A fizzy ocean on Enceladus

Jan 27, 2011 By Dauna Coulter
A Cassini image of vaporous, icy jets emerging from fissures on Enceladus. Credit: NASA/JPL/SSI

For years researchers have been debating whether Enceladus, a tiny moon floating just outside Saturn's rings, is home to a vast underground ocean. Is it wet--or not? Now, new evidence is tipping the scales. Not only does Enceladus likely have an ocean, that ocean is probably fizzy like a soft drink and could be friendly to microbial life.

The story begins in 2005 when NASA's Cassini probe flew past Enceladus for a close encounter.

"Geophysicists expected this little world to be a lump of ice, cold, dead, and uninteresting," says Dennis Matson of NASA's Jet Propulsion Laboratory. "Boy, were we surprised!"

Cassini found the little moon busily puffing plumes of water vapor, icy particles, and out through fissures (now known as "tiger stripes") in its frozen carapace. Mimas, a nearby moon about the same size, was as dead as researchers expected, but Enceladus was precociously active.

Many researchers viewed the icy jets as proof of a large subterranean body of water. Near-surface pockets of with temperatures near 32o F could explain the watery plumes. But there were problems with this theory. For one thing, where was the salt?

A close-up view of a Tiger stripe on Enceladus obtained by Cassini in 2008. Does a fizzy ocean lie underneath? Credit: Cassini Imaging Team/ISS/JPL/ESA/NASA

In initial flybys, Cassini's instruments detected carbon, hydrogen, oxygen, nitrogen, and various hydrocarbons in the plume gasses. But there were none of the elements of salt that ocean water should contain.

In 2009 Cassini's cosmic dust analyzer located the missing salt – in a surprising place.

"It wasn't in the plume gasses where we'd been looking for it," says Matson. "Instead, sodium and potassium salts and carbonates were locked up in the plumes' icy particles.* And the source of these substances has to be an ocean. Stuff dissolved in an ocean is similar to the contents of these grains."

The latest Cassini observations presented another intriguing discovery: thermal measurements revealed fissures with temperatures as high as -120o Fahrenheit (190 Kelvin).

"This discovery resets our clocks!" says Matson. "Temperatures this high have to be volcanic in origin. Heat must be flowing from the interior, enough to melt some of the underground ice, creating an underground waterworks."

The finding has led the scientists to ponder how contents of an ocean capped by a crust of ice as much as tens of miles thick could reach the surface.
"Have you ever been sprayed when you popped the top of a soda can?" asks Matson.

The model he and his colleagues propose suggests that gasses dissolved in water deep below the surface form bubbles. Since the density of the resulting "sparkling water" is less than that of the ice, the liquid ascends quickly up through the ice to the surface.**

"Most of the water spreads out sideways and 'warms' a thin surface ice lid, which is about 300 feet thick," explains Matson. "But some of it collects in subsurface chambers, builds up pressure, and then blasts out through small holes in the ground, like soda spewing out of that can you opened. As the remaining cools, it percolates back down to replenish the ocean and start the process all over again."

Another mystery remains: "Where's the heat coming from on this tiny body?" wonders Larry Esposito of the University of Colorado. "We think tidal heating could be contributing."

Saturn's powerful tides actually cause the shape of Enceladus to change slightly as it orbits. Flexing motions in the moon's interior generate heat--like the heat you feel in a paperclip when you rapidly bend it back and forth. In this model, internal friction powers volcanic activity, which warms and melts the ice.

"It's clear now that, whatever is producing the heat, meets many requirements for life," says Esposito. "We know it has a liquid ocean, organics, and an energy source. And to top it off, we know of organisms on Earth in similar environments."

No one knows for sure what's going on under the ice, but it seems this little moon has quite a story to tell: erupting jets, an underground , the possibility for life.

And they thought this place was dull.

Explore further: SDO captures images of two mid-level flares

More information:
*Cassini's Cosmic Dust Analyzer Principal Investigator Ralf Srama of the Max Planck Institute for Nuclear Physics in Heidelberg, Germany, led the study.
**In their 1988 study of Europa, Crawford and Stevenson introduced the term "Perrier Ocean" for this model. See G. D. Crawford, D. J. Stevenson, Icarus, 73, 66-79 (1988).

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Quantum_Conundrum
2 / 5 (8) Jan 27, 2011
Another mystery remains: "Where's the heat coming from on this tiny body?" wonders Larry Esposito of the University of Colorado. "We think tidal heating could be contributing."


Perhaps the rock/metal core has radio isotopes...

190k is around 45k higher than what wikipedia is reporting as the previous known max temperature. That's a pretty big deal which is around 81F difference. That large of a temperature spike seems to require a radioactive heat energy source.

*ahem*
LKD
1 / 5 (1) Jan 27, 2011
Yes, since the moon has a nice magnetosphere and water does a wonderful job of absorbing radiation, I wouldn't be surprised if it did.

The only thing holding that hypothesis back, would be the preconceived notion that this moon is all water and light elements that Saturn didn't absorb.
GSwift7
1.7 / 5 (6) Jan 27, 2011
Cassini's instruments detected carbon, hydrogen, oxygen, nitrogen, and various hydrocarbons in the plume gasses


carbon

it's global warming, silly.

Just kidding. On the serious side, wouldn't radioactive decay be more widely distributed around the moon, rather than just at the south pole? This moon gets yanked pretty hard by tidal force. Look how fast it orbits; 1.37 days, and it gets pulled between Saturn and Dione every time it passes Dione. Look how much it is deformed. It's more than 3% wider in one direction that it is the other at the equator, when it should be widest at the equator and shortest at the poles, like the earth.

On another track entirely, imagine what it would be like to stand on the surface anywhere near one of those jets.
barakn
3.9 / 5 (7) Jan 27, 2011
Preconceived? How would you explain Enceladus's extremely low average density of 1.6 g/cc without invoking water and light elements? How about the fact that the entire surface is water ice?
GSwift7
2.3 / 5 (3) Jan 27, 2011
A further thought about what it would be like on the surface near those jets. Sunlight would be dim of course, but there should be a steady slow snow of ice crystals all over the south pole there. The ice from the plumes should fall back down all over that region. That would also mean that the surface should also have a coating of ice, unless it sublimates too fast, but I doubt that with the limited sunlight. It must be a stunning view.
barakn
4.1 / 5 (13) Jan 27, 2011
That large of a temperature spike seems to require a radioactive heat energy source.

Because you developed a series of realistic models of Enceladus covering the entire range of possible structures, and applied to all of them the kind of tidal force Enceladus experiences, and none of the models developed the heated pockets of water you were looking for. Oh, wait, no you didn't...
Quantum_Conundrum
1.4 / 5 (9) Jan 27, 2011
You realize how fast and hard you have to rub something to increase temperature by 81F?!

That's the equivalent of starting a fire by rubbing two sticks together...

Maybe I'm wrong, but I don't see how tidal forces could possibly be generating that much friction, because even at that orbital velocity the action isn't happening fast enough. Enceladus only has about 1/100th of g for surface gravity, so it's not like there's a lot of gravitational potential energy available to make heat from a collapsing tidal bulge...

I calculated that if all of the change in gravitation potential energy was converted to heat it would make 1.328E20J per orbit.

This is only enough to affect the temperature of the moon by less than 1/4000th of a degree Kelvin per orbit.

If we assume the entire lunar mass is deflected by 7.566KM per orbit, that's a lot more energy,but still would only be 1/120th of a K per orbit, even if 100% was converted to heat...
Quantum_Conundrum
1 / 5 (5) Jan 27, 2011
If all of the potential energy gets converted to heat, and all of the heat is in a tight pocket, and somehow ideally insulated from the rest of the moon, you could increase 1/9000th of the Moon's mass by 45K per orbit...

But that is a lot of "ifs", such as requiring coefficent of friction near 100% and ideal insulation...and we still haven't managed to get any isolated pockets up to 373 kelvin to boil in a geyser..

190K is a very long way from 373K...
Paljor
1 / 5 (1) Jan 27, 2011
Is that for the entire planet? or just one particle. perhaps it's icy crust reflects heat back? building up over the years.
GSwift7
3.7 / 5 (6) Jan 27, 2011
I don't think you are conceptualizing it correctly.

To correctly view the tidal forces, try to imaging the position of the center of mass of a multibody system. In this case the primary actors are Dione, Enceladus and Saturn. If you look at the Earth, Sun and Moon system and observe the power of gravity to move the entire oceans by several meters twice a day then you start to apreciate the forces involved. You are looking at the gravity of Enceladus, but it's the gravity of Saturn and Dione that matter here. The effect of Dione was strong enough that Enceladus has ended up in an orbit that matches the orbit of Dione on a 2:1 ratio. That's a strong influence.

I'm no expert, but you can easily find expert opinions from people who know how to do the math that I don't.
Quantum_Conundrum
1 / 5 (6) Jan 27, 2011
Is that for the entire planet? or just one particle. perhaps it's icy crust reflects heat back? building up over the years.


All figures assum 99.99r% coefficient of friction to convert all potential energy to heat.

The first part of the first post was assuming only 3% of the planets mass bulged by 3%, giving the 1/4000th figure for mean temperature input per orbit distributed across all Enceladus mass.

The 1/120th figure is assuming the entire moon's mass deflected by the 3% of the moon's radius, and then heat is distributed evenly throughout the moon.

The 1/9000th figure tells what fraction of the moon could be increased by 45K per orbit, if it was ideally insulated and all of the heat was concentrated in that fraction of mass...IF the coefficient of friction is arbitrarily close to 100% and 100% of Enceladus' mass was in the Tidal bulge...

Even with the "Myth Busters" style overkill approach, it is very hard to produce energy to accomplish this from tidal action
Quantum_Conundrum
2.3 / 5 (6) Jan 27, 2011
Gswift:

The gravity of Saturn matters, but the maximum influence from Saturn cannot be greater than the influence from Enceladus itself, else Enceladus would be torn apart. Saturn's influence is limited by the surface gravity and escape velocity of Enceladus. If saturns influence exceeded Enceladus' influence the moon would not even exist...
Paljor
not rated yet Jan 27, 2011
why don't you compare masses and then see how strong their gravity is. like saturn for instance. it must shift the moon a lot while closer dione shifts it back. focus on gravity! not on what the planet looks like.
GSwift7
4.4 / 5 (14) Jan 27, 2011
and we still haven't managed to get any isolated pockets up to 373 kelvin to boil in a geyser


what do you mean by that? I think you are totally confused here.

You do understand that they aren't talking about anything on this moon reaching that kind of temperature, don't you? I think you are very confused about what's happening here. The liquid water we are talking about here is actually at a very cold temperature (190 K) and is only semi-liquid because of the pressure from being underground. Then it gets squeezed up out of the ground and insta-freezes when it is ejected because it's already supercooled. There's no boiling going on here. This isn't like a gyser at yellowstone. There's no steam pressure from "hot" things. We're still talking about cryogenically cold temperatures, even at the warmest part of this moon.
GSwift7
4.2 / 5 (10) Jan 27, 2011
If you look up the Roche limit, you'll see that you are not correctly figuring the tidal forces and that you are not really understanding why an object can be torn apart by tidal force if it gets too close to a larger object. wow. just stop and admit that you are over your head here.

Let me reapeat part of what I just said, in case you missed it: The tidal forces can be SO STRONG that they can TEAR AN ENTIRE MOON OR PLANET APART. That's a lot of force. I also notice that you are treating this moon as a solid object. An object held together by gravity like this moon can be viewed more as a fluid (not a liquid, but a fluid similar to window glass.)
GSwift7
3.4 / 5 (5) Jan 27, 2011
Keep in mind that the extremely low density of Enceladus will amplify tidal effects compared to what they would be if it was more dense. Its low density means that it deforms more under the same force.
Quantum_Conundrum
1 / 5 (6) Jan 27, 2011
Paljor:

What the heck do you think I was doing?

Anyway, I did focus on gravity, and when you plug in numbers for Saturn and Enceladus it doesn't make sense anyway.

For example, if you plug in Enceladus' Estimated mass and use Newton's law for it's average radius, you will find surface gravity Acceleration of only 0.02859m/s^2, vs the NASA number of 0.111m/s^2, which is 4 times larger than Newton's law says it should be, given the mass THEY report.

If you plug in Saturn's mass and calculate gravitational acceleration at Enceladus' distance, you will get 0.669m/s^2, which is impossible, because the Saturn-side face of Enceladus would simply fall off into the host planet, as the surface of Enceladus would literally be attracted to Saturn 23.4 times more strongly than it's own self, which means Enceladus shoudln't even exist, and should fall apart in a matter of minutes...
Quantum_Conundrum
1 / 5 (13) Jan 27, 2011
If you look up the Roche limit, you'll see that you are not correctly figuring the tidal forces and that you are not really understanding why an object can be torn apart by tidal force if it gets too close to a larger object. wow. just stop and admit that you are over your head here.

Let me reapeat part of what I just said, in case you missed it: The tidal forces can be SO STRONG that they can TEAR AN ENTIRE MOON OR PLANET APART.


Wow you are dense...I know this guy. My god read my previous post. You're fricken clueless.

As I was saying, the numbers do not add up at all, because NASA's numbers on Enceladus' mass, radius, and surface gravity are off from Newton's law by nearly 400%, and Saturn is 23 times more massive than would be needed to destroy Enceladus in one second...
GSwift7
4.3 / 5 (11) Jan 27, 2011
Your math doesn't make any sense at all, and you are totally falling off the cliff when it comes to understanding the mechanism of the Roche limit.

The Roche limit is based on the gradient of gravity over distance, not on the absolute acceleration of gravity at the center of mass. It's the difference in the pull of gravity between the inward and outward faceing sides of the smaller object. As it gets closer, the delta in force increases between the near side and the far side.
Quantum_Conundrum
1 / 5 (7) Jan 27, 2011
By comparison, the Earth's Moon's surface is influenced by it's own mass at 1.622m/s^2, but is influenced by the Earth's gravity at a miniscule 0.003m/s^2.

I hope you see what I am talking about.

I can post the equations of Newton's law for you, or you can do them yourself.

And yes, I realize technically you're supposed to use relativity, but the difference isn't even supposed to matter at these velocities and acceleration rates...
Quantum_Conundrum
1.4 / 5 (10) Jan 27, 2011
Your math doesn't make any sense at all, and you are totally falling off the cliff when it comes to understanding the mechanism of the Roche limit.


DO THE FRICKEN NEWTON'S LAW AND LOOK AT HOW INCONSISTENT IT IS WITH NASAS NUMBERS.

If you have 3 objects: Planet A, Moon B, and particle C:

If particle C is somewhere between A and B it will accelerate towards whichever one has the strongest gravitational acceleration, as per Newton's law, because the net acceleration will be in that direction, assuming it isn't at precise equilibrium.

If you do the math for youself, you will find that it doesn't work:

Enceladus has 4 times more surface gravity than it is supposed to given NASA's mass estimate.

Meanwhile, for a particle C on Enceladu's surface, Saturn's gravity accelerates that particle toward Saturn at 0.669m/s^2 while Enceladus accelerates it towards Enceladus at a mere 0.028m/s^2 (By Newton's Law. Either NASA's mass number is wrong, or their surface gravity is.)
GSwift7
4.3 / 5 (11) Jan 27, 2011
As I am trying to point out. You are using the entirely wrong set of laws.

that's why it doesn't seem to add up to you. You shouldn't be using newton's laws to figure tidal force. You're totally not getting the concept of what tidal force is.
Quantum_Conundrum
1 / 5 (7) Jan 27, 2011
The point is, given NASA's numbers, or anything even in the ballpark with those numbers, none of it works anyway.

Enceladus' own surface gravity is off by 400%, so when I saw that I pretty much stopped caring about the tidally generated heat issue, as none of it matters anyway if they can't even measure it's mass and surface gravity more accurately than half an order of magnitude...
GSwift7
4.2 / 5 (10) Jan 27, 2011
If you look up the wiki page on tidal force, you can see the full explaination and the proper equations you should be using. Then you can start to understand how the interior of Enceladus might reach 190 K under the influence of both internal pressure from gravity and from tidal forces. I should also mention, because it seems like you don't understand this either, that you can get heat from physical stress without having noticable deformation of shape. You can see this when you squeeze a diamond in a high pressure press.
LKD
2 / 5 (1) Jan 27, 2011
barakn,

You can't create a moon sized magnetic field with water or carbon or most light elements. There has to be more there.

Preconceived is not a slight by me, it's a statement of the truth. My first inclination would be there is no core, and it's all light molecules frozen in ball form. But something tells me, this is wrong because of the magnetic forces present. Only way to prove it is to go there, hopefully the rover they are working on now, will answer this question.
GSwift7
3.4 / 5 (5) Jan 27, 2011
Are you sure you're using the right equation for the force of gravity and then using the right units when you do the conversions? It looks fine to me, but if you are using the wiki page numbers then you need to convert the units to compatible units to get a meaningful result.

Quantum_Conundrum
1.6 / 5 (7) Jan 27, 2011
As I am trying to point out. You are using the entirely wrong set of laws.

that's why it doesn't seem to add up to you. You shouldn't be using newton's laws to figure tidal force. You're totally not getting the concept of what tidal force is.


Tidal force is the difference in gravitational attraction from one side of a body to another, or more particularly from the center of mass to the near side of the planet. I very much get that. Get over yourself guy...

If you want to get exactly what it is, on average, for Enceladus, the difference is:

A = GM/(R^2) - GM/((R-r)^2)

R = 238,000km from Saturn CoG to Enceladus CoG

r = 504.2km = Enceladus average radius.

This gives:

0.690621248499m/s^2 - 0.693556718155m/s^2 =...

0.002935469656m/s^2 difference between CoG and near side to saturn. now while it isn't exactly correct to do so, you could double this to get the difference from one side to the other, within about 10^-4 precision...
AlwaysRight
5 / 5 (9) Jan 27, 2011
QC you should go tell those idiots at NASA their calculations are way off. Tell them to stop using all those fancy computers when all they need for any calculation is good ol F=ma.
pauljpease
5 / 5 (11) Jan 27, 2011

If you plug in Saturn's mass and calculate gravitational acceleration at Enceladus' distance, you will get 0.669m/s^2, which is impossible, because the Saturn-side face of Enceladus would simply fall off into the host planet, as the surface of Enceladus would literally be attracted to Saturn 23.4 times more strongly than it's own self, which means Enceladus shoudln't even exist, and should fall apart in a matter of minutes...


No, it shouldn't. Because every particle of Enceladus is accelerated at roughly the same amount, and the whole moon IS falling into Saturn. It is also moving really fast, so it doesn't every fall into Saturn, it is in ORBIT! The tidal force is generated by the small difference in acceleration experienced by the near and far side of Enceladus. This difference causes enceladus to stretch a little bit. That stretching causes internal friction which is converted to heat. The heat in the interior has a hard time escaping to the surface, warming the center.
Quantum_Conundrum
1 / 5 (7) Jan 27, 2011
Are you sure you're using the right equation for the force of gravity and then using the right units when you do the conversions? It looks fine to me.


Newton's Law:

G = 6.67E-11

NASA's Enceladus info:
Average radius: 504.2km.
surface gravity: 0.111m/s^2
mass: 1.08022E20kg

What Newton's Law predicts:

A = GM/(r^2)

A = 6.67E-11 * 1.08022E20 / (504000^2)

A = 0.028364620339m/s^2...

Any other quetions? I got this off wikipedia, and they got it directly from NASA, and it's wrong by 391%...
AlwaysRight
5 / 5 (4) Jan 27, 2011
How can you possibly get a 10^-4 precision from a radius that is obviously rounded off at 10^3. Clearly you have gotten into significant digits in your highschool physics course yet.
GSwift7
4 / 5 (8) Jan 27, 2011
radius of enceladus is 251 km. You're using the cross-sectional width or diameter.

There's your first problem. I am still looking at your other math.
GSwift7
3.9 / 5 (7) Jan 27, 2011
Next, did you convert the units above to account for the mismatch between kilometers and kilograms? You know you need to use meters, right?
Quantum_Conundrum
4.2 / 5 (10) Jan 27, 2011
Ah crap. I've been reading diameter as radius, so that explains the discrepancy in the gravity for enceladus. my bad.

I don't know exactly why I did that, but that part is clearly my mistake.

A = GM/(r^2)

is supposed to be r = 252.1, and instead i was using average diameter because I kept getting my numbers mixed up.

In that case Newton's Law predicts g should be:

0.113368m/s^2

Which is much better, but still off by 2%...

So I was using the wrong fricken number.
GSwift7
3.9 / 5 (7) Jan 27, 2011
that's okay, I seem to recall making a mistake once but my memory is so bad that I could be wrong about that.

That 2% difference could get kinda technical as it probably is related to the fact that a body like Enceladus isn't uniform in density, but you should be able to understand the effect of the center being more dense. That would explain your figure, based on uniform density assumption being higher.
Quantum_Conundrum
1.7 / 5 (6) Jan 27, 2011
Next, did you convert the units above to account for the mismatch between kilometers and kilograms? You know you need to use meters, right?


Yes, and I know I was using the wrong number, that was my fault, as I said, and I apologize for that.

It still doesn't make a big enoug difference to explain how tidal influence could raise any significant fraction of the planet's mass by 45kelvin.

The pure "tidal" force is still only a mere fraction of even Enceladus' gravity, regardless of margins of error, around 1/40th.

So if the tidal force is only 1/40th of Enceladus' own gravity it would be even harder to explain thermal build ups, because I was using a worst case scenario earlier.

This would make the amount of thermal energy available at least 40 times smaller than my worst case estimates using Enceladus' own mass and gravitational potential energy maximums.

GSwift7
4 / 5 (8) Jan 27, 2011
I'll repeat. You are not calculating tidal force correctly. It's not a function of the absolute gravity of Saturn. It's a static force caused by the difference of the pull of gravity on one part of Enceladus versus the pull of gravity on another part of Enceladus. Every time it passes Dione, that difference is amplified and then subsides. It's not the magnitude of Saturn or Dione's gravity in comparison to Enceladus' gravity that generates the tidal force or determines the heat generated by the stress. You can see a demonstration of the tidal forces around Enceladus by looking at the ring it sits in. That's the shape the tidal force wants Enceladus to be; flat and spread out over billions of miles. Enceladus has to resist that pull.
Quantum_Conundrum
1 / 5 (4) Jan 27, 2011
For example, the gravitational potential energy due to tidal differences from one side of the moon to the other can be found by the number above, being: (since my original radisu number was actually accidentally the diameter...)

0.002935m/s^2

Now if we figured the maximum distance this could act in is the diameter itself,a nd if we figure the entire mass maoves by this much, which it doesn't, but just for "worst case imaginable".

The potential energy from one side to the other with respect to Saturn is certainly less than that given by moving the entire mass:

U = mah

a is difference in acceleration.)
h is normally height, but here it's the maximum tidal displacement, which we were using 3% of planet radius.

This works because we're doing work on one side of moon compared to the other.

This gives a change in potential energy of around 4.795E21 joules.

We need 4200 joule to change a kilogram of water by 1K or 1C.

Again, I made 100% as heat and 100% of mass.
Quantum_Conundrum
1 / 5 (4) Jan 27, 2011
You do realize the gravity of hte other moons in this system is so tiny as to be completely insignificant?

At closest approach, Dionne would be 139,396km from Enceladus, which gives average attraction of just about 3.7E-6m/s^s.

This is already completely insignificant, and the difference from one side of moon to other would be even smaller than this number, and that difference is the only thing that could do "work" or be converted to heat. It's already several orders of magnitude too small to even matter, even over geological or cosmological time...
GSwift7
4 / 5 (8) Jan 27, 2011
Okay, couple things. First the specific heat of water is unusually high, and a rocky body like Enceladus will have a much lower specific heat at room temperature. HOWEVER, wer're not talking about room temperature. At cryogenic temperatures specific heat of materials drop a lot. It's not as hard to raise the temp from 0 to 190 K as it is to raise it from 200 to 390 K. You are comparing apples to oranges there.

Next. You are still not getting the math involved in figuring out the energy of tidal force. You are grossly oversimplifying it. You aren't even close. The actual math to figure out the tidal force is an angular momentum problem, like figuring the force on a part of a suspension bridge and then you have to take the sum of the differences of all the places in Enceladus with a differential equation to get a meaningful answer. It's a hugely complex math problem. I wrote a Fortran program to run on the Auburn Univax that solved this kind of problem when I was there.
Quantum_Conundrum
1 / 5 (4) Jan 27, 2011
The reason potential energy works, is because it tells you the maximum difference in kinetic energy from one place in a gravity well to another place in a gravity well.

The tidal influence cannot produce more energy than the potential energy, or the amount of work that would be done between those two energy levels.

Then, for the sake of simplicity, I converted ALL of this energy to heat, and assumed none of the heat radiated away.

In reality, a much smaller fraction of the energy is converted to heat, as most of it rebounds at different points in the orbit, and also, much of it will simply radiate away through thermal radiation.

So only something like 1/40000th of a Kelvin of Enceladus average temperature could conceiably be attributed to tidal forces.

Then trying to explain how all of this could get concentrated to produce isolated pockets of +45K violates any thermodynamics, excluding radioactivity. Energy from friction would conduct through the moon, not concentrate.
GSwift7
3.9 / 5 (7) Jan 27, 2011
I should note that my program solved for a simplified 2 dimensional case, not this three dimensional spherical problem with three bodies. This is some REALLY complicated math. My example isn't even close to what this problem is. I just want to make sure I'm not sounding like I could figure out this tidal problem. I think it's over my head.
Quantum_Conundrum
1 / 5 (4) Jan 27, 2011
Yes it's a differential in respect to change in gravity, multiplied by the mass affected by each partition under the gravity curve. (hard to say exactly in 1000 words plain language.)

Being a non-uniform body makes that impossible anyway. But you could approximate it by at least assuming shells that are close to uniform, but it would only be exactly right if you had exact mass and composition of the moon.

An no, I don't have the resources to solve that absolutely.

What I did was compute the absolute maximum change in energy, as if the entire moon's mass was vibrating or oscillating by 3% of radius per orbit, which is far more than the equation would provide, because in reality only a fraction of the moon's mass is moving that much.

I then converted 100% of this maximum change in energy to heat...
Quantum_Conundrum
1 / 5 (2) Jan 27, 2011
Gswift:

The real formula involves several integrals and derivatives, and would involve nesting shells of infinitesmal thickness (i.e. the smaller the thickness of the shells, the more precise the estimate,) inside of Enceladus to approximate density and composition of the moon at slices of these shells more precisely, to determine what portion of which "shell" is being affected by how much gravity from Saturn, and what the realtive differences are from one side of enceladus to the other, as that's what a tide represents.

It actually involves at least a triple integral (assuming density transitions inside Enceladus are arbitrarily "smooth",) and a derivative...
sorg
not rated yet Jan 27, 2011
I think coke or pepsi should sponser a scientific expedition there......i mean common,,,,fizzy!
Its a marketing dream.
Pyle
4.3 / 5 (12) Jan 27, 2011
Wow! I'm impressed.
GSwift7 - You are one patient fellow. Good on ya!
GSwift7
3 / 5 (4) Jan 28, 2011
You are also neglecting the fact that you are dealing with vectors and that the direction of the vectors is different accross the volume of the body.
jselin
5 / 5 (2) Jan 28, 2011
QC-

Have you considered that the energy can be concentrated locally at the cracks much like how earthquakes relieve stress in our rocky crust? Combining what are likely different measurement locations with the uncertainty of the surface emissivity and its variance and you are easily arguing over nothing. I'm sure there are 10 other influencial factors not being considered for the sake of simplicity that have a significant impact on all of this as well. With so many important things that need to be solved, it seems to be a waste of time to hold half informed arguements. If something is worth doing, its worth doing right. Right?

I'd also like to add that I understand you are trying to bound the problem. Thats a good first stab at things, but it leaves a lot of wiggle room when things are inhomogeneous (compositional, thermal, etc)
GSwift7
3.4 / 5 (5) Jan 31, 2011
Have you considered that the energy can be concentrated locally at the cracks much like how earthquakes relieve stress in our rocky crust?


That's a really good point. Since the surface is geologically active, there should be friction from tectonic plate movement. The deformation of those thick layers of ice that produced the wavy surface features we can observe from space surely generate some heat above ambient.
barakn
2.3 / 5 (3) Feb 26, 2011
You can't create a moon sized magnetic field with water or carbon or most light elements. There has to be more there. -LKD

Really? I hadn't heard there was a law. Generally dynamos need a conductive fluid (salt water) moving in a turbulent (convective ) manner. Perhaps more importantly, no one said that Enceladus had a magnetic field. Cassini did measure a magnetic field deflection but this was due to to electrical currents created by the interaction of Enceladus's atmosphere with Saturn's magnetic field. This is far different than Enceladus having a dynamo under its surface.
barakn
2.3 / 5 (3) Feb 26, 2011
You realize how fast and hard you have to rub something to increase temperature by 81F?!

That's the equivalent of starting a fire by rubbing two sticks together...

I calculated that if all of the change in gravitation potential energy was converted to heat it would make 1.328E20J per orbit.

This is only enough to affect the temperature of the moon by less than 1/4000th of a degree Kelvin per orbit.

How ridiculous. Heat loss is somewhat proportional to the surface area to volume ratio which for simply shaped objects means it scales as 1/R. Considering that a stick is about a million times smaller than Enceladus, it'll lose heat (per unit volume) a million times faster (assuming similar temperatures). Also the tidal friction is distributed throughout the bulk of the moon where it is efficiently trapped, whereas the friction when rubbing sticks is limited to small contact zones on the surface where the heat is free to radiate away. (cont'd)
barakn
2.3 / 5 (3) Feb 26, 2011
Also power radiated varies with the fourth power of temperature, and QC would have us believe that the two situations - a moon with a surface at 77 K and two sticks just under the ignition temp of wood, say 693 K - are similar, when the sticks are radiating energy at 9^4 = 6561 times the rate of the moon (per unit surface area, and we're ignoring minor differences in emissivity).

To rise 45 degrees using his estimate of 1/4000 degree per orbit would require only 180,000 orbits, or 675 Earth years. This is small even considering QC's status as a YEC. QC has produced yet another crap flood where he demonstrates that he can plug values into formulae but often uses the wrong values and doesn't demonstrate an understanding of the results.

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