I Comparison of tidal forces acting on the Moon vs Enceladus

[Vrbic]

I have heard about a moon Enceladus. Which is powered by tidal force. I suppose this force press back and forth on the moon and friction in the core causes heat. I hope I'm right :-)
Now tidal force $F_t=\frac{2GMmr}{R^3}$, where $G$ is Gravitation constant, $M$ is mass of planet causes gravitational field, $m$ and $r$ is mass and radius of body where we looking for tidal force and $R$ is distance of both objects.
I took data from Wikipedia about Enceladus, Saturn, Moon and Earth and put it to this formula ($M_S$~$100M_E$,
$m_{enc}$~$m_m/1000$, $r_{enc}$~$r_m/10$, $R_{enc-S}$~$R_{m-E}$). I find out that force acting on Moon is 100 times stronger. My question is why our Moon is cold and on Enceladus is warm water? Or ok, probably there are other factors (radioactive decay in core of Enceladus and on Moon it is not) but I would expect something more than cold stone :-)
Can anyone explain it?

 

[haruspex]

I have heard about a moon Enceladus. Which is powered by tidal force. I suppose this force press back and forth on the moon and friction in the core causes heat. I hope I'm right :-)
Now tidal force $F_t=\frac{2GMmr}{R^3}$, where $G$ is Gravitation constant, $M$ is mass of planet causes gravitational field, $m$ and $r$ is mass and radius of body where we looking for tidal force and $R$ is distance of both objects.
I took data from wikipedia about Enceladus, Saturn, Moon and Earth and put it to this formula ($M_S$~$100M_E$,
$m_{enc}$~$m_m/1000$, $r_{enc}$~$r_m/10$, $R_{enc-S}$~$R_{m-E}$). I find out that force acting on Moon is 100 times stronger. My question is why our Moon is cold and on Enceladus is warm water? Or ok, probably there are other factors (radioactive decay in core of Enceladus and on Moon it is not) but I would expect something more than cold stone :-)
Can anyone explane it?

Force is not the same as power. What relates the two? How is that different between our moon and Enceladus?

 

[Vrbic]

Force is not the same as power. What relates the two? How is that different between our moon and Enceladus?

$P=\frac{F s}{t}$, where $s$ is a path induced by the force and $t$ is time duration of acting force.
Honestly, I'm not sure, how find out the path (prolongation or relative prolongation). If I use Hooke's law , I don't know Young's modulus E. And time is also tough nut to crack. Moon and Earth rotate in captured rotation. So Moon is still stretched in same direction. On the other hand now I read that Enceladus also rotates in captured rotation...
Please kick me again :-)

 

[haruspex]

where s is a path induced

Well, it does not need to be induced by the force, but certainly influenced by the force.

Moon and Earth rotate in captured rotation

Right, and that was certainly something to consider. What else about the orbit might cause the force distribution within the satellite to vary over time?

 

[Vrbic]

Right, and that was certainly something to consider. What else about the orbit might cause the force distribution within the satellite to vary over time?

Ok, I know what you point out. Eccentricity. A change in radial distance from center of a planet. Ok so, then some mean power could find out as $<P>=\frac{\Delta F_t}{T/2}$ where $\Delta F_t=F_t(R_p)-F_t(R_a)$ and $T$ is period.
Do you agree?

 

[haruspex]

Ok, I know what you point out. Eccentricity. A change in radial distance from center of a planet. Ok so, then some mean power could find out as $<P>=\frac{\Delta F_t}{T/2}$ where $\Delta F_t=F_t(R_p)-F_t(R_a)$ and $T$ is period.
Do you agree?

I'm unsure of your notation.
Consider a rod length 2r, always pointing at the planet (from some constraint), with a mass at each end. Orbiting independently, the distance between the masses would vary, so the force in the rod varies.

 

[Vrbic]

I'm unsure of your notation.
Consider a rod length 2r, always pointing at the planet (from some constraint), with a mass at each end. Orbiting independently, the distance between the masses would vary, so the force in the rod varies.

Ok, I found a mistake in my logic :-) There is missing $\Delta r$ the change in distance (the change in diameter of a moon) between pericenter $R_p$ and apocenter $R_a$.
$<P>=\frac{\Delta F_t *\Delta r}{T/2}$.

So rod with masses at each end... I understand that distance between them and force varies with distance from the center of planet. But I don't know how. What is the change in distance?

 

[haruspex]

What is the change in distance?

That depends on the elasticity. If the satellite were perfectly rigid it would not be warmed.

 

[Vrbic]

That depends on the elasticity. If the satellite were perfectly rigid it would not be warmed.

Yes, that's what I mentioned above, in second message where I wrote about Hooke's law and Young's modulus.
But back to my question, why the heat induced by tidal friction is not in our Moon so big?
I realized that important isn't force but power, ok. But you asked my what power is induced in both moons but I still don't know how to resolve it.
Is eccentricity important? Or elasticity? I guess both, but I'm not capable to guess what is more important.

 

[haruspex]

Is eccentricity important? Or elasticity? I guess both, but I'm not capable to guess what is more important.

It's the combination. Both are necessary (given the locked rotation).

 

[Vrbic]

It's the combination. Both are necessary (given the locked rotation).

Ok and what about power which you mentioned? And global question why our Moon isn't heated by tidal friction and Enceladus is?

 

[haruspex]

Ok and what about power which you mentioned? And global question why our Moon isn't heated by tidal friction and Enceladus is?

Develop some equations.
Suppose the moon has radius r, the orbital radius varies between R_min and R_max, the period is T, the elasticity is k, and whatever other unknowns you think you need. How much energy is stored and released each orbit?

 

[willem2]

There still seems to be tidal heating on the moon as well, but since the moon has little water, you'll only get some rocks that are a bit hotter, and no cryovulcanism.
https://www.nao.ac.jp/en/news/science/2014/20140807-rise.html
It's much easier to melt ice to get vulcanoes than to melt rock.

 

[Vrbic]

Develop some equations.
Suppose the moon has radius r, the orbital radius varies between R_min and R_max, the period is T, the elasticity is k, and whatever other unknowns you think you need. How much energy is stored and released each orbit?

Ok, may I suppose that moon is a kind of dumbbell with massless bar? Or should I choose another approximation?

 

[Vrbic]

Ok so I used dumbbell approximation:
$r \textrm{(radius of moon)}<<R\textrm{(distance earth-moon)}$ I use approximation $\frac{1}{(R+r)^2}\doteq \frac{1}{R^2}-\frac{2r}{R^3}...$ I hope I may.
In $R_{max}$
$$F_{max}=GMm(\frac{1}{R_{max}^2}-\frac{1}{R_{max}^2}+\frac{2r_{max}}{R_{max}^3})=\frac{2GMmr_{max}}{R_{max}^3}$$
In $R_{min}$
$$F_{min}=GMm(\frac{1}{R_{min}^2}-\frac{1}{R_{min}^2}+\frac{2r_{min}}{R_{min}^3})=\frac{2GMmr_{min}}{R_{min}^3}$$
I use eccentricity $e$:
$$R_{min}=\frac{1-e}{1+e}R_{max}=yR_{max}, y=y(e) $$
and relative prolongation $l$:
$$l=\frac{r_{max}-r_{min}}{r_{max}} =>r_{min}=(1-e)r_{max}=xr_{max}, x=x(l)$$
So diference of forces on opposite side of orbit is: $$\Delta F=2GMm(\frac{r_{max}}{R_{max}^3}-\frac{r_{min}}{R_{min}^3})=2GMmr_{max}\frac{y^3-x}{(yR_{max})^3}$$ or $$\Delta F=\frac{2GMmr_{max}}{R_{max}^3}(1-\frac{x}{y^3}) $$
Maybe I can suppose that $e$ and $l$ are small so: $x=1-l$ and $y^{-3}\doteq1+6e$ than
$$\Delta F=\frac{2GMmr_{max}}{R_{max}^3}(6e-l)$$
What do you mean about that? Is it ok as a approximation?

 

[Vrbic]

Continuation:
Potential energy stored due to half orbit should be: $U=\int{F(r)dr}$, where $r$ is prolongation from $r_{max}$ (now I see that in my case it's more like a shortening). Integration with substitution $l_0=\frac{r}{r_{max}}$ than is:
$$U=\frac{2GMmr_{max}^2}{R_{max}^3}\int_0^l(6e-l_0)dl_0=\frac{GMmr_{max}^2}{R_{max}^3}(12e-l)l$$
Is it good idea to deal with potential energy of this process like that?

 

[haruspex]

Ok so I used dumbbell approximation:
$r \textrm{(radius of moon)}<<R\textrm{(distance earth-moon)}$ I use approximation $\frac{1}{(R+r)^2}\doteq \frac{1}{R^2}-\frac{2r}{R^3}...$ I hope I may.
In $R_{max}$
$$F_{max}=GMm(\frac{1}{R_{max}^2}-\frac{1}{R_{max}^2}+\frac{2r_{max}}{R_{max}^3})=\frac{2GMmr_{max}}{R_{max}^3}$$
In $R_{min}$
$$F_{min}=GMm(\frac{1}{R_{min}^2}-\frac{1}{R_{min}^2}+\frac{2r_{min}}{R_{min}^3})=\frac{2GMmr_{min}}{R_{min}^3}$$
I use eccentricity $e$:
$$R_{min}=\frac{1-e}{1+e}R_{max}=yR_{max}, y=y(e) $$
and relative prolongation $l$:
$$l=\frac{r_{max}-r_{min}}{r_{max}} =>r_{min}=(1-e)r_{max}=xr_{max}, x=x(l)$$
So diference of forces on opposite side of orbit is: $$\Delta F=2GMm(\frac{r_{max}}{R_{max}^3}-\frac{r_{min}}{R_{min}^3})=2GMmr_{max}\frac{y^3-x}{(yR_{max})^3}$$ or $$\Delta F=\frac{2GMmr_{max}}{R_{max}^3}(1-\frac{x}{y^3}) $$
Maybe I can suppose that $e$ and $l$ are small so: $x=1-l$ and $y^{-3}\doteq1+6e$ than
$$\Delta F=\frac{2GMmr_{max}}{R_{max}^3}(6e-l)$$
What do you mean about that? Is it ok as a approximation?

The way you involve r_max and r_min in your equations unnecessarily mixes up two things.
For the purpose of finding the differential gravitational forces between min and max R, the variation in r is too small to be relevant. Take r as constant at that stage.
Having found the variation in stress on the satellite, use the elasticity to find the variation in stored PE.

Note that this still does not directly translate into heat generated. When the stress on an elastic body is varied, so that it alternately stretches and contracts, it has two different coefficients. It is the difference between them that causes mechanical work to turn into heat. But the important thing here is that for a given type of body the heat will be proportional to the total turnover of work.

 

[Vrbic]

The way you involve r_max and r_min in your equations unnecessarily mixes up two things.
For the purpose of finding the differential gravitational forces between min and max R, the variation in r is too small to be relevant. Take r as constant at that stage.
Having found the variation in stress on the satellite, use the elasticity to find the variation in stored PE.

Note that this still does not directly translate into heat generated. When the stress on an elastic body is varied, so that it alternately stretches and contracts, it has two different coefficients. It is the difference between them that causes mechanical work to turn into heat. But the important thing here is that for a given type of body the heat will be proportional to the total turnover of work.

Thank you for leading. I have few question if I may:
1) Is introducing of eccentricity and everything around it alright?

2) How now I find the potential energy due to orbiting? The path is now the distance $R$ from a planet? It is my integration variable? (I don't mean by it energy for a heating, I mean, energy coming from effect of rotation - a changing distance.)

3) Will I find the heat $Q$ in way I subtract my potential energy $U$ and kinetic energy $E$ due to vibration of moon? $Q=U-E$?

 

[haruspex]

Thank you for leading. I have few question if I may:
1) Is introducing of eccentricity and everything around it alright?

2) How now I find the potential energy due to orbiting? The path is now the distance $R$ from a planet? It is my integration variable? (I don't mean by it energy for a heating, I mean, energy coming from effect of rotation - a changing distance.)

3) Will I find the heat $Q$ in way I subtract my potential energy $U$ and kinetic energy $E$ due to vibration of moon? $Q=U-E$?

Your expressions for F_min and F_max in post #15 are fine, except that you can use just average radius r of the satellite in both.
If the elasticity of the satellite is k, you can find the corresponding min and max elastic PE. The turnover of the elastic work each orbit is the difference between the two. The work turned to heat will be some fixed fraction of that. You can assume the fraction is the same for bodies of similar composition.

 

[Vrbic]

Your expressions for F_min and F_max in post #15 are fine, except that you can use just average radius r of the satellite in both.
If the elasticity of the satellite is k, you can find the corresponding min and max elastic PE. The turnover of the elastic work each orbit is the difference between the two. The work turned to heat will be some fixed fraction of that. You can assume the fraction is the same for bodies of similar composition.

The EP is my $U$? I'm a bit lost in terminology. Now I don't know if my concept of potential energy from post #16 is right i.e what do you mean about question 2 from post #18 .
Also you are mentioned a elasticity k. So you mean to use Hooke's law. Or you mean k use as a spring constant (or find some conection between them)?
What do you mean about question 3 from post #18? Is it good idea?

 

[Vrbic]

I was thinking about elasticity: A moon isn't any "material". It is gravitationally bound object. So I believe that main factor acting against stretching is gravity. What do you mean? May I equal gravitational force and force describing spring or Hook's law? And extract elasticity?

 

[A.T.]

So I believe that main factor acting against stretching is gravity.

Gravity is opposing the stretching by gravity?

 

[Vrbic]

Ok I tried it (probably you were talking about that). If the moon is in all position of orbit stable it should be true:
$F$ is spring force, $G$ is gravitational force, $x$ is change in diameter
$F_0-F_1=G_0-G_1$
$0+kx=-2Gm(\frac{1}{r^2}-\frac{1}{(r+x)^2}), x>>r => kx=\frac{4Gmx}{r^3}$
$k=\frac{4Gm}{r^3}$
What do you mean about it?

 

[Vrbic]

Gravity is opposing the stretching by gravity?

The gravity of the moon is against the gravity of the planet which stretches the moon. No? Wrong assumption?

 

[A.T.]

The gravity of the moon is against the gravity of the planet which stretches the moon.

If you decompose gravity like that, it makes sense.

 

[haruspex]

The EP is my $U$? I'm a bit lost in terminology. Now I don't know if my concept of potential energy from post #16 is right i.e what do you mean about question 2 from post #18 .
Also you are mentioned a elasticity k. So you mean to use Hooke's law. Or you mean k use as a spring constant (or find some conection between them)?
What do you mean about question 3 from post #18? Is it good idea?

The gravitational PE is not relevant.
In the dumbell model, the end nearest the planet is trying to go in one orbit, but the far end is trying to go in a different one. The distance between the two orbits varies, stretching the dumbell. The work that ends as heat is a portion of the work goes into stretching and relaxing the dumbell. So it is the variation in elastic PE that matters.

 

[Vrbic]

The gravitational PE is not relevant.
In the dumbell model, the end nearest the planet is trying to go in one orbit, but the far end is trying to go in a different one. The distance between the two orbits varies, stretching the dumbell. The work that ends as heat is a portion of the work goes into stretching and relaxing the dumbell. So it is the variation in elastic PE that matters.

Ok, I understand but I don't know where are you pointing on. Gravitational potential energy is not relevant. Ok but potential energy caused by stretching important is. No? And I hope it these which I tried to find in #18. Or why I dealed with a tidal force? It was useless? Could you be more specific how to find a power which drives a heating? I can't recognize what I have done good and what was useless.

 

[Janus]

I have heard about a moon Enceladus. Which is powered by tidal force. I suppose this force press back and forth on the moon and friction in the core causes heat. I hope I'm right :-)
Now tidal force $F_t=\frac{2GMmr}{R^3}$, where $G$ is Gravitation constant, $M$ is mass of planet causes gravitational field, $m$ and $r$ is mass and radius of body where we looking for tidal force and $R$ is distance of both objects.
I took data from Wikipedia about Enceladus, Saturn, Moon and Earth and put it to this formula ($M_S$~$100M_E$,
$m_{enc}$~$m_m/1000$, $r_{enc}$~$r_m/10$, $R_{enc-S}$~$R_{m-E}$). I find out that force acting on Moon is 100 times stronger. My question is why our Moon is cold and on Enceladus is warm water? Or ok, probably there are other factors (radioactive decay in core of Enceladus and on Moon it is not) but I would expect something more than cold stone :-)
Can anyone explain it?

There are a number a factors to consider. Enceladus is less dense than the Moon and thus likely not as rigid. As a comparison, consider what would happen if you were rapidly squeezing and releasing a foam rubber ball, vs. doing the same, with equal force to the ball made of a rigid material. The rubber ball will get warmer, while the rigid ball not so much.
Also consider the difference between repeatedly squeezing the ball in quick succession vs. doing so at a slower pace. The ball will warm more in the first case than in the second. This comes into play when you consider that Enceladus orbits Saturn in just 33 hrs, vs. the 27 1/3 days it take for the Moon to orbit the Earth.

 

[haruspex]

Ok, I understand but I don't know where are you pointing on. Gravitational potential energy is not relevant. Ok but potential energy caused by stretching important is. No? And I hope it these which I tried to find in #18. Or why I dealed with a tidal force? It was useless? Could you be more specific how to find a power which drives a heating? I can't recognize what I have done good and what was useless.

As I wrote, your equations for F_min and F_max in post #15 are nearly right. (I assume these are what you mean by tidal force.) The only change needed is to replace r_min and r_max in those with the mean satellite radius, r.
(Actually, I think the tension in the satellite/dumbbell would be half of these, but it does not matter for present purposes.)

We don't know what the elasticity is, and it may different for different satellites. And the work lost to heat depends on the difference between that for the stretching phase and that for the relaxation phase. Let these be k_s, k_r respectively.

If you plot force against extension this means you get two lines of different slopes, k_s above k_r. A stretch/relax cycle consists of starting at some point on the upper line, moving to the right and up as it is stretched, up to the max extension. As soon as relaxation starts, you can suppose it drops almost vertically to the lower slope, then follows that line down and to the left until directly under the point where it started. On going back into extension, it rises amost vertically until back on the upper slope. And so the cycle repeats.
This is an example of hysteresis. The work lost to heat each cycle is the area inside the quadrilateral.

Can you compute that area in terms of F_min, F_max, k_r and k_s?

 

[Vrbic]

As I wrote, your equations for F_min and F_max in post #15 are nearly right. (I assume these are what you mean by tidal force.) The only change needed is to replace r_min and r_max in those with the mean satellite radius, r.
(Actually, I think the tension in the satellite/dumbbell would be half of these, but it does not matter for present purposes.)

You are right, or $m$ is a weight only one part of dumbbell.

 

[Vrbic]

We don't know what the elasticity is, and it may different for different satellites. And the work lost to heat depends on the difference between that for the stretching phase and that for the relaxation phase. Let these be ks, kr respectively.

I understand what you're saying, but I can't imagine how or why is there different elasticity for stretching and relaxing phase.

If you plot force against extension this means you get two lines of different slopes, ks above kr. A stretch/relax cycle consists of starting at some point on the upper line, moving to the right and up as it is stretched, up to the max extension. As soon as relaxation starts, you can suppose it drops almost vertically to the lower slope, then follows that line down and to the left until directly under the point where it started. On going back into extension, it rises amost vertically until back on the upper slope. And so the cycle repeats.
This is an example of hysteresis. The work lost to heat each cycle is the area inside the quadrilateral.

Can you compute that area in terms of Fmin, Fmax, kr and ks?

Again, mathematically I understand two linear function, quadrilateral, area of quadrilateral is $A=ef\sin(\phi)/2$, where $e$ and $f$ are diagonals and $\phi$ is angle between them, no problem, but in #15 isn't any elasticity. Formula for a force of spring is F=-kx ok here is k and I suppose $2GMm/R^3$ play a role of $k$, but still I'm not sure how do you mean it. Or how to compose $k$ in there.

 

[Vrbic]

Again, mathematically I understand two linear function, quadrilateral, area of quadrilateral is $A=ef\sin(\phi)/2$, where $e$ and $f$ are diagonals and $\phi$ is angle between them, no problem, but in #15 isn't any elasticity. Formula for a force of spring is F=-kx ok here is k and I suppose $2GMm/R^3$ play a role of $k$, but still I'm not sure how do you mean it. Or how to compose $k$ in there.

Or do you mean that $k_s=2GMm/R_{min}^3$ and $k_r=2GMm/R_{max}^3$??

 

[haruspex]

Or do you mean that $k_s=2GMm/R_{min}^3$ and $k_r=2GMm/R_{max}^3$??

No.
If the elasticity coefficients were the same for deformation and relaxation then all the work done in compressing or stretching a spring or piece of rubber etc. would be recoverable in the relaxation phase. But we know this is never true. Balls never bounce to the original height. Rubber bands warm up when repeatedly stretched and released.

 

[Vrbic]

No.
If the elasticity coefficients were the same for deformation and relaxation then all the work done in compressing or stretching a spring or piece of rubber etc. would be recoverable in the relaxation phase. But we know this is never true. Balls never bounce to the original height. Rubber bands warm up when repeatedly stretched and released.

Ok,
and now back

Can you compute that area in terms of F_min, F_max, k_r and k_s?

How I get elasticity to my formula for $F_{min}$ and $F_{max}$?

 

[haruspex]

area of quadrilateral is $A=ef\sin(\phi)/2$

Not the most convenient formula here. The sides are vertical, which makes it much simpler.

How I get elasticity to my formula for $F_{min}$ and $F_{max}$?

The two elasticities give the slopes at top and bottom of the quadrilateral, and you know both lines pass through the origin.
You need to decide which vertex corresponds to F_min and which to F_max. Any thoughts?

 

[Vrbic]

Not the most convenient formula here. The sides are vertical, which makes it much simpler.

The two elasticities give the slopes at top and bottom of the quadrilateral, and you know both lines pass through the origin.
You need to decide which vertex corresponds to F_min and which to F_max. Any thoughts?

Ok, I want to draw an energy. So for a stretching phase I have to put more energy more strength to change r than in relaxing phase. Stretching starts at Rmax and finishes at Rmin and here is acting Fmax. A relaxing phase starts at Rmin and finishes at Rmax where I guess is acting Fmin. i.e. I have to have greater value for ks than kr. I'm attaching a figure.

https://www.dropbox.com/s/hxuetrfgcim92j3/F vs r.jpg?dl=0
It is a trapezoid so area $A=\frac{(F_s(r)-F_r(r))+(F_s(r+dr)-F_r(r+dr))*dr}{2}$, where $F_r(r)=F_{min}$ in Rmax and $F_s(r+dr)=F_{max}$ in Rmin.
Right?

 

[Vrbic]

Not the most convenient formula here. The sides are vertical, which makes it much simpler.

The two elasticities give the slopes at top and bottom of the quadrilateral, and you know both lines pass through the origin.
You need to decide which vertex corresponds to F_min and which to F_max. Any thoughts?

I know, it is hard with me, but could you summarize it briefly?

 

[haruspex]

Ok, I want to draw an energy. So for a stretching phase I have to put more energy more strength to change r than in relaxing phase. Stretching starts at Rmax and finishes at Rmin and here is acting Fmax. A relaxing phase starts at Rmin and finishes at Rmax where I guess is acting Fmin. i.e. I have to have greater value for ks than kr. I'm attaching a figure.

https://www.dropbox.com/s/hxuetrfgcim92j3/F vs r.jpg?dl=0
It is a trapezoid so area $A=\frac{(F_s(r)-F_r(r))+(F_s(r+dr)-F_r(r+dr))*dr}{2}$, where $F_r(r)=F_{min}$ in Rmax and $F_s(r+dr)=F_{max}$ in Rmin.
Right?

Sorry, I forgot to reply to this. Your diagram is good, but I needed more time to go through the rest... and I don't have time right now. Will get to it when I can... maybe in a few hours.

 

[Vrbic]

Sorry, I forgot to reply to this. Your diagram is good, but I needed more time to go through the rest... and I don't have time right now. Will get to it when I can... maybe in a few hours.

No problem, I'm happy for any help and advice.
Thnaks and keep in touch :-)

 

[haruspex]

No problem, I'm happy for any help and advice.
Thnaks and keep in touch :-)

Ok, your equation is right, but it would be better to have k_s and k_r in there instead of needing dr.
r=F_min/k_r and r+dr=F_max/k_s.

 

[Vrbic]

Ok, your equation is right, but it would be better to have k_s and k_r in there instead of needing dr.
r=F_min/k_r and r+dr=F_max/k_s.

My problem is, that I can't see where are you pointing on, and only do exactly what you say.
So I try to summarize what I know about it till now, please correct the wrong.

When I study such problem, I have to expect, that a energy loss is due to different elasticity. If I have some observing data, I determine a apocenter and a pericenter and difference in radius of a moon and calculate forces in these places. Then I "guess" a elasticity and plot similar graph as I attached last time. And then the closed area - trapezoid - says a deposited energy due a vibration.
Do I understand right? If not, please summarize this procces better.

 

[Dr Wu]

I understand that Enceladus is also gravitationally affected by its neighbouring satellite, the fairly large moon of Tethys, as well as Saturn itself.

 

[haruspex]

My problem is, that I can't see where are you pointing on, and only do exactly what you say.
So I try to summarize what I know about it till now, please correct the wrong.

When I study such problem, I have to expect, that a energy loss is due to different elasticity. If I have some observing data, I determine a apocenter and a pericenter and difference in radius of a moon and calculate forces in these places. Then I "guess" a elasticity and plot similar graph as I attached last time. And then the closed area - trapezoid - says a deposited energy due a vibration.
Do I understand right? If not, please summarize this procces better.

That is how I understand it.
Are you familiar with hysteresis in other contexts? It arises in solenoids, e.g., and in PV diagrams in thermodynamics. In all these cases, you can draw a graph in which ∫y.dx represents energy, and the process forms a closed loop.

 

[Vrbic]

That is how I understand it.
Are you familiar with hysteresis in other contexts? It arises in solenoids, e.g., and in PV diagrams in thermodynamics. In all these cases, you can draw a graph in which ∫y.dx represents energy, and the process forms a closed loop.

First i remember a Carnot cycle. Is this case of hysteresis? But in general not :-( I just read some texts in wikipedia when you mentioned first.

 

[haruspex]

First i remember a Carnot cycle. Is this case of hysteresis? But in general not :-( I just read some texts in wikipedia when you mentioned first.

Actually, I might be out on a limb describing a Carnot cycle as an example of hysteresis. Seems most authors restrict the use of hysteresis to refer to the dissipative losses in the cycle, rather than the cycle as a whole. E.g. https://en.wikipedia.org/wiki/Hysteresis.
However, I did spot one paper that suggested the whole cycle could be considered hysteresis: https://books.google.com.au/books?i...PjAD#v=onepage&q=hysteresis carnot PV&f=false

 

[Vrbic]

Actually, I might be out on a limb describing a Carnot cycle as an example of hysteresis. Seems most authors restrict the use of hysteresis to refer to the dissipative losses in the cycle, rather than the cycle as a whole. E.g. https://en.wikipedia.org/wiki/Hysteresis.
However, I did spot one paper that suggested the whole cycle could be considered hysteresis: https://books.google.com.au/books?id=I7a7BQAAQBAJ&pg=PA178&lpg=PA178&dq=hysteresis+carnot+PV&source=bl&ots=Yry3eBUVnK&sig=Gz4OH9DiKeAMOQqUVyAmKvrclEg&hl=en&sa=X&ved=0ahUKEwjoyufE6qjVAhWBerwKHS3mB8QQ6AEIPjAD#v=onepage&q=hysteresis carnot PV&f=false

I'm probably starting to understand. I have seen this video , very intuitive. And with combination with text on wiki I believe I understand what's going on. Changing one variable changes the other but even more, changing something in deep (let's say inner structure of environment of problem) and it causes different behivor in reverzing process. Yes?

 

[haruspex]

I'm probably starting to understand. I have seen this video , very intuitive. And with combination with text on wiki I believe I understand what's going on. Changing one variable changes the other but even more, changing something in deep (let's say inner structure of environment of problem) and it causes different behivor in reverzing process. Yes?

I think of it as a bit like friction. Suppose a weight rests on such a steep slope that it will slide down. The force required to push it up the slope is greater than the force it provides when you let it slide down, because friction acts against the motion in both directions.

 

[Vrbic]

I think of it as a bit like friction. Suppose a weight rests on such a steep slope that it will slide down. The force required to push it up the slope is greater than the force it provides when you let it slide down, because friction acts against the motion in both directions.

Excuse me, I had a little holiday.
Your example with friction is a bit peculiar for me. I suppose that friction has same coefficient and doesn't matter on a direction up or down. It is mostly dependent on speed. Here I feel here the biggist differnce in gravity. I can't realize that a sledge has different friction when I go up or down.

But generally I mean I understand what is the hysteresis. How may I apply it to in my problem to find deposited energy?

 

[Charles Link]

I have just read through this discussion somewhat quickly, but I think a factor that would be involved in tidal forces is that in a moon's orbit, the ratio $R^3/T^2$ is not precisely obeyed for the entire moon since it is finite in size, and $T$ is the same for every point on the moon. The result is there is a lot of tugging internally to keep every particle in this orbit with period $T$. In the event the moon is too near to the planet, these forces can actually rip the moon apart, as is what is believed to happen in cases of ring formation where the moon gets inside the Roche limit for where a moon can be held together by its own gravity.

 

[haruspex]

I have just read through this discussion somewhat quickly, but I think a factor that would be involved in tidal forces is that in a moon's orbit, the ratio $R^3/T^2$ is not precisely obeyed for the entire moon since it is finite in size, and $T$ is the same for every point on the moon. The result is there is a lot of tugging internally to keep every particle in this orbit with period $T$. In the event the moon is too near to the planet, these forces can actually rip the moon apart, as is what is believed to happen in cases of ring formation where the moon gets inside the Roche limit for where a moon can be held together by its own gravity.

Not sure how that is different from the mechanism discussed in posts #8, #10, ...