# Tag Info

15

A system projected into the sky Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction. First, ...

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The equatorial coordinate system is very similar to the system used on a globe or on maps. To specify a point on a sphere or a globe you need only two numbers. These are the longitudes and latitudes. On a globe of the earth you have longitudes from -180º to +180º and latitudes from -90º to +90º. The south pole lies at -90º latitude, the north pole at +90º ...

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The easiest way to think of equatorial coordinates is by extending lines of latitude/longitude (the geographic coordinate system we usually learn in school) out into space. The earth's equator becomes the celestial equator, and the north and south poles become the celestial north and south poles, respectively. By doing this you have a fairly simple way to ...

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The answers to this question are very clear and much better than I could have written. I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your ...

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The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees. This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on ...

2

The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time. (Alternatively, the local sidereal time is RA of the observer's meridian.)

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Imagine you are patching a plane with circles, instead of patching the sky sphere with circles (which is the same as filling the space with cones). Now, for every circle of diameter D define a square of diagonal D (D=2d). It is easier to patch the plane with squares, isn't it? So you have squares of side D/2*sqrt(2), and that is exactly your DEC_step, for ...

2

If you are calculating the angular distance between planets, as viewed from Earth, then of course you need to include the retrograde motion. The natural way of doing this would be to first use heliocentric coordinates to calculate the positions of both planets and the Earth relative to the sun (You could do this using Kepler's laws). Then change your ...

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In our moving Earth-based point of view, retrograde apparent motion means a planet's geocentric ecliptic longitude is temporarily decreasing instead of increasing as usual. The word "apparent" avoids confusion with a retrograde orbit. All major planets' orbits are prograde, with heliocentric longitude increasing at all times1. Most retrograde-orbiting Solar ...

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Right ascension at Zenith should always be your local sidereal time.

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In the Southern hemisphere, the maximum declination you can see is 90-L where L is your latitude. The minimum is -90, since you can see the south celestial pole. As the earth rotates and revolves, you can see any right ascension within those declinations. The only exception is that you ordinarily can't see stars that are too close to the Sun. However, ...

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If you mean solar time midnight than the answer is simple (as commented by @rgettman). The Sun should then be at the opposite side of the sky at $\alpha = 13^{\textrm{h}}$, where $\alpha$ is the right ascension. There is a problem, however, when you use local time (using a timezone). If, for instance, you mean midnight GMT in Liverpool, which is at 3° west ...

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Use Stellarium. You can set your location and date as needed and simulate all the conditions (ignoring weather...) :) http://www.stellarium.org/

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