New answers tagged coordinate
0
I found an equation on this website,
$\sin{\theta_p}=\sin{\theta_{az}} \cos{\theta_{lat}}/ \cos{\delta}$,
with $\theta_p$ being the parllactic angle, $\theta_{az}$ being the azimuth, $\theta_{lat}$ being the latitude of the observing stations, and $\delta$ being the declination. The website cites this equation as being from "Spherical astronomy, Small pg. ...
3
You are correct that M81 transits the meridian at upper culmination just past mid-February in the eastern US. As long as the Moon is well below the horizon, as it will be in Feb. 2020, this will be the best time to observe M81, and its neighbor, M82. Since these galaxies are relatively bright, good views are available from late December through early May ...
1
This can be done using the Python package Skyfield fairly easily.
Here is a shell of a script; I used some dictionaries to hold data, you may want to do something else. I've stored a lot of goodies but only printed azimuth, altitude and distance in kilometers.
By default Skyfield's .altaz() method calculates atmospheric refraction for anything higher than -...
4
Yes. The observer is "flipped". In northern hemisphere the sun in in the southern half of the sky. So the bottom of the sun is toward the south horizon. The top of the sun is toward the north. By convention this side is the north of the sun (or moon). When facing the sun (at noon) north is up toward the top of your head.
In the southern hemisphere the ...
6
Yes, the orientation of the Sun will be different from the Earth's northern and southern hemisphere, just like your example of the Moon.
I would not say that a sunspot in the "northern" hemisphere would appear to be in the "southern" hemisphere just because of the change in orientation. North is fixed on the Sun and Moon, just like they are on the Earth. (...
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