3
$\begingroup$

I would like to project the Moon's position on to the surface of the Earth, starting with its position expressed as Right Ascension in degrees.

I know that the declination of the Moon corresponds to the terrestrial latitude but how can I "convert" the right ascension of the moon to find my longitude?

I'm not an astronomer, so I'm hoping for a fairly simple answer.

EDIT

My friend used this formula a long time ago:

For the time at which I want to calculate the longitude,

time_hrs = hours + minutes / 60

then using right ascension of Sun,

delta_x = time_hrs * 15 + right_ascension_sun - 180;

The coordinates of the Moon's ground track at this time are then:

longitude = Right_ascension_moon - delta_x;
latitude  = Declination_of_moon

but i don't know if is correct, or how precise it will be.

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Are you asking "what latitude and longitude on the Earth is directly underneath the Moon"? The longitude can be calculated by determining at which longitude the local sidereal time (LST, or LMST) has the same right ascension as the Moon. $\endgroup$
    – JohnHoltz
    Commented Apr 11, 2019 at 16:25
  • $\begingroup$ And if i know even right ascension of sun is possible knowing these data (right ascension for both) calculate longitude directly underneath the Moon ? $\endgroup$
    – Borja
    Commented Apr 11, 2019 at 20:30
  • $\begingroup$ How do you propose to get the longitude of the Moon from knowing the right ascension of the Sun? I cannot think of a procedure that is any easier than calculating the LST to get the answer directly. $\endgroup$
    – JohnHoltz
    Commented Apr 11, 2019 at 21:22
  • $\begingroup$ @JohnHoltz i edited the question adding formula that my friend suggested me a lot of years ago nut i don't know if is correct or the same $\endgroup$
    – Borja
    Commented Apr 12, 2019 at 0:07
  • 1
    $\begingroup$ I will provide a more formal answer when I get time, but your method is approximately correct. It assumes the Sun is above 0 longitude at 12:00 UT, but that is incorrect by up to 16 minutes of time (= 4 degrees of longitude). I assume that you are calculating the position of the Moon (instead of having a known position from a table). If you have to calculate the position of the Sun in order to use your approximation, you should instead calculate the LST and avoid the approximation. See astronomy.stackexchange.com/questions/24859/local-sidereal-time $\endgroup$
    – JohnHoltz
    Commented Apr 12, 2019 at 16:43

1 Answer 1

Reset to default
4
$\begingroup$

Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows:

  • latitude = declination of the Moon
  • longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ascension. (LMST dependents on the date, time, and longitude.) Everything is known except for the longitude.

Accurate Method (using GMST)

Calculate the Local Mean Sidereal Time (LMST) from $LMST=GMST+long_{east}$, where GMST is the Greenwich Mean Sidereal Time and $long_{east}$ is the longitude with positive values in the eastern hemisphere. From the post Local Sidereal Time, $\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H $ where D is the number of days (including the fraction of days) from J2000 (Jan 1, 2000 at 12 h UT = Julian Day 2,451,545.0) and H is the Universal Time (UT) in hours.

For example, at H=17 hour UT on Nov 1, 2000, I calculate the following values: $$ RA_{Moon}=18h\;49m\;35s = 282.400° \\ Julian Day = 2,451,850.208 \\ \therefore D=2,451,850.208 - 2,451,545=305.209 \\ \therefore GMST = 296.288° \\ $$ Then $$ RA_{Moon}=LMST=GMST+long_{East} \\ 282.400=296.288+long_{East} \\ \therefore long_{East}=-13.888° $$

Approximate Method (using Sun's position)

Your method of using the Sun's right ascension assumes that the Sun is over 0° longitude at 12 h UT. Written as a formula and allowing for different times and longitudes, it would be $LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east}$. Setting this LMST to the right ascension of the Moon gives the following:

$$ RA_{Moon}=LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east} \\ \therefore long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h) $$

The reason this is an approximation is because of the following:

  1. The Sun is not at 0° longitude at 12:00:00 UT everyday because of the Equation of Time. The Sun can be off by up to 16 minutes of time (which is equivalent to 4° of longitude).
  2. The term 15(Time in UT - 12h) should be 15*1.002738*(Time in UT - 12 h), but this difference is small compared to the first approximation.

Continuing the example of Nov 1, 2000, I calculate the following: $$ RA_{sun}=14h\;28m\;46s = 217.19° \\ long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h) \\ long_{East}=282.400-217.19-15(17-12) \\ \therefore long_{East}=-9.79° $$

This is different by 4° because I chose Nov 1 since it is close to the date of the largest Equation of Time.

Likewise, if you wanted to calculate the sub-point of the Sun, it would be better to use the GMST and the Sun's right ascension by solving for the longitude from $RA_{sun}=LMST=GMST+long_{East}$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ If instead I have an object that is not the moon or the sun and of which I know right ascension and declination, I can use these formulas to calculate the sub-point on earth of this object ? $\endgroup$
    – Borja
    Commented Apr 15, 2019 at 10:50
  • 1
    $\begingroup$ That is correct. The formulas only depend on the position of the object, not on the name of the object. Be sure to mark my message as an answer if it answers your original question. :-) $\endgroup$
    – JohnHoltz
    Commented Apr 15, 2019 at 13:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .