<p>The widely cited figure of 384400 km is the Moon's mean inverse sine parallax value. This is not the same as the semi-major axis length, which is closer to 385000 km. An even better value is 384748 km, which is the value specified in <a href="http://articles.adsabs.harvard.edu/pdf/1983A%26A...124...50C" rel="noreferrer">The lunar ephemeris ELP2000</a>. I generally use 385000 km unless I need a more precise figure.</p>
<p>A nice easily memorizable value for the mass of the Moon is 0.012305 Earth masses. Think of 0.012345 and remember to change the 4 to a 0. The Earth's gravitational parameter is <span class="math-container">$3.986004418\times10^{14}\,\text{m}^3/\text{s}^2$</span>. Applying Kepler's laws, this results in an orbital period of
<span class="math-container">$$\frac{2\pi}{\sqrt{(3.986004418\times10^{14}\,\text{m}^3/\text{s}^2) * (1.012305) / (384748\,\text{km})^3}} = 655.72\,\text{hours}$$</span></p>

CC BY-SA 4.0
2021-07-30T10:34:00.060Z
Solving for the semi-major axis length using a sidereal rotation period of 655.728 hours yields 384749 km as opposed to 384748 km. This is akin to arguing about the number of angels that can dance on the point of a pin.
David Hammen
CC BY-SA 4.0
2021-08-01T07:52:48.637Z
Yes, I worked backwards and got that figure, which mysteriously appears on a Wikipedia page, but with no explanation.  I also eventually found some notes on Geophysical Gravity from eps McGill, CA, at https://www.eps.mcgill.ca/~courses/c510/gravity.pdf, which confirmed the figure.
JerryFrog
CC BY-SA 4.0
2021-08-02T05:06:08.623Z
Where do you get your figures for G' and the relative mass ?  The G' figure is mildly different from the one in the ELP2000 paper and both are different from those in another paper I found at http://proba2.sidc.be/aux/data/spice/docs/DE430_Lunar_Ephemeris_and_Orientation.pdf.  Not that I need such accuracy, but I do like to get a consistent set of data.
JerryFrog
CC BY-SA 4.0
2021-08-02T07:21:08.407Z
@JerryFrog Don't use G, don't use M. Instead use the [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) mu, which is the product of the two.
David Hammen
CC BY-SA 4.0
2021-08-02T09:48:09.860Z
agreed, that was why it was G' instead of G.
JerryFrog
CC BY-SA 4.0
2021-07-30T10:09:05.263Z
2021-07-30T10:22:16.650Z
2021-07-30T10:22:16.650Z
David Hammen
<p>For the general case of two masses interacting according to Newton's Law of Universal Gravitation, the two masses actually orbit about the center of mass of the system, not necessarily the center of the more massive body.
Recall the equation for center of mass. For a two mass system, we will refer to the separation of the two masses as a = r1 + r2 , where r1 is the distance of mass m1 from the center, and r2 is the distance of mass m2 from the center. Consider the case when the two masses are in circular orbits. During their motion, the two planets must be acted on by a centripetal force.
Now, by Newton's third law, these two forces must be equal in magnitude (and opposite in direction), which means m1r1 = m2r2 . This actually proves that the center of the circular motion is the center of mass.
P<em>P = [4</em>pi<em>pi /G(m1+ m2)]a</em>a*a,
which is the expression corresponding to Kepler's third law.</p>

CC BY-SA 4.0
2021-07-30T19:56:12.653Z
This doesn't answer the OP's question, where the commonly cited value of 384400 km as the semi-major axis length results in a conflict with the Moon's sidereal rotation rate.
David Hammen
CC BY-SA 4.0
2021-07-30T15:30:09.603Z
2021-07-30T15:30:09.603Z
Lariliss
<p>According to the sidereal period of the Moon's orbit around the Earth of 27.32166 earth days of 86400 seconds we get an angular velocity of <span class="math-container">$\frac{2\pi}{27.32166×86400} = 2.6617×10^{-6}$</span>.  (NASA's factsheet has the Moon's sidereal period at 655.728 hrs.)</p>
<p>If we use the NASA factsheet figures for the Earth mass of <span class="math-container">$5.9724×10^{24}$</span> kg, the Moon mass of <span class="math-container">$7.346×10^{22}$</span> kg, the semimajor axis of 384,400,000 m, and using a figure for 'Big G' of <span class="math-container">$6.6743×10^{-11}$</span>, then we get an angular velocity of <span class="math-container">$2.6654×10^{-6}$</span>.  Kepler's angular velocity implies that the combined mass of the Earth and the Moon are too high.</p>
<p>Assuming that the figures are all accurate to the last significant figure, as stated, then there are possible errors of 0.0015% in Big G, 0.0017% in the Earth mass, 0.014% in the Moon mass, and probably as small as 0.0003% in the semimajor axis, but the error in omega is 0.139%.</p>
<p>How to reconcile ... what have I missed ?  Or is it something to do with a time-weighted average separation that is somewhat closer to 385,000,000 m ?</p>

CC BY-SA 4.0
2021-07-30T08:20:07.700Z
@James K.  Many thanks.
JerryFrog
CC BY-SA 4.0
2021-07-30T10:36:43.073Z
There's some useful info about MathJax at https://math.stackexchange.com/help/notation and a brief beginner's guide here: https://math.meta.stackexchange.com/q/33179/207316
PM 2Ring
CC BY-SA 4.0
2021-07-30T10:48:06.550Z
The value you're calculating is generally referred to as  the [mean motion](https://en.wikipedia.org/wiki/Mean_motion); For non-circular elliptical orbits, there are only two points where it's equal to the angular velocity.
notovny
CC BY-SA 4.0
2021-07-30T10:54:31.660Z
You should avoid multiplying G & M, when you can. [Wikipedia](https://en.wikipedia.org/wiki/Standard_gravitational_parameter has current GM values, but also see https://space.stackexchange.com/q/22948/38535
PM 2Ring
CC BY-SA 4.0
2021-07-30T18:11:53.043Z
How does one pronounce sidereal?  Is it "sii-der-ee-al" or "side-reel"?
Tony Ennis
CC BY-SA 4.0
2021-07-30T20:07:27.170Z
/saɪˈdɪə.ri.əl/  sigh-DEAR-ree-al
James K
CC BY-SA 4.0
2021-08-02T05:18:32.307Z
@PM 2Ring.  Many thanks.
JerryFrog
<the-moon><earth>
How to resolve the Moon's observed angular velocity with that given by Kepler's Third Law
JerryFrog
# How to resolve the Moon's observed angular velocity with that given by Kepler's Third Law

(Asked by: username_0 on 2021-07-30)

According to the sidereal period of the Moon's orbit around the Earth of 27.32166 earth days of 86400 seconds we get an angular velocity of $\frac{2\pi}{27.32166×86400} = 2.6617×10^{-6}$. (NASA's factsheet has the Moon's sidereal period at 655.728 hrs.)

If we use the NASA factsheet figures for the Earth mass of $5.9724×10^{24}$ kg, the Moon mass of $7.346×10^{22}$ kg, the semimajor axis of 384,400,000 m, and using a figure for 'Big G' of $6.6743×10^{-11}$, then we get an angular velocity of $2.6654×10^{-6}$. Kepler's angular velocity implies that the combined mass of the Earth and the Moon are too high.

Assuming that the figures are all accurate to the last significant figure, as stated, then there are possible errors of 0.0015% in Big G, 0.0017% in the Earth mass, 0.014% in the Moon mass, and probably as small as 0.0003% in the semimajor axis, but the error in omega is 0.139%.

How to reconcile ... what have I missed ? Or is it something to do with a time-weighted average separation that is somewhat closer to 385,000,000 m ?

**Tags:** the-moon, earth
> **Comments:**
> **username_0**: @James K.  Many thanks.  
> **username_1**: There's some useful info about MathJax at https://math.stackexchange.com/help/notation and a brief beginner's guide here: https://math.meta.stackexchange.com/q/33179/207316  
> **username_2**: The value you're calculating is generally referred to as  the [mean motion](https://en.wikipedia.org/wiki/Mean_motion); For non-circular elliptical orbits, there are only two points where it's equal to the angular velocity.  
> **username_1**: You should avoid multiplying G & M, when you can. [Wikipedia](https://en.wikipedia.org/wiki/Standard_gravitational_parameter has current GM values, but also see https://space.stackexchange.com/q/22948/38535  
> **username_3**: How does one pronounce sidereal?  Is it "sii-der-ee-al" or "side-reel"?  
> **username_4**: /saɪˈdɪə.ri.əl/  sigh-DEAR-ree-al  
> **username_0**: @PM 2Ring.  Many thanks.  

---

## 2 answers

### Answer by username_5 on 2021-07-30. Score: 8

The widely cited figure of 384400 km is the Moon's mean inverse sine parallax value. This is not the same as the semi-major axis length, which is closer to 385000 km. An even better value is 384748 km, which is the value specified in [The lunar ephemeris ELP2000](http://articles.adsabs.harvard.edu/pdf/1983A%26A...124...50C). I generally use 385000 km unless I need a more precise figure.

A nice easily memorizable value for the mass of the Moon is 0.012305 Earth masses. Think of 0.012345 and remember to change the 4 to a 0. The Earth's gravitational parameter is $3.986004418\times10^{14}\,\text{m}^3/\text{s}^2$. Applying Kepler's laws, this results in an orbital period of
$$\frac{2\pi}{\sqrt{(3.986004418\times10^{14}\,\text{m}^3/\text{s}^2) \* (1.012305) / (384748\,\text{km})^3}} = 655.72\,\text{hours}$$

> **Comments:**
> **username_5**: Solving for the semi-major axis length using a sidereal rotation period of 655.728 hours yields 384749 km as opposed to 384748 km. This is akin to arguing about the number of angels that can dance on the point of a pin.  
> **username_0**: Yes, I worked backwards and got that figure, which mysteriously appears on a Wikipedia page, but with no explanation.  I also eventually found some notes on Geophysical Gravity from eps McGill, CA, at https://www.eps.mcgill.ca/~courses/c510/gravity.pdf, which confirmed the figure.  
> **username_0**: Where do you get your figures for G' and the relative mass ?  The G' figure is mildly different from the one in the ELP2000 paper and both are different from those in another paper I found at http://proba2.sidc.be/aux/data/spice/docs/DE430_Lunar_Ephemeris_and_Orientation.pdf.  Not that I need such accuracy, but I do like to get a consistent set of data.  
> **username_5**: @JerryFrog Don't use G, don't use M. Instead use the [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) mu, which is the product of the two.  
> **username_0**: agreed, that was why it was G' instead of G.  

---

### Answer by username_6 on 2021-07-30. Score: 0

For the general case of two masses interacting according to Newton's Law of Universal Gravitation, the two masses actually orbit about the center of mass of the system, not necessarily the center of the more massive body.
Recall the equation for center of mass. For a two mass system, we will refer to the separation of the two masses as a = r1 + r2 , where r1 is the distance of mass m1 from the center, and r2 is the distance of mass m2 from the center. Consider the case when the two masses are in circular orbits. During their motion, the two planets must be acted on by a centripetal force.
Now, by Newton's third law, these two forces must be equal in magnitude (and opposite in direction), which means m1r1 = m2r2 . This actually proves that the center of the circular motion is the center of mass.
P*P = [4*pi*pi /G(m1+ m2)]a*a\*a,
which is the expression corresponding to Kepler's third law.

> **Comments:**
> **username_5**: This doesn't answer the OP's question, where the commonly cited value of 384400 km as the semi-major axis length results in a conflict with the Moon's sidereal rotation rate.  

---

