some help with astronomy coordinate systems?

Question

i have homework to do and these questions are puzzling me. im not so much interested in the right answers as the method of finding them.

theres just too many new words flying around at the minute for me, so im confused with this stuff. can anybody help with the following questions? if not all of it then some of it?

i know what the sine and cosine rules are though.

at 18:00 on october 25th:
Venus had right ascension 17h 11m 41s and declination -21 degrees, 8.1'
The Sun has right ascension 14h 0m 53s and declination -12 degrees 18.7'

how do i find what azimuths does venus rise and set on the 25th of october, as seen at greenwich (latitude 51 degrees, 28.6')????

also, how do i find the hour angle when it rises and sets, and the apparent solar time at greenwich when venus rises & sets?

Answer 1

It is a very visual problem. The words are confusing, see the picture in the source (or find other pictures that help you). Look it A LOT.

I think the azimuth is the easy part of the problem. The date, time, coordinates of the Sun, and right ascension of Venus are not needed to solve for azimuth, only the declination of Venus and the latitude of Greenwich are needed for that.

You need to draw a celestial sphere that looks something like the one in the source. Draw the horizon and the local meridian (labeled as the "celestial meridian in the picture, but I don't call it that), then the north celestial pole as a tic mark on that meridian at 51 degrees above the horizon (because Greenwich is at 51 degrees north latitude). Draw the celestial equator so that it is a great circle perpendicular to the meridian and crossing the meridian 90 degrees from the pole. The equator will cross the horizon at an angle. Venus will be on another meridian that is not perpendicular to the horizon but that is perpendicular to the celestial equator. You want to draw it so that it crosses the horizon 21 degrees, 8.1' south of the celestial equator, That distance is measured along that meridian. The drawing does not have to be exactly accurate, but label all the angles. So you have a spherical triangle made up of a segment of the horizon, a segment of the celestial meridian and segment of that other meridian that Venus is on that crosses the horizon some distance after is crosses the celestial equator. The length of the side of the triangle made by the celestial meridian in 51 degrees, because you drew the pole 51 degrees above the horizon. The length of the side made by part of the horizon is the azimuth you want to find. The length of the side made by the other meridian is 90 + 21 deg. 8.1', because it is 90 degrees from the pole to the equator and another 21 deg. 8.1' more to Venus. The horizon and local meridian meet at 90 degrees below the pole. So you have two sides and one angle in the spherical triangle. I THINK that is enough to solve for the 3rd side, which is the azimuth. It will be that distance from the north at both rising and setting time, but in a negative direction, or 360 - that distance, at the other time.

And now MY head is spinning and I cannot help more with the rest of the problem. But basically you draw all the lines you can on the celestial sphere and LOOK at it and write the values of all the sides and angles that the statement of the problem gives you and use spherical trig to solve for the side or angle you want. And remember that only great circles are useful. The spherical trig rules do not work with small circles. If there is something that is not a great circle that you need to know, you have to find other lines that ARE great circles to solve the problem. You may need to solve several different triangles to get the final value you want in a particular problem. There is no real set formula or cookbook procedure that works in every case. You just have to draw and label and look and figure it out. As I say, it is a very visual problem.

Answer 2

Astronomical co ordinates are similar to earth co ordinates like Latitude and longitude. Celestial globe like earth globe is marked with co ordinates. Right ascension is measured from a point where celestial equator meets the ecliptic.Like we measure longitude from Greenwich as zero longitude,R A is marked as 24 hours each hour 15degrees.Declination is marked north or south off celestial equator. Towards north as positive towards south as negative
Other things you can find from astronomical emepheris
Details about RA/DA can be send to you by E mail with sketch. For astronomical calculations you can down load free astronomical software from net and install. Send your e mail ID I will send you a map of celestail sphere with RA/DA
[e-mail]
[e-mail]

Answer 3

1. For horizon the observer is the origins (zero point), while for equatorial and ecliptic it is the Earth (Centre). 2. Horizon is static, while equatorial and ecliptic vary with time (Sun's diurnal and yearly cycles). 3. Casual observer doesn't care for any other system than 'horizon'. 4. These three are 3 dimensional representations with time having no role. If time is also to be depicted thn we need to have the co-ordinate numbers along with time in brackets in UT & Julian date.