<p>You use rounded values for orbital period and geostationary radius with only three significant digits. And your result differs from the actual value just by 1 in the third significant digit. As a rule of thumb, you cannot expect the output to have more correct digits than you used for the input.</p>
<p>Then, the geostationary orbit isn't 360° per 24 hours, but a bit more, as 24h is the time until some point on our surface again points to the sun, and (as the earth moves around its orbit 1/365 in a day) the rotation angle for 24h is closer to 361°. And that is followed by geostationary satellites, so your reference isn't correct - you have to count it with a bit less than 24h.</p>

CC BY-SA 4.0
2022-03-18T07:56:00.640Z
Many thanks! Only sorry I can't tick both of you.  I had spotted the four minute difference between sidereal and solar day, but it didn't seem to make enough difference to account for the discrepancy. Many thanks for the other points.
Mike Stone
CC BY-SA 4.0
2022-03-18T08:00:46.023Z
Re the radius of GSO, I have seen this given as 26199 on some sites and 26200 on others. Since I seemed to be underestimating I used the higher figure to be "on the safe side".
Mike Stone
CC BY-SA 4.0
2022-03-18T10:46:49.857Z
Small point. Does the figure of 27.32166- - - relate to sidereal days or solar ones?
Mike Stone
CC BY-SA 4.0
2022-03-18T13:54:22.127Z
@MikeStone Unless qualified otherwise, a "day" refers to the length of a mean solar day around the year 1900, or 86400 seconds as ticked by a perfect atomic clock at sea level. The concept of the length of a 1900 mean solar day as a standard of time was developed once people realized that the length of a mean solar day is not constant due to the transfer of angular momentum from the Earth's rotation to the Moon's orbit.
David Hammen
CC BY-SA 4.0
2022-03-18T15:26:32.433Z
Many thanks. That's actually what I thought., but I felt I'd better make sure. Cheers.
Mike Stone
CC BY-SA 4.0
2022-03-18T15:27:28.577Z
@DavidHammen Sort of. I have some info & links here: https://physics.stackexchange.com/a/677946/123208 "the ephemeris second was redefined in terms of the mean tropical year of 1900, [...] However, the mean solar second length used in those definition was calculated using data gathered from 1750 to 1892, and corresponds to the mean solar second from around the middle of that period, i.e., ~1820".
PM 2Ring
CC BY-SA 4.0
2022-03-18T15:45:27.323Z
@PM2Ring That data gathered from 1750 to 1892 was extrapolated to noon GMT on January 0 1900 (aka noon GMT December 31 1899) to define the ephemeris second. Time is messy.
David Hammen
CC BY-SA 4.0
2022-03-18T16:06:06.947Z
@David Very messy! of course, Wikipedia isn't exactly a great reference, but I tend to trust Steve Allen's info on the Lick Observatory site.  We have a post from Steve himself on the topic: https://physics.stackexchange.com/a/402062/123208
PM 2Ring
CC BY-SA 4.0
2022-03-17T10:03:17.963Z
2022-03-17T10:03:17.963Z
Ralf Kleberhoff
<p>In addition to the issues raised in Ralf Kleberhoff's answer, you need to account for the mass of the Moon. The correct form for Kepler's third law is <span class="math-container">$$a^3 = G(M_1+M_2) \left(\frac T {2\pi}\right)^2$$</span>
With regard to calculating the Moon's semi major axis given its orbital period and given geostationary orbit data (period of 1 sidereal day and semi major axis of 26199 miles) this becomes
<span class="math-container">$$a_\text{Moon} = a_{GEO}\left(\left(\frac{1 \text{sidereal month}}{1 \text{sidereal day}}\right)^2 \left(1+\frac{M_\text{Moon}}{M_\text{Earth}}\right)\right)^{1/3}$$</span>
An easy to remember value for the Moon/Earth mass ratio is 0.0123. This is accurate to several decimal places. (A more precise value is 0.012300037.) Do this calculation right and you'll get <a href="https://www.wolframalpha.com/input?i=%2826199+miles%29*%28%28%28one+sidereal+month%29%2F%28one+sidereal+day%29%29%5E2*1.0123%29%5E%281%2F3%29" rel="noreferrer">239066 miles</a>. (The published value is 239071 miles.)</p>
<p><strong>Error analysis</strong><br />
Not accounting for the mass of the Moon makes your result small by about 0.41%. This is the largest source of error in your calculation. Using one day instead of one sidereal day makes your result small by about 0.18%. Using 27.3 days instead of 27.321661477 days makes your result small by about 0.08%. Using 26200 miles instead of 26199 miles makes your result too large by about 0.0038%, which is negligible. All of the non-negligible error sources make your result a bit smaller than my calculated value of 239066 miles or the published value of 239071 miles.</p>

CC BY-SA 4.0
2022-03-17T15:07:34.243Z
I used 384748 km (239071 miles) as the Moon's semi major axis length. Two other widely cited values are 385000 km (239228 miles) and 384400 km (238855 miles). Those other two values are actually the mean distance (which is a bit larger than the semi major axis length) and the inverse sine parallax (which is a bit smaller than the semi major axis length). The value of 384748 km comes from the ELP-2000 ephemeris.
David Hammen
CC BY-SA 4.0
2022-03-19T02:08:08.853Z
And of course your result has an even smaller error as it doesn't take into account the mass of the satellite. :)
Acccumulation
CC BY-SA 4.0
2022-03-19T03:55:23.993Z
@Acccumulation That is a 1 out of $10^{20}$ error, or even smaller. My 5 mile error (1 out of ~50000 error) most likely arises from perturbations due to the non-spherical nature of the Earth, the Sun, and general relativity.
David Hammen
CC BY-SA 4.0
2022-03-17T12:58:34.060Z
2022-03-17T13:36:18.907Z
2022-03-17T13:36:18.907Z
David Hammen
<p>The Moon's orbital period is generally given as 27.3 days, and the radius of Geostationary orbit as 26,200 miles. However, using Kepler's Third Law, if I raise 27.3 to the power of 2/3, and multiply the result by 26200, I get a value of a bit over 237,500 miles - about 1000 miles less than the actual value - for the distance from Earth to Moon.</p>
<p>Am I making a silly mistake somewhere?</p>

<distances>
Am I miscalculating distance to Moon?
Mike Stone
# Am I miscalculating distance to Moon?

(Asked by: username_0 on 2022-03-17)

The Moon's orbital period is generally given as 27.3 days, and the radius of Geostationary orbit as 26,200 miles. However, using Kepler's Third Law, if I raise 27.3 to the power of 2/3, and multiply the result by 26200, I get a value of a bit over 237,500 miles - about 1000 miles less than the actual value - for the distance from Earth to Moon.

Am I making a silly mistake somewhere?

**Tags:** distances


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## 2 answers

### Answer by username_2 on 2022-03-17. Score: 29 | Accepted answer

In addition to the issues raised in username_1's answer, you need to account for the mass of the Moon. The correct form for Kepler's third law is $$a^3 = G(M\_1+M\_2) \left(\frac T {2\pi}\right)^2$$
With regard to calculating the Moon's semi major axis given its orbital period and given geostationary orbit data (period of 1 sidereal day and semi major axis of 26199 miles) this becomes
$$a\_\text{Moon} = a\_{GEO}\left(\left(\frac{1 \text{sidereal month}}{1 \text{sidereal day}}\right)^2 \left(1+\frac{M\_\text{Moon}}{M\_\text{Earth}}\right)\right)^{1/3}$$
An easy to remember value for the Moon/Earth mass ratio is 0.0123. This is accurate to several decimal places. (A more precise value is 0.012300037.) Do this calculation right and you'll get [239066 miles](https://www.wolframalpha.com/input?i=%2826199+miles%29*%28%28%28one+sidereal+month%29%2F%28one+sidereal+day%29%29%5E2*1.0123%29%5E%281%2F3%29). (The published value is 239071 miles.)

**Error analysis**  
Not accounting for the mass of the Moon makes your result small by about 0.41%. This is the largest source of error in your calculation. Using one day instead of one sidereal day makes your result small by about 0.18%. Using 27.3 days instead of 27.321661477 days makes your result small by about 0.08%. Using 26200 miles instead of 26199 miles makes your result too large by about 0.0038%, which is negligible. All of the non-negligible error sources make your result a bit smaller than my calculated value of 239066 miles or the published value of 239071 miles.

> **Comments:**
> **username_2**: I used 384748 km (239071 miles) as the Moon's semi major axis length. Two other widely cited values are 385000 km (239228 miles) and 384400 km (238855 miles). Those other two values are actually the mean distance (which is a bit larger than the semi major axis length) and the inverse sine parallax (which is a bit smaller than the semi major axis length). The value of 384748 km comes from the ELP-2000 ephemeris.  
> **username_4**: And of course your result has an even smaller error as it doesn't take into account the mass of the satellite. :)  
> **username_2**: @username_4 That is a 1 out of $10^{20}$ error, or even smaller. My 5 mile error (1 out of ~50000 error) most likely arises from perturbations due to the non-spherical nature of the Earth, the Sun, and general relativity.  

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### Answer by username_1 on 2022-03-17. Score: 20

You use rounded values for orbital period and geostationary radius with only three significant digits. And your result differs from the actual value just by 1 in the third significant digit. As a rule of thumb, you cannot expect the output to have more correct digits than you used for the input.

Then, the geostationary orbit isn't 360° per 24 hours, but a bit more, as 24h is the time until some point on our surface again points to the sun, and (as the earth moves around its orbit 1/365 in a day) the rotation angle for 24h is closer to 361°. And that is followed by geostationary satellites, so your reference isn't correct - you have to count it with a bit less than 24h.

> **Comments:**
> **username_0**: Many thanks! Only sorry I can't tick both of you.  I had spotted the four minute difference between sidereal and solar day, but it didn't seem to make enough difference to account for the discrepancy. Many thanks for the other points.  
> **username_0**: Re the radius of GSO, I have seen this given as 26199 on some sites and 26200 on others. Since I seemed to be underestimating I used the higher figure to be "on the safe side".  
> **username_0**: Small point. Does the figure of 27.32166- - - relate to sidereal days or solar ones?  
> **username_2**: @MikeStone Unless qualified otherwise, a "day" refers to the length of a mean solar day around the year 1900, or 86400 seconds as ticked by a perfect atomic clock at sea level. The concept of the length of a 1900 mean solar day as a standard of time was developed once people realized that the length of a mean solar day is not constant due to the transfer of angular momentum from the Earth's rotation to the Moon's orbit.  
> **username_0**: Many thanks. That's actually what I thought., but I felt I'd better make sure. Cheers.  
> **username_3**: @DavidHammen Sort of. I have some info & links here: https://physics.stackexchange.com/a/677946/123208 "the ephemeris second was redefined in terms of the mean tropical year of 1900, [...] However, the mean solar second length used in those definition was calculated using data gathered from 1750 to 1892, and corresponds to the mean solar second from around the middle of that period, i.e., ~1820".  
> **username_2**: @PM2Ring That data gathered from 1750 to 1892 was extrapolated to noon GMT on January 0 1900 (aka noon GMT December 31 1899) to define the ephemeris second. Time is messy.  
> **username_3**: @David Very messy! of course, Wikipedia isn't exactly a great reference, but I tend to trust Steve Allen's info on the Lick Observatory site.  We have a post from Steve himself on the topic: https://physics.stackexchange.com/a/402062/123208  

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