I Comparison of tidal forces acting on the Moon vs Enceladus

[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, ...

 

[Canis Lupus]

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?

Vrbic, If the calculations are still astray you might want to consider this paper https://www.nature.com/articles/srep37740

It's possible that the same process which is the subject of the paper is occurring on Enceladus (and a few other masses in the solar system).

It's against the mainstream, I know, but is peer reviewed. One of the paper's implications is that the present definition of a planet is incorrect concerning the lack of a nuclear reaction, unless you exempt Earth from being a planet, which would seem highly irregular.

I cannot say with technical expertise whether I think the paper is correct, anymore than I can help with the details of your calculations as others far better qualified than me have attempted to do thus far, but I would be less than honest if I stated I believe the paper is incorrect.

Hopefully, it may prove helpful with you calculations if you find it impossible to resolve your calculations using reasonable assumptions.

 

[Charles Link]

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

The OP may have already done a similar derivation, and/or used the results of this type of derivation, but if I may, I will show you how I like to derive the tidal forces: There is a difference between the centripetal force that is necessary to create circular motion around the larger body for the parts of the moon that are away from the center and the gravitational force from the larger body at that point. This difference will be the tidal force. To write this out algebraically $F_t=\frac{mv^2}{(r+r_o)}-\frac{GMm}{(r+r_o)^2}$. $\\$ Now $v=\frac{2 \pi(r+r_o)}{T}$. Also, we have $T^2=\frac{4 \pi^2 r_o^3}{GM}$ by Kepler's 3rd law.($T$ is the same for every point on the orbiting moon). $\\$ A little algebra gives $F_t=GMm( \frac{1}{r_o^2}+\frac{r}{r_o^3}-\frac{1}{r_o^2} \frac{1}{(1+\frac{r}{r_o})^2}) \approx 3\, GMm \frac{r}{r_o^3}$. $\\$ Here $r_o$ is the distance from the larger body to the center of the moon, and $r$ is the difference in distance (e.g. extra distance from $r_o$) to a point on the moon that is not at its center. $\\$ This difference in forces is made up to some extent by the moon's own gravity, but if the interior of the moon is part liquid, some motion due to unbalanced forces could certainly result. $\\$ Additional note: Other than the factor of 3 vs. 2, this is in agreement with the equation that the OP @Vrbic presented in post #1.

 

[haruspex]

I can't realize that a sledge has different friction when I go up or down.

The direction of the friction reverses. How different can you get?
(Oh, and it does not depend on speed.)

 

[Vrbic]

Vrbic, If the calculations are still astray you might want to consider this paper https://www.nature.com/articles/srep37740

It's possible that the same process which is the subject of the paper is occurring on Enceladus (and a few other masses in the solar system).

It's against the mainstream, I know, but is peer reviewed. One of the paper's implications is that the present definition of a planet is incorrect concerning the lack of a nuclear reaction, unless you exempt Earth from being a planet, which would seem highly irregular.

I cannot say with technical expertise whether I think the paper is correct, anymore than I can help with the details of your calculations as others far better qualified than me have attempted to do thus far, but I would be less than honest if I stated I believe the paper is incorrect.

Hopefully, it may prove helpful with you calculations if you find it impossible to resolve your calculations using reasonable assumptions.

Interesting paper, but I'm not specialist on nuclear reactions. I have read just abstract but:
1) Because of expectations of high pressures and temperatures in this paper, I'm not sure whether it is applicable on small moon.
2) Peculiarity for me is the catalysis by pions. As far as I know pions exist only in very high pressures and temperaturs only in cores of massive neutron stars (or some colliders).

But as I said I'm not expert. For me was most interisting question why some objects has a potential of tidal heating and some not. So I tried to understand a process of tidal heating.

 

[Charles Link]

One of the factors in tidal heating would seem to be the rotation of the moon. (This was probably previously mentioned.) If the moon is always facing the same way as the Earth's moon with period of rotation equal to period of revolution, the tidal forces=difference in the two forces that I mentioned in post #52, would reach an equilibrium and there were be little motion in the interior. Alternatively, if the moon were rotating, this equilibrium could not occur because the tidal forces, as calculated by the equation $F_t=3 \, GMm \frac{r}{r_o^3}$, at each location keep changing. $\\$ If the heating that occurred were sufficient to cause a molten state, it would allow for increased tidal action and heating inside the moon.

 

[haruspex]

One of the factors in tidal heating would seem to be the rotation of the moon. (This was probably previously mentioned.) If the moon is always facing the same way as the Earth's moon with period of rotation equal to period of revolution, the tidal forces=difference in the two forces that I mentioned in post #52, would reach an equilibrium and there were be little motion in the interior. Alternatively, if the moon were rotating, this equilibrium could not occur because the tidal forces, as calculated by the equation $F_t=3 \, GMm \frac{r}{r_o^3}$, at each location keep changing. $\\$ If the heating that occurred were sufficient to cause a molten state, it would allow for increased tidal action and heating inside the moon.

According to post #3, Enceladus is rotationally locked. See also post #5.

 

[Charles Link]

According to post #3, Enceladus is rotationally locked. See also post #5.

A google of the moon Enceladus and tidal heating that I just did shows Wikipedia has an article that explains the tidal heating of Enceladus apparently comes from an eccentric resonance orbit with another moon. Looks like it's not as simple as what I proposed in post #55. $\\$ Editing: Yes, and I see you covered some of that already in post#5, etc.