where did the rotational energy and angular momentum go? Some claim that the angular momentum went to the Moon. Astronauts put a corner reflecting mirror on the moon and reflected a Laser and timed the reflection and found the Moon to be moving away at a couple centimeters a year. This increases the lever arm part of angular momentum but the orbits slow down at further distance. Could this account for the lost angular moment and rotational energy over 100 million years?

  • $\begingroup$ The energy went into tidal motion and the angular momentum is still there (minus small changes due to similar orbital angular momentum coupling to the sun and other planets). The mechanism is exactly what you said it is. You can look up the moment of inertia for the Earth and calculate the total angular momentum for both configurations - fast Earth rotation and a closer Moon vs. slower rotation today with a more distant Moon. $\endgroup$
    – CuriousOne
    May 2, 2015 at 16:18
  • $\begingroup$ where did you read 17 hours? $\endgroup$
    – innisfree
    May 2, 2015 at 17:16
  • $\begingroup$ To the best of my recollection it was an exercise in a Physics text. Cutnel and Johnson or Halliday and Resnick or Serway and Faughn, they might have worked backwards. $\endgroup$
    – user78939
    May 2, 2015 at 17:26
  • $\begingroup$ That 17 hours is rather short for the length of day "in the time of the dinosaurs". That's the length of day about 2.45 billion years ago (Williams, "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit," Reviews of Geophysics 38.1 (2000): 37-59.) $\endgroup$ May 2, 2015 at 19:34

2 Answers 2


First, let us look at the current rate at which the moon slows down.

I have a few different sources, and they don't all give me the same answer.

First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change corresponds to $1.6\cdot 10^{-10}$ or about 1 part in 6 billion.

Next, there is the NASA Moon factsheet which tells us that the rate at which the Moon-Earth distance is increasing is $3.8 \mathrm{\;cm/year}$. We know from Kepler's 3rd Law (the Law of Periods) that the square of the period scales with the cube of the semimajor axis, so if the axis increases by $\Delta r$ and the semimajor axis is $r$, the period scales as

$$T=\alpha r^{\frac32}$$

Which, with a little manipulation, can be turned into

$$\frac{\Delta T}{T} = \frac32 \frac{\Delta r}{r}$$

With a semimajor axis of $3.84\cdot10^{8}\mathrm{\;m}$, we find

$$\frac{\Delta T}{T} = 1.5 \cdot \frac{0.038}{3.8\cdot 10^8} = 1.5 \cdot 10^{-10}$$

This is almost the same rate of change.

Next there is the Wikipedia article on the subject; it gives a current value of about 2.3 ms/century (based on moon ranging experiments - the Moon recedes from the Earth at 3.8 cm/year as given in this NASA moon factsheet). That value is a little high; over the past 2700 years it is believed the average was about 1.7 ms/day/century (although how they can use observations over 2700 years to conclude this is something that I cannot fathom...). It is believed that the difference is caused by the disappearance of polar ice after the last Ice Age - as ice is no longer pressing down on the poles, the Earth becomes a little less flattened and this increases the rate of rotation which counters the effect of the Moon.

Finally, there are the numbers quoted by David Hammen in the comments; he gives a value of 1.7 ms/day/century. Converting this to ms/year/year, I get 0.006 ms/year/year. This is a slightly higher value than the 0.005 I used above, but agrees well with the number from the Wiki article.

With a simple linear extrapolation, we could multiply this rate of change by the number of years to conclude what the length of a day must have been at the time of the dinosaurs. If we do that, then if we say the "time of the dinosaurs" started about 200 million years ago, then the years has lengthened by 1.2 million seconds in that time - or just under an hour a day less than today. To get all the way to a day of 17 hours, you need to go back a little before the dinosaurs - about 2 billion years.

With that said, it is indeed the tidal forces between the earth and the moon that cause the earth to slow down and the moon to speed up: this preserves the angular momentum, but energy is lost in the process. This follows from the expression for energy in terms of angular momentum: knowing that $L=I\omega$ and $E = \frac12 I \omega^2$, we see that $$E = \frac{L^2}{2I}$$

The simplified calculation above ignores two important effects. The first of these is that the effect of tidal drag scales with the sixth power of the distance. The effect of the moon on Earth's tides scales with the third power of the distance (the differential of the inverse square law), and the tidal bulge can be thought of as a "mass dipole" whose effect drops off with the inverse third power of distance - that's how that sixth power comes about. To estimate how important this might be, let us see how far the moon would have moved in 200 million years, moving at 3.8 cm/year: it is about 7600 km, or 2% of the distance. This means that the tidal drag "at the time of the dinosaurs" started out about 12% stronger than it is now. The second effect, that I briefly mentioned above, is that ice formation on the polar caps (and the attending changes in the amount of liquid water in the ocean) have a rather large effect on the moment of inertia of the Earth, and therefore can affect the rate of rotation strongly. I am ignoring both these effects here.

Instead, I would like to finish with a calculation of the energy levels involved. Since $L$ is preserved in the earth-moon system, an increase in $I$ means that the energy will be less. The difference is lost as friction in the earth's surface: mostly heating, some fracturing (erosion) of the earth's crust. A small amount of energy is transferred to the moon - but this is quite small, given that the moment of inertia of the moon (relative to the center of rotation of the Earth-Moon system) is so much larger than that of Earth.

The amount of energy lost in this way is staggering (just think about the energy that could be harvested if you could capture all the energy of the tides) - but it is spread over a very large surface, and a very long time.

Let's imagine that the heat is lost due to radiation into space. In that case, we should be able to calculate how much hotter the surface is because of this mechanism.

First - the moment of inertia of the earth is $8\cdot 10^{37} \mathrm{kg\; m^3}$ source, and the moment of inertia of the moon (about the center of the earth) is $7.3\cdot 10^{23} \mathrm{kg} \cdot 3.8\cdot 10^8 \mathrm{\;m^2}=1.1\cdot10^{41}\mathrm{\;kg\;m^2}$ source

As the Earth slows down in its rotation, it loses energy. 1.7 ms/day/century is roughly 1 part in five billion. The rotational energy of the earth is

$$\frac12 I \omega^2 = 0.5 * 8\cdot 10^{37} * \left(\frac{2\pi}{24*3600}\right)^2 = 2.1\cdot 10^{29}J$$

Slowing down by 1 parts per five billion ($2\cdot 10^{-10}$) means the energy (which goes as rate of rotation squared) will twice as fast, or $4\cdot 10^{-10}$, so

$$\Delta E = 8\cdot 10^{19}\mathrm{\;J/yr}$$

This is per year, so the power dissipated per second is

$$P = 2.7 TW$$

This answer is quite close to the Wikipedia value of 3.3 TW. I am willing to say the difference is due to slightly different rounding during calculations.

Assuming that this energy is evenly distributed over the surface of the earth ($5.1\cdot10^14\mathrm{\; m^2}$), it means the energy per square meter per second is 5 mW.

Now black body radiation goes as temperature to the fourth power:

$$S = \epsilon \sigma T^4$$

The sun irradiates half of the earth with about $1370 \mathrm{\;W/m^2}$; the area "seen" by the sun is $\pi R^$ which is 1/4 of the actual surface area, so the average power of sunlight per unit area is 343 W. Increasing this by 5 mW we expect the temperature rise to be given by

$$\frac{T+\Delta T}{T} = \sqrt[4]{\frac{E + \Delta E}{E}}\\ 1+\frac{\Delta T}{T} = 1 + \frac{\Delta E}{4E}\\ \Delta T = \frac{\Delta E}{4E} T$$

Putting in the numbers, we find the earth's temperature to be around 1 mK higher because of the tidal friction.

Note that I ignored the extent to which some of this energy ends up in the moon - as the moon's moment of inertia is so much higher, the energy it gains will be very small compared to the energy lost by the earth.

  • $\begingroup$ Its worth noting that the rate of slowing depends ocean levels and continental configurations (through the $Q$ of the tidal forcing). I don't recall if the value you quote is the current rate or the long term average. $\endgroup$ May 2, 2015 at 18:43
  • $\begingroup$ @dmckee fair point. I will check into the instantaneous vs long term question. $\endgroup$
    – Floris
    May 2, 2015 at 18:46
  • $\begingroup$ it seems obvious the rate of slowing depends on the distance between the moon and the earth and their difference of angular velocities. For short distance and greater velocities difference the torque is greater. So with time this deceleration decreases ... till the moon is in phase with the earth and would not seem to move anymore. By the way the numbers are wrong: 0.005s/year/year = 1.5 10^-10 / year $\endgroup$
    – ceillac
    May 2, 2015 at 19:05
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    $\begingroup$ With regard to the 2700 years (actually, longer), the Chinese, Babylonians, and Egyptians were scared witless by solar eclipses. They recorded them meticulously. Those eclipse records provide a nice ancient anchor for solar system dynamics and for the Earth's rotation rate. $\endgroup$ May 3, 2015 at 4:26
  • 1
    $\begingroup$ To expand on David's comment about historical eclipse records: the East-West location of the affected areas on the surface of the Earth depend the integrated slowing between now and then. $\endgroup$ May 3, 2015 at 17:09

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving.
The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward itself. As a result the tide will have an equilibrium position as in this picture.
Then you can see the tidal bulge will constantly accelerate the rotation of the Moon and its total energy. And here you have a transfer of energy.
Because the energy of the Moon increases it goes away from the Earth just as a result.


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