Estimating Planetary Cooling Times

1. Dec 13, 2008

Widdekind

All rocky planetoids are born as balls of molten magma, and take millions of years to cool, crust over, & solidify. As a rule-of-thumb, a planetoid's Cooling Time (tcool) scales as its Surface Gravity (g):
$$t_{cool} \approx \frac{Heat}{Dissipation Rate}$$
$$\approx \frac{Mass}{Surface Area}$$
$$\approx \frac{M}{R^{2}}$$
$$\approx g_{surface}$$​
Now, Mars appears cold & dead, lacking a Dynamo-Driven (?) central magnetic field. But, there is evidence of relatively recent volcanic activity on Olympus Mons (w/in the last few million years*). Thus, Mars, whose Surface Gravity is roughly 0.4, seems to have lived roughly 4.0 billion years**.
* http://en.wikipedia.org/wiki/Olympus_Mons
** The Solar System, and the Earth, are around 4.5 billion years old.
This suggests the following Order-of-Magnitude estimate for planetoids' Cooling Times:
tcool ~ g x 10 billion years
where g is measured in Earth Gravities (~9.81 m/s2). This yields the following estimates:
Code (Text):

Planetoid       g        Cooling Time (Gy)
Mercury        0.376           4
Venus          0.949           9
Earth          1              10
Moon           0.17            2

For comparison, the Moon (probably) solidified something like ~2.5 billion years ago, about the time that Earth's Continents were forming, and atmospheric concentrations of Molecular Oxygen (O2) were at long last increasing*.

2. Dec 13, 2008

Staff Emeritus
Neither Venus nor Mars has been observed undergoing any geologic activity. We know Venus did ~500 million years ago, and Mars did 2-30 million years ago. Based on that, it seems contrary to the evidence to claim that Venus is still geologically active and Mars is not.

Of course, there are a number of other, smaller bodies that also exhibit volcanism. Io is the classic example.

As I said before, you can't use your theory to decide what data to consider to support your theory. This isn't science.

3. Dec 13, 2008

Widdekind

The European Space Agency (ESA) is currently analyzing whether Venus has active volcanoes. Their Venus Express probe has found evidence consistent w/ active Volcanism, but there are other possible interpretations*.
If both Venus & Mars are geologically dead, then planetoid Cooling Times would be significantly shorter than I suggested.

However, the underlying Physics, that tcool ~ gsurface, is sound. It was told to me by an Astronomy Professor at UCSD, for whom I was TAing in 2001, and to an auditorium full of ~200 Freshman. It relies on basic, and simple, scaling laws.

MORE DETAIL: That Professor said, that all rocky planets:
• began life in the same molten phase
Thus, for an Order-of-Magnitude estimate, all planetoids begin life at the same Temperature, (just under) the Boiling Point of rock.

Then, since all planetoids begin life at roughly the same Temperature, and have roughly the same (bulk) Heat Capacities, their heat energy content scales as their Mass (M). And, since they start at the same Temperature, their Black Body radiation rate varies only according to their Surface Area. The rest is algebra (M / 4 pi R2) ~ gsurface.

I attempted to estimate the constant of proportionality, which was my only original contribution (to my knowledge). I could be wrong, about that constant of proportionality.

However, since Venus is nearly the same size as Earth, often being called "Earth's Twin", and since there is active Volcanism on Earth, it stands to reason that there should be Volcanism on Venus. In fact, Venus is much hotter (on the surface), so why would it be colder (in its core) ?

But, if my Astronomy Professor at UCSD was wrong, then I, and hundreds of other students of his, have been misinformed. If the Venus Express, & other probes, prove my professor wrong, then I will acknowledge it, if there is really no correlation between Cooling Times and M / R2.

ADDENDUM: I recall asking that Professor, why initial heat energy does not scale as its Gravitational Binding Energy, ~ M2 / R. I recall him telling me, that that would be too much energy, boiling the rock away. By the time a planet condenses into a molten magma ball, it must be below the melting point of rock. He told me, that the difference between those 2 energies, between the Gravitational Binding Energy (~ M2 / R ) and its actual initial heat energy (~ M), must have been radiated away before the planetoid could finally collapse.

Thus, this rubric for estimating tcool applies only to the actual planet phase, and does not include the millions of years (??) that the "proto-planet" had been in some kind of nebular phase. If that was a source of confusion, then I appologize.

Last edited: Dec 13, 2008
4. Dec 14, 2008

Staff Emeritus
I notice you didn't actually address any of my complaints.

1) You ignored Io.

2) You pick which data to include in testing your model based on how well it agrees with your model. There is no independent evidence that Venus is active and Mars is dead - they are most likely both active, although there is stronger evidence for Mars than Venus.

This is not science.

5. Dec 14, 2008

Eimacman

I believe that Io should be ignored because it is continuously heated by gravitational effects of the Jovian system, any cooling of that body would be constantly replaced by heating due to tidal flexing of Io.

As for the activity of Venus and Mars, it is hard to tell without seismic data. Such data would reveal more precisely the composition of the crust, mantle, and core. As far as Venus goes we will probably never in the next 50 years be able to get seismometers onto Venus that will be able to withstand the environment for the time required. Knowing the composition of a planetary body will go toward knowing the cooling rate for that body.

There is another problem with Mars in that it was impacted with possibly Pluto sized body at Hellas Planitia, this could have caused a massive release of Martian magma over a large area of the martian surface. This, in turn, could have caused a rapid cooling of the martian core resulting in less activity. It will be necessary to send geologists there for a few decades to finally find out for sure. One would also have to take into account the composition of the impacting body. An icy body;water ice, methane ice etc., would have a very different effect as compared to a rocky body, silicate, or even a iron body.

Widdekind:

It is also necessary to add the calculation the possibility of transuranic elements U235, U236, U237, and U238 in the core that would add heat to such a planetary body, the half life's of which can be many billions of years. The formula could give one a 'ballpark guesstimation' of the cooling of a planetary body, if you ignore any of the effects, properties here mentioned. It is theorized that the Earth's internal heat may persist for as long as 30 billion years due to nuclear heating. This is longer than the time that the Sun will be on the main sequence. Any planetary bodies of similar mass but of different composition will have different heating rates due to initial gravitational compression. Lighter element bodies will not compress and heat up like a rocky or a metallic body composed of heaver elements. Also the heaver the element composition the more likely there will be radioactive elements that will maintain heat longer than could be achieved by compression alone.

6. Dec 15, 2008

Widdekind

That is surely a valid point. And, according to the National Geographic documentary Naked Science -- The Moon (TV), the Moon used to be about 15x closer to Earth than it is today. It pulled up 10,000' ocean tides that dragged backwards against the then-rapid rotation of the planet (the day was about 6 hours long).

These fierce tidal forces eventually phase-locked the Moon in its orbit, and spun it out an additional ~350,000 km.

Assuming that all the Rotational Kinetic Energy:
$$K = \frac{1}{2} I \omega^{2}$$​
which the Earth has lost as it has spun down, has been converted into heat, then that Tidal Heating Budget is:
$$\Delta K = \frac{1}{2} I \Delta\omega^{2}$$
$$= \frac{0.3315}{2} M R^{2} \left( \omega_{i}^{2} - \omega_{f}^{2} \right)$$
$$= \frac{0.3315}{2} M R^{2} \left( \frac{2 \pi}{T_{i}^{2}} - \frac{2 \pi}{T_{f}^{2}} \right)$$
$$= 6.544 M R^{2} \left( \frac{1}{T_{i}^{2}} - \frac{1}{T_{f}^{2}} \right)$$
$$= 3.195 \times 10^{30} J$$​
Now, the Earth is currently venting heat at the rate of ~4 x 1013 W*. Thus, the above Heat Budget could power that dissipation rate for ~2.5 billion years.
* Carroll & Ostlie. Introduction to Modern Astrophysics, pp. 818 & 797.

CONCLUSION: These fierce tidal forces have generated gargantuan quantities of heat. Perhaps that explains why Earth seems more active, even than Venus.

Alternatively, the current Tidal Heating Rate is:
$$\frac{\partial K}{\partial t} = -4 \pi^{2} I \frac{1}{T^{3}} \frac{\partial T}{\partial t}$$
$$\approx 2.5 \times 10^{12} W$$​
since the current rate of spin-down is 0.0016 s century-1*. Thus, Tidal Heating currently accounts for only 6% of the Earth's rate of heat loss.
* Carroll & Ostlie. ibid., pg. ~763.
This gives a Tidal Heating Time Scale of:
$$\frac{\Delta K}{\frac{\partial K}{\partial t}} \approx 40 billion years$$​
Since that is ~10x the age of the Earth, surely Tidal Heating was more important in the past. This suggests, that Tidal Heating could have once been a primary factor for the early Earth.
That is true. Please consider a group of planetoids of comparable composition, which of course varies by their formation zone*:
• near zone (Mercury)
• Habitable Zone (Venus, Earth, Moon, Mars, Asteroids)
• Ice Zone (Jupiter, Saturn)
• Methane-Ice Zone (Uranus, Neptune)
For all those bodies, the basic physics that tcool ~ M / R2 is probably valid. However, the constant of proportionality will change, possibly dramatically, according to bulk composition. For, Lord Kelvin calculated that the Earth's tcool ~ 80 million years (w/o nuclear fission heating). Thus, planetoids far out in the Solar System, where there may be relatively fewer heavy elements, would have less fission heating, and so cool down much more quickly. But, for 2 such similar bodies, their relative Cooling Times are probably dictated by the quantity M / R2. (Unless there are other sources of heat, like Tidal Friction.)
* Carroll & Ostlie. Introduction to Modern Astrophysics, pg. 893 ; National Geographic Naked Science -- Formation of the Solar System (TV)

Last edited: Dec 15, 2008
7. Dec 15, 2008

Staff Emeritus
1) Io is rocky and not icy.

2) You excluded Mercury before, when it didn't match your model. Now you include it when it does.

3) You're still deciding on the status of Mars and Venus based on how well they fit your model, not on anything external.

This isn't science.

8. Dec 15, 2008

Widdekind

How important is Tidal Heating for Io ?

If I quoted Carroll & Ostlie (Intro. Mod. Astrophy.), would you accept their authority?

And, do you understand, that this is not my model ? My UCSD Astronomy tenured Professor told me this, in 2001 AD. Should I quote his name ?

The M / R2 law is (to my knowledge) UCSD Astronomy Tenured Prof. David _. ___________'s model. Do you understand this ?

You're putting all kinds of words in my mouth, alleging that I've said all kinds of things I haven't said.

Last edited: Dec 15, 2008
9. Dec 15, 2008

Widdekind

I have emailed my former Prof., and asked for his opinion.

If I have misquoted my former Prof., or if V_50 is a better science authority than a Tenured Professor at UCSD, then I will retract my suggestion.

10. Dec 15, 2008

Eimacman

Widdekind:

Your mathematics is sound, however as pertaining to Io and the Jovian system there is a flaw in that there are more massive moons and the fact that Jupiter its self is many times more massive than the the Earth. The many gravitational interactions between Jupiter the other Galilean moons and Io would cause heating greater than what could be caused by a planetary system like the Earth Moon system. Also because Io is a body consisting of lighter rocky elements like silicates, sulfur, and sodium would have to be taken into account as these elements dissipate heat at a different rate, the elastic modulus of these elements would also have to be taken into consideration as they would directly relate to the amount of heating caused by such tidal forces.

I apologize that I am not a good enough mathematician to describe this mathematically.

11. Dec 15, 2008

Integral

Staff Emeritus
In order for this topic to continue, you need to post a link to the peer reviewed journal in which your prof published his work.

Please see our site guide lines concerning speculative posts.

12. Dec 15, 2008

Staff Emeritus
Of course Io is different because of tidal heating. The Earth and Moon are different because of the event that formed the Earth-Moon system. Mercury is different because of the event that formed it, as well. Venus is different because its atmosphere acts as a blanket to slow heating. And Mars is, well, normal, but as the only normal world, is itself unique.

The idea that you can ignore a planetary body's history to calculate the cooling rate - as proposed - is clearly false, as evidenced by Io. If you want instead to argue that this model predicts some sort of ideal cooling rate - the cooling rate of planetary bodies if they had a different history than they did - I suppose one can, but in this case the theory is completely unpredictive.

We'd have ham and eggs - if only we had some ham, and if only we had some eggs.

13. Dec 16, 2008

Eimacman

As you said:

"Of course Io is different because of tidal heating. The Earth and Moon are different because of the event that formed the Earth-Moon system. Mercury is different because of the event that formed it, as well. Venus is different because its atmosphere acts as a blanket to slow heating. And Mars is, well, normal, but as the only normal world, is itself unique."

Your comment is on the mark. The history as well as the composition of a planetary body must be known (or as close to it as we can by observations) to make any accurate estimation of cooling by such a planetary body.

14. Dec 16, 2008

cristo

Staff Emeritus
Indeed: I'm locking this until that time. Widdekind, please PM me with a peer-reviewed reference to this work, and the thread will be reopened.