(Edit:Please see the comments, for they would change the meaning of my question)

How can one calculate the pure orbital speed of the moon around Earth by removing the effect of the Sun and keeping only the effect of the Earth (which is also considered stationary because it is the reference of the calculation). The calculation I am trying to do is for a sidereal month.

Somewhere I saw that it is equal to Vcos(A) where to calculate A they used (in a way I did not understand) the numbers:

X=365.25636 days is the number of days in a solar year

Y=27.321662 days is the number of days in a sidereal month

I think they converted the solar year into degrees then introduced the number of days of the sidereal month in some way.

The result they reached is cos(A)=0.89157 . Is it right? If yes, how is it reached?


1- I am trying to do some exact calculations, so sorry for the many digits

2- V=3682.07 km/h is the orbital speed of the moon

3- I am not majoring in astronomy or physics, so my knowledge might seem so little (which it is), but I am interested in this calculation

Thanks in advance

  • $\begingroup$ @JohnForkosh I think I misunderstand the initial question so here is the part I was checking: "The vector "E" (beginning from the Moon) perpendicular to the segment Earth-Sun points to a different direction every synodic month. Thus from this "triangle" of vectors, can we calculate mean Moon's speed E' around a hypothetic heliocentrically motionless Earth E=square root [ (E sin phi)² + (E')² ] thus E'= E * cos phi = current Moon's speed * cos phi = 3680km/hour * cos 27° E'=3279 km/hour So without Sun influence on Earth-Moon system, would Moon revolve around the Earth at 3279 km/hour" $\endgroup$
    – Lame Grass
    Jun 3, 2017 at 10:36
  • $\begingroup$ phi being about 27 degrees, which they say is equivalent to the arc the Earth-moon system cover relative to the terrestrial orbit around the sun in a lunar month (I have to point out that this is mostly Chinese for me, but I only care whether the final number is true or not) They also mentioned something about the usual speed (3682 km/h) being using a non-inertial system and that if you convert it into an inertial one, you get this 3279 km/h, and that it is equivalent to removing some effect of the Sun (and again sorry for the ambiguity but I myself don't really understand what I'm asking :/) $\endgroup$
    – Lame Grass
    Jun 3, 2017 at 10:54
  • $\begingroup$ Please do not edit to tell users to look at comments, which are temporary, but edit the relevant points from the comments into your question. Also, please do not let posts look like revision histories. $\endgroup$
    – ACuriousMind
    Jun 3, 2017 at 14:02

1 Answer 1


Even though this question has so far received a downward voting score, it is something that occasionally crops up, so it seems worth offering an answer that might perhaps help the questioner and others similarly placed.

It looks as if the questioner is trying to get just one single figure for the speed of the moon in its orbit, relative to the earth
(treating the earth as if stationary, as the question says).

One should bear in mind that in reality the moon is continually varying in its speed and in its distance relative to the earth.

So what the question would be aiming for, with just one single speed result, has to be some kind of a mean or average value, as the example figures suggest.

There are many different ways of getting to mean values, of different kinds, for quantities that change in intricate ways. Any such mean or average must be considered a kind of approximation, and despite the possible appearance of many digits (perhaps just for the sake of maintaining numerical consistency with a source), there is no real exactness about it. Here is a suggestion for such an approximation:--

Perhaps the simplest and most drastic simplification to make, to approximate the moon's orbital speed, is to treat the problem as if the moon moves around the earth at its mean angular rate, and at its mean distance. That would be motion in a circle.

On this basis, the calculation is much simpler and easier than the details given in the question.

The two basic essential numbers can both be taken from a useful paper from researchers at the Paris Observatory, ("ELP 2000-85 - A semi-analytical lunar ephemeris adequate for historical times", by Michelle Chapront-Touzé and Jean Chapront, [http://adsabs.harvard.edu/abs/1988A%26A...190..342C]).

The mean distance earth-moon in the cited source is 385000.52899 km,

and the moon's mean angular rate of motion around the earth, in the reference-frame of the fixed stars (not that of the slowly-moving equinox point), was (at 1 Jan 2000):

1732559343.19572 arc-seconds per century of 36525 days.

Converting the units, that gives 0.549014926 degrees per hour, or in radians (units of the radius) per hour 0.00958211810, and then the corresponding speed (km/h) in the circular orbit would just be

radius (km) x speed (radians per hour)

i.e. about 3689.12 km/hr, pretty close to the number quoted by the questioner, by whatever route that was obtained, which I can't identify from the question.

If anything more exact is wanted, then there seems little point in looking for some differently-composed mean, because the different simplifications and approximations needed to reach it would in practice be just as artificial, as the simplifying supposition that the moon moves in a circle.

To illustrate that, one need only consider the following --

The real motion of the moon is neither exactly in a circle nor an ellipse: but it can be approximately modeled by a kind of movable and plastic ellipse, one that -- apart from its further perturbations -- basically would have the earth at one focus and the moon's angular speed inversely proportional to the square of the varying earth-moon distance. On top of that, the motion shows mainly the following kinds of fluctuation:-

-- the mean major-axis direction of the ellipse slowly rotates, once in about 8.8 years,

-- but with librations in the moving direction of the axis, so that it continually wags to and fro in advance and in retard of the average rotation, on a cycle of nearly 7 months,

-- and an eccentricity that fluctuates, again on a cycle of the same period of nearly 7 months, but about a quarter-turn out of phase with the librations in the direction of axis,

-- and then a further speed-up and slow-down, so that the speed increment is maximum at new- and full-moon, and the slowing is maximal at the quarters, while the figure of the moon's orbital path is a little 'squashed' along the line joining earth and sun, so that the 'squashing' of the path brings the moon closer to the earth at the new and full, and farther away at the quarters.

Beyond that, the motion is as if confined to a slowly-fluctuating plane, but that has little effect on the orbital speed the subject of this question.

In light of all the fluctuations, it is apparent that for anything seriously approaching exactness in stating the moon's orbital speed, one would need to specify a date and time, or times, and make a calculation of the moon's position and motion relative to the earth, valid for the calculated times only.

For many purposes, the information in the ELP2000-85 paper cited and linked above would be entirely adequate for estimation of every aspect of the moon's motion around the earth for several centuries, though the calculation would be laborious.

Another and most exact approach would be to read out Chebyshev coefficients enabling calculation of the moon's positions and velocities over time, from recent datafiles of solar-system ephemerides placed online at the Jet Propulsion Laboratory [e.g. the set at ftp://ssd.jpl.nasa.gov/pub/eph/planets/Linux/de430/]. Although much of the calculation is already done with those data, the business of correctly handling the data-file format and processing the Chebyshev coefficients might be hardly less laborious in practice than doing the full calculation from the analytical paper.

I hope that might help and also place the simplified problem in a useful perspective.


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