# FeaturedI Comparison of tidal forces acting on the Moon vs Enceladus

1. Jul 18, 2017

### 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?

Last edited by a moderator: Jul 20, 2017
2. Jul 18, 2017

### haruspex

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

3. Jul 19, 2017

### Vrbic

$P=\frac{F s}{t}$, 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...

4. Jul 19, 2017

### haruspex

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

5. Jul 19, 2017

### Vrbic

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?

6. Jul 19, 2017

### haruspex

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.

7. Jul 19, 2017

### Vrbic

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?

8. Jul 19, 2017

### haruspex

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

9. Jul 19, 2017

### Vrbic

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.

10. Jul 19, 2017

### haruspex

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

11. Jul 19, 2017

### Vrbic

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

12. Jul 19, 2017

### haruspex

Develop some equations.
Suppose the moon has radius r, the orbital radius varies between Rmin and Rmax, 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?

13. Jul 19, 2017

### willem2

14. Jul 19, 2017

### Vrbic

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

15. Jul 19, 2017

### 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?

Last edited: Jul 19, 2017
16. Jul 19, 2017

### 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?

17. Jul 19, 2017

### haruspex

The way you involve rmax and rmin 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.

18. Jul 20, 2017

### Vrbic

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? $Q=U-E$?

19. Jul 20, 2017

### haruspex

Your expressions for Fmin and Fmax 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.

20. Jul 20, 2017

### Vrbic

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?