In this question it proposes Earth be put in a closed system box. My question is... What if we first stop the planet from moving through the universe... We're just going to ignore that aspect of it... and then throw this closed system box around Earth, lets say out as far as the 1.25x the distance of from the center of the Earth to the Moon's further distance in it's orbit...

So each side of the box is 1,220,100 km making it have a volume of 1,820,000,000,000,000,000 km3.

How much energy would this box roughly contain? Ignoring the force pushing the moon away and throwing things into chaos, how long till, if ever, the majority of the energy is in the radiation surrounding the Earth/Moon rather than on the surface or in the system?

Matter of Earth/Moon? Light Energy that is captured in this area? Heat energy radiating from Earth/Moon? Rotational energy of the Earth/Moon?

  • $\begingroup$ Why do you think the majority of the energy would ever be in the radiation surrounding the Earth/Moon? I ask because it's an unusual question and understanding why you thought to ask it may help guide an answer. $\endgroup$
    – Cort Ammon
    Jan 17 '17 at 23:39
  • $\begingroup$ I don't think it would ever be, but doesn't hurt to ask since it's already part of the equations one will likely do to answer this, or at least probably an extension to them by a small bit ^.^ $\endgroup$
    – Durakken
    Jan 17 '17 at 23:49
  • $\begingroup$ So I would expect 99.99999% of the energy to remain in the Earth/Moon, if not more than that (probably several more nines). The next question would be what types of energy are you interested in quantifying? The mass-energy of the entire system will be orders of magnitude greater than just the kinetic energy. $\endgroup$
    – Cort Ammon
    Jan 18 '17 at 0:03
  • $\begingroup$ You might be interested in one of my favorite pages on Wikipedia: Orders of magnitude(energy) $\endgroup$
    – Cort Ammon
    Jan 18 '17 at 0:03
  • 2
    $\begingroup$ Really belongs in physics or similar $\endgroup$
    – Murphy
    Jan 18 '17 at 11:09

My new favorite question! And I have all night off! Lets begin.


Calculated by $E = mc^2$.


$$6.0\times10^{24} \cdot (3.0\times10^8)^2 = 5.4\times10^{41} \text{ J}$$


$$7.3\times10^{22} \cdot (3.0\times10^8)^2 = 6.6\times10^{39} \text{ J}$$

Gravitional potential energy

Calculated for a sphere by $$\frac{3GM^2}{5R},$$ where $M$ is mass of the planet, $G = 6.67\times10^{-11}$, and $R$ is radius.

Gravitational binding energy of Earth: $$\frac{3\cdot6.67\times10^{-11}\cdot\left(6.0\times10^{24}\right)^2}{5\cdot 6371000} = 2.3\times10^{32} \text{ J}$$

Gravitational binding energy of the Moon: $$\frac{3\cdot6.67\times10^{-11}\cdot\left(7.3\times10^{22}\right)^2}{5\cdot 1737000} = 1.2\times10^{29} \text{ J}$$

Turns out both of those should be considerably less, since the densest parts are in the center. I should go integrate each layer individually.

Gravitational potential between the Earth and the Moon can be estimated using $$\frac{GmM}{r} = \frac{6.67\times10^{-11}\cdot7.3\times10^{22}\cdot6.0\times10^{24}}{384000000} = 7.6\times10^{28} \text{ J}$$

Kinetic Energy

First, let us assume that the box is moving with the Earth in orbit, or whatever the Earth is doing, so the Earth has zero velocity relative to the frame of the box, and thus zero kinetic energy.

Kinetic Energy of the Moon's orbit around the Earth is due to about 1 km/s velocity and is $$\frac{1}{2}mv^2 = 7.3\times10^{22}\cdot 1000^2 = 3.7\times10^{28} \text{J}$$

Rotational energy is $$\frac{1}{2}I\omega^2$$ and the moment of inertia for a sphere is $$\frac{2mR^2}{5}.$$

Rotational energy of the Earth is $$\frac{2}{10}mR^2\omega^2 = \frac{1}{5}\,6.0\times10^{24}\left(6371000\right)^2\left(7.3\times10^{-5}\right)^2 = 2.6\times10^{29} \text{ J}$$

Rotational energy of the Moon is $$\frac{2}{10}mR^2\omega^2 = \frac{1}{5}\,7.3\times10^{22}\left(1737000\right)^2\left(2.7\times10^{-6}\right)^2 = 3.2\times10^{23} \text{ J}$$

There are a variety of smaller effects, like apsidal precession and eccentricity of the moon's orbit, but I'm going to call those insignificant.

Thermal energy

We will divide the Earth and moon into parts and estimate the thermal energy of each part.

The entire core of the Earth is modeled as solid iron at 5700 K. This calculation, and the related simplifications, are too extensive to reproduce in full. I used Vocadlo, et. al., 2003 to get heat capacity curves at core temperatures and pressures, and Desai, 1985 for lower temperatures. Basically, I model specific heat capacity of Iron at 54 J mol$^{-1}$K$^{-1}$ to 46 between 5700K and 3200 K; and then following Table 2 of Desai from there.

The simplifications are significant, since I can't find a good way to account for the phase change at such high pressures. In fact, some of the references seem to indicate that the enthalpy of fusion is negligible at such high pressures. So take this one with a grain of salt.

Using radius and density numbers from Wikipedia, mass of the inner core is $$\frac{4}{3}\pi r^3\rho=\frac{4}{3}(3.1415)(1200000)^3(13000) = 9.4\times10^{22} \text{ kg}.$$ For the outer core $$\frac{4}{3}\pi \left(r_{outer}^3-r_{inner}^3\right)\rho = \frac{4}{3}(3.14)\left(3450000^3-1200000^3 \right)(11000) = 1.8\times10^{24} \text{ kg}.$$

The total is $1.9\times10^{24} \text{ kg}$. Integrating that over the heat capacity numbers above, I get that each mol of iron at 5700K has 250 kJ. The total thermal energy of the core is then $$1.9\times10^{24} \text{ kg} \cdot \frac{1 \text{ mol}}{0.058 \text{ kg}}\cdot \frac{250000 \text{J}}{\text{mol}} = 8.2\times10^{30} \text{J}$$


  • $\begingroup$ I think a good takeaway here is that the mass-equivalent energy is literally 1,000,000,000 times greater than all other sources of energy put together. $\endgroup$
    – Cort Ammon
    Jan 18 '17 at 4:03
  • $\begingroup$ @CortAmmon Well, the point I'm going to make eventually, is that if you take the massive, thermodynamically closed box, and fast forward to the entropy endstate, you will have a gas. Since the mass isn't going to miraculosly transform into energy, I'm going to try to calculate the energy of the gas. I don't think the energy will be in the 'radiation surrounding the Earth/Moon system', I think all the energy will be the thermal energy of the gas. $\endgroup$
    – kingledion
    Jan 18 '17 at 4:07
  • $\begingroup$ Ok, timeout, I'll be back to this tomorrow. Thermal energies are hard. $\endgroup$
    – kingledion
    Jan 18 '17 at 4:09

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