# Thermodynamics of the Earth’s core

Wikipedia presents conflicting data in the article “Tidal Acceleration”. Hopefully one of you experts can clarify this.

The torque on the Earth is measured at $$3.9 \cdot 10^{16}$$ yet the tidal friction is only $$3.75\cdot 10^{12}$$. Wikipedia say that the Earth’s days will be an hour longer in 180 million years. Using the standard formula $$KE_{rot} = \frac{1}{2} T \omega^2 = \frac{1}{2} \cdot (\frac{2}{5}M R^2) \cdot \omega^2 = \frac{1}{5}M R^2 \cdot \omega^2$$ rotational kinetic energy of the Earth is $$2.5765\cdot 10^{29}$$ joules. If you do the math this requires $$1.3\cdot 10^{15}$$ Watts over the $$180$$ million years. This is $$1290$$ times the friction generated by the tide. It seems to me that somebody dropped the ball. The $$3.75 \cdot 10^{12}$$ watts of tidal friction is a joke! That doesn’t even begin to slow the Earth's rotation at the rate require to fit the geological evidence. If you assume a 30 to 1 ratio of friction to energy applied $$3.9 \cdot 10^{16}$$ watts is perfect but this generates $$3.77\cdot 10^{16}$$ watts of tidal friction roughly $$10,000$$ times the afore mentioned $$3.75\cdot 10^{12}$$ watts. I have spent quite a bit of time calculating how this could be. It seems that the people who calculated the tidal friction at $$3.75\cdot 10^{12}$$ watts incorrectly assumed only the water in the ocean moves and there is no need to calculate the friction of the remainder of the Earth commonly refer to as "Earth-Tide". The Earth is $$\frac{1}{4400}$$ water by mass. This makes some sense but this would mean the rest of the mass of the Earth is generating heat at $$2.28$$ times the rate of the water. Actually that does seem plausible however if this is right the calculations surrounding the thermodynamics of the Earth’s core should be changed. It is stated that the Earth core is not heated by tidal friction which is true if your calculations are based on water. Earth's core it is heated almost completely by "Earth-tide" and that should be pointed out!

I doubt I am the only one to figure this out, what is the explanation for this? Is there something I am over looking?

• It is a little hard to untangle exactly what your issue with the math is. One hour longer day is about 8% less rotational energy. So, the 3E12 Watts for tidal friction losses is pretty much spot on. – Jon Custer Jun 21 '19 at 19:18
• Units. The exercise of always writing your units out explicitly will make you a better physicist, and make your writing clearer. – dmckee --- ex-moderator kitten Jun 22 '19 at 15:32

If you do the math this requires $$1.3⋅10^{15}$$ Watts over the 180 million years.
You did a number of things wrong there. First off, you didn't show how the math. It appears you calculated the power needed to make the Earth's rotation come to a complete stop in 6 million years. A much smaller mistake is that you used $$\frac25 MR^2$$ for the Earth's moment of inertia. It's a bit smaller than that due to the fact that the Earth's core is significantly more dense than is the mantle. A better value is $$0.33MR^2$$, or $$\mathrm I =8\cdot10^{37}\,\text{kg m}^2$$.
The correct calculation for the change in rotational energy is $$\Delta T = \frac12\mathrm I \left(\left(\frac{2\pi}{\text{sidereal day}}\right)^2 - \left(\frac{2\pi}{1\,\text{sidereal day}+1\,\text{hour}}\right)^2\right)$$ With this expression, the change in energy is $$1.7\cdot10^{28}$$ joules. Dividing this by 180 million years yields $$3.0\cdot10^{12}$$ watts, which is consistent with the $$3.75\cdot10^{12}$$ watts currently dissipated by the ocean tides and with the decrease over time in the power dissipation as the Moon recedes from the Earth.