I once saw on TV that the Moon is slowly drifting away from the Earth, something like an inch a year. In relation to that the day on Earth what also increase in time. I wonder why is that?


3 Answers 3


This says it concisely, when describing the effect of tides:

Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon into a higher orbit with a longer period. As a result, the distance between the Earth and Moon is increasing, and the Earth's spin slowing down.

In fewer words: it is the tides.

Edit: I am copying from a comment:

To show the right sign, one must show that the orbital angular momentum of the Moon actually increases with the radius - despite the decreasing velocity as the function of the radius For a $1/r$ potential, $mv^2\propto m/r$ says $v\propto 1/\sqrt{r}$, so the angular momentum $L=rp=mrv=mr/\sqrt{r}\propto \sqrt{r}$ which increases with $r$. – Luboš Motl

In addition I found this better link by googling.

  • $\begingroup$ Hi Anna, just a detail. To show the right sign, one must show that the orbital angular momentum of the Moon actually increases with the radius - despite the decreasing velocity as the function of the radius - which is not obvious but true after a short calculation. $\endgroup$ Commented Apr 30, 2011 at 9:44
  • $\begingroup$ For a $1/r$ potential, $mv^2\sim m/r$ says $v\sim 1/\sqrt{r}$, so the angular momentum $L\sim rp\sim mrv \sim mr / \sqrt{r}\sim \sqrt{r}$ which increases with $r$. $\endgroup$ Commented Apr 30, 2011 at 10:03
  • $\begingroup$ Yes ok that I think is now clear, but what if the moon drifts away, will the effect of the tide not be less (as the distance increase increases), and what does result from that? $\endgroup$ Commented Apr 30, 2011 at 12:34
  • 1
    $\begingroup$ That's right: over time this effect becomes weaker. The final stage will be "tidal locking," in which the Earth's rotational period is the same as the Moon's orbital period (one day = one lunar month). At that point, the tidal bulge on Earth will always be directly below the Moon, and there'll be no torque. This tidal locking has already happened for the Moon: presumably at one point the Moon rotated faster than it does today, but tidal torques on the Moon (just like the tidal torques on the Earth considered here) slowed down its rotation until it kept the same face toward the Earth. $\endgroup$
    – Ted Bunn
    Commented Apr 30, 2011 at 14:17
  • 1
    $\begingroup$ @Ted. If we only had the earth and the moon and we had enough time that would happen. But the lunar orbit can't get too large, before it begins wondering around the solar system, as other gravitational effects, such as the sun and other planets would separate the earth and moon. But IIRC tidal dissapation scales like the inverse 6th power of distance, so the time scale gets very long, its taken 4plus billion years to get as far as it is at present, and by the time it gets 10-20% further out the sun will reach its red giant phase, and the game will be over. $\endgroup$ Commented May 1, 2011 at 0:00

from Astrometric Solar-System Anomalies of Anderson and Nieto, 2009, page 9,
Increase in the eccentricity of the Moon's orbit

While the mean motion and semi-major axis rates of the lunar orbit are consistent with physical models for dissipation in Earth and Moon, LLR orbital solutions consistently reveal an anomalous secular eccentricity variation. After accounting for tides on the Earth that produce an eccentricity change of $1.3*10^{-11} yr^{-1}$ and tides on the Moon that produce a change of $-0.6*10^{-11} yr^{-1}$, there is an anomalous rate of $(0.9\pm0.3)*10^{-11} yr^{-1}$ , equivalent to an extra 3.5 mm $yr^{-1}$ in perigee and apogee distance (Williams \& Boggs 2009). This anomalous eccentricity rate is not understood and it presents a problem.

On the anomalous secular increase of the eccentricity of the orbit of the Moon, by L. Iorio, 2011, explores several alternatives to explain the problem. All of them were inviable, concluding:

Thus, the issue of finding a satisfactorily explanation for the anomalous behavior of the Moon’s eccentricity remains open.


Let us assume that there is a small radial extra-acceleration of the form...

This procedure is called 'data fit' and obviously it is not an EXPLANATION at all.

I based my other answer on the decreasing LOD FACT, and all other answers are saying the contrary.
With this anser I make notice to the fact that an anomaly is present and that the present physics is unable to model it.

Again I point to a MODEL where the reported anomaly is not present because data is along with theory (eq 35 and 36).

I'm expecting a reception to this answer in line with the reception to the other answer.
May be the case that some doubt in your minds can rise the need to read 'out-of-the-box' and try to see if the Sun-Earth anomaly (AU increase) or the Pioneer anomaly can be explained by this model.
When we have the chance to make measures with more precision we have to report an anomaly.
A bunch of anomalies is a call for a new model.
I'm expecting that someday, somebody, will read that paper. It is a cosmological model that do not need DM, DE, Cosmological Constant, it goes well with GR, and agrees with data both at local scale and cosmological scale.


The tidal effect was supposed to deccelerate the Earth rotation and the lost angular momentum should be transferred to the receding Moon.

But facts go on the contrary: LOD - Length Of Day is decreasing.

The Moon-Earth distance displays an increase of 3.8 cm/yr of the semimajor axis of 384,399 km. ([Williams J.G. et al, 2008]) There is an increase of orbital radius at a ratio of $2H_{0}$ as modeled here (eqs 35 and 36).

This is a surprise to many.

EDIT add (to address some concerns evident in the comments)

Statement 1: Established theory: Any transfer of momentum is delayed only by the speed of gravity (the usual light 'c' speed). It happens in a gravitationally bound system, like the Moon-Earth, that the angular momentum is conserved.
Statement 2: It is a widespread beleif (a wish is not theory, Ok?): the Earth slows down the rotation due to tidal effects. The paper with the computations is ..., please?
Statement 3: Facts: We are recording the most precise direct measurements ever made:
- The Moon is receding.
- The LOD ( Length of Day ) is shorter and shorter, implying that the the Earth's angular momentum is increasing!
A choice is mandatory because it exists an internal contradition between the Statements 1, 2 and 3.
My choice is that the tidal effect must be of a low order.
The question remains: Why is the Moon receding?

All previous studies about the past, based in proxies, shall be ignored because the inconsistency exist now, and we must seek a credible answer to explain the present time.
Those studies deserve a few words:
Eclipses - The lack of correct time keeping in the past, and imprecisions in the recorded time and location of eclipses can put to trash those studies.
Tides - the tide amplitude can be affected by several localized factors as coastal line configuration, how deep is the ocean here and there, etc,... Today, as in the past, the tide values are not uniform along the globe because the Earth is a dynamical system. It is not enough to line up some values.

I wonder why persists the spreading of the misconception about the LOD?
(IMO, the unpleasant answer to this is out of the scope of PSE and I will keep it to myself)

LOD - from maia.usno.navy.mil/lplot1.png

  • $\begingroup$ Some comments about this answer can be found on an earlier post containing the same plot. $\endgroup$ Commented Apr 30, 2011 at 15:31
  • $\begingroup$ @dmckee : I wonder why the spreading of misconceptions about the LOD? Fact: The Moon is receding. Established theory: the angular momentum lost by Earth is transferred to the receding Moon. Surprsing Fact: the Earth is increasing the angular momentum ! This goes on contrary to Established. To be Explained: Why is the Moon receding? (instead of downvote give a reasonable LOGICAL explanation, pls) $\endgroup$ Commented Apr 30, 2011 at 16:07
  • 1
    $\begingroup$ Not my field, but wikipedia references Williams, George E. (2000). "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit". Reviews of Geophysics 38 (1): 37–60. Bibcode 2000RvGeo..38...37W. doi:10.1029/1999RG900016 as evidence of changing day and month over the geological time scale. Something about the layers in ancient tidal flats. $\endgroup$ Commented Apr 30, 2011 at 16:15
  • $\begingroup$ @dmckee : also in your link: "The average recession rate of the Moon between then and now has been 2.17±0.31 cm/year, which is about half the present rate." How come? We know, according to the theory, that this rate must be decreasing. I accept that the recent satelite measures are more precise then ever. The Earth is accelerating, and I accept it as a fact. Because any transfer of moment is immediate no one can logically maintain the usual interpretation. $\endgroup$ Commented Apr 30, 2011 at 16:49
  • 2
    $\begingroup$ You can also tell that the length of the day has been changing over historical time scales by looking at ancient eclipse records: cambridge.org/aus/catalogue/catalogue.asp?isbn=9780521461948 . The idea is that we can calculate how long ago (in seconds) an eclipse should have occurred, and compare that with the calendar date on which it was observed. Even though the change in the length of the day is very small, when integrated over a couple of thousand years it adds up to multiple days. This method confirms the standard picture of a slowing rotation rate. $\endgroup$
    – Ted Bunn
    Commented Apr 30, 2011 at 18:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.