BGoodForGoodSake wrote:Lets look only at the heat generated by the motion of the crust.
Earlier I attempted to roughly calculate the kinetic energy assuming the continents moved as specified in the Russian site reference by Jbuza: http://www.cnt.ru/users/chas/tectonic.htm
. I assumed an average speed based on the data given and only tried to estimate the kinetic energy. This effort completely ignored the effects of friction.
We know from Newton's first Law of Motion that "Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it". This is intuitively unreasonable. We know that if we slide a book across a table it slows down and stops rather than continuing on at the initial velocity. The explanation of course is that friction is a force acting on the book; the contact between book and table slows down the book, dissipating kinetic energy as heat.
When the continents slide across the earth, there must be some frictional resistance. This was not considered in the earlier calculation. The equations necessary to estimate this energy are fairly simple and can be found in many places, e.g., the website referenced by Jbuza earlier http://www.roymech.co.uk/Useful_Tables/ ... _frict.htm
or for a worked example http://www.physicsclassroom.com/Class/n ... U2L3c.html
In the case of the continents (as explained also by Jbuza earlier), there is a downward force caused by gravity, simply F=mg (force, mass, g is the gravitational acceleration of 9.8m/s^2). The force needed to push an object sideways (perpendicular to gravity, overcoming the frictional resistance) is proportional to the downward force, here, F(friction) = k x F(gravity), where k is the coefficient of friction. The energy used in moving an object is just the force times the distance. So let's evaluate these equations, first considering the values of the different variables:
1) The mass of the continents used in the earlier post was 2.23x10^22 kg. I only found one reference for that, but here is another estimate at 2.6x10^22 kg http://www.nineplanets.org/earth.html
, so I am a bit happier about the validity of this number.
2) The coefficient of friction is a bigger guess; perhaps we can use a range and see what the corresponding range of results is. The Russian website proposes "friction coefficient between molten basalts is about k~0.001". Examining tables of friction coefficients (referenced by Jbuza and Google: http://www.roymech.co.uk/Useful_Tables/ ... _frict.htm
and also http://www.engineershandbook.com/Tables ... cients.htm
, we see values of 1.4 for dry aluminum sliding on dry aluminum, 0.62 (static) for wood on concrete, and 0.04 for teflon sliding on lubricated teflon or lubricated steel. The latter seems like it should be at the low end of friction coefficients. Additionally, http://www.geosci.usyd.edu.au/research/ ... i-etal.pdf
has "In their study of plate generation in a convecting mantle layer, Moresi and Solomatov found that an effective friction coefficient between 0.03-0.13 allowed for plate-like behavior and associated lithospheric subduction. As discussed in Moresi and Solomatov  this value, although low relative to laboratory values, is consistent with seismic field studies, with the lack of heat flow anomalies associated with major faults, and with effective fault friction values deduced from studies of trench topography. The value is also not unphysical as it can result from the development of a fault gouge layer or through pore fluid pressure effects".
3) To calculate the energy, the distance of 2,300,000 meters from the Russian site will be used. [This is based on the separation between South America and Africa and is used as the distance traveled by all continental masses.]
So F(gravity)=m x g = 2.23x10^22 kg x 9.8 m/s/s = 2.19x10^23 N (Newtons).
Then using a range for k of 0.001 to 0.1, F(friction)= k x F(gravity) =2.19x10^20 N for k=0.001 and 2.19x10^22 N for k=0.1.
Note that the kinetic energy of an object depends on its velocity but not the distance traveled whereas the frictional energy expended depends on the distance but not (to a first approximation) the velocity.
So Energy(friction)=F(friction) x d and Energy = 2.19x10^20 N x 2,300,000 meters = 5.03x10^26 J for k=0.001 and 5.03x10^28 J for k=0.1.
From the prior post, the estimated energy to vaporize the oceans was 3.62x10^27 J.
We see the the frictional energy is a bit larger to much larger than the kinetic energy depending on the coefficient of friction. It is certainly quite possible that the energy released by continental motion in 6 hours as described could have generated enough heat to vaporize the oceans (although only a fraction (1/3 ?) of the heat energy released would have transferred to the oceans).