0
$\begingroup$

I apologize if this is similar to existing questions asked by others, but I was asked to post it instead of asking for clarification of the 'answers' given on silimar questions asked by others...

How do I calculate myself [not by using existing astonomy apps or programs that do the calculating] the celestial [R.A. and DEC] coordinates for my Zenith and Nadir if the following is known:

My location on Earth is known. The date and time are known [a date in the past].

Of course, my location on the surface of the Earth would be in Longitude and Latitude. The trick is to determine the celestial Right Ascension and Declination [these values are stuck/glued/static plots on the ever apparently-rotating Celestial Sphere. As I stand, unmoving, at my location on Earth [thus, my Longitude and Latitude are unmoving as well], overhead, the R.A. and DEC of my Zenith Point and Nadir Point beneath my feet are constantly changing as time passes (assuming I'm not standing at the north or south poles).

So, is there an equation (likely complex geometry or more-so) I can use to do the calculation myself? Those software apps, including those used by the USNO can do it (the old Celestia software could even do it for locations out-in-Space) ... so, what's the formula [equation]?

$\endgroup$
3
  • $\begingroup$ Related. $\endgroup$
    – rob
    Feb 27, 2020 at 2:24
  • $\begingroup$ How much precision are you hoping for? There's a few-paragraph explanation that would let you get within few degrees, but to do better than that you have to know things about the calendar and the eccentricity of Earth's orbit and a bunch of weird edge cases. Ephemeris is a field of study rather than "an equation." You might also search on Astronomy. $\endgroup$
    – rob
    Feb 27, 2020 at 2:35
  • 1
    $\begingroup$ @rob I disagree. Only the Earth's spin is relevant here, but not the Earth's orbit. Because the spin angular velocity is constant, the solution is very simple. $\endgroup$ Feb 27, 2020 at 4:25

1 Answer 1

2
$\begingroup$

Your problem is a special case of converting horizontal coordinates (azimuth, altitude) to equatorial coordinates (right ascension, declination). See for example at Positional astronomy - Conversion between horizontal and equatorial systems.

For the zenith the altitude is $90°$. Its equatorial coordinates are $$\begin{align} \delta_Z&=\phi \\ \alpha_Z&=t_0-\lambda\frac{24^h}{360°} \end{align}$$ where
$\delta_Z$ is the declination of the zenith,
$\alpha_Z$ is the right ascension of the zenith,
$\phi$ is the geographical latitude of your place on earth,
$\lambda$ is the geographical longitude of your place on earth,
$t_0$ is the Greenwich sidereal time (GST), i.e. the sidereal time at $\lambda=0°$.

The nadir is diametrally opposite from the zenith. Hence its equatorial coordinates are $$\begin{align} \delta_N&=-\phi \\ \alpha_N&=t_0-\lambda\frac{24^h}{360°}+12^h \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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