00772547 can represents the real accuracy of the measurements.
However, if for example they have used a clock with accuracy of ± 0.03 seconds per year, then the total real uncertainty in the measurement must be:
± 0.00772547 ± 0.03.
Do you agree that in this case it could change dramatically the meaning of the measurement?
Unless, the ± 0.00772547 represents the uncertainty in the measurement plus the uncertainty in the clock.

If that was the case, then they would have said that the uncertainty was ± 0.03772547 seconds per year. Do you really not think they know how to take these kinds of things into account?

In any case, if we focus in the solar system:
I understand why our scientists assume that Triton is drifting inwards.
However, would you kindly explain why they also assume that Phobos is drifting inwards?
Is it based on the same idea as Triton?

Yes, because it orbits faster than Mars rotates. However, we do actually have measurements of Phobos' orbital decay from Martian probes. It loses about 0.8 seconds from its orbital period every year: http://www-geodyn.mit.edu/bills_phobos05.pdf