8

What can you tell from a picture of some stars? At the very least, you need a recognisable asterism If the exposure is too low or the light pollution too high to identify unambiguously an asterism of three or more stars, you're simply not going to be able to tell much from the photo. An asterism on its own can only tell you where the picture wasn't shot. ...


8

In other phases, the angular distance of Sun from Moon is equal to the phase of the moon, in the direction of the lit side. One of the methods could be to estimate the position of Sun based on this, and then you can, assuming time is known, estimate North. If you don't know time, use constellations (which can form a separate question) to estimate North and ...


6

Here are some answers to your various comments and questions: you can estimate a direction of south True, but it would be a very rough approximation at best. In order for it to be accurate, the bright limb of the Moon would need to face either due east or west on the celestial sphere. (In astronomy terminology, the position angle of the bright limb would ...


6

There are many indeed. It may be specified from South westwards(0° to 360°), or from North eastwards and westwards(180° to -180°). The most popular one, as used by Roy and Clarke in "Astronomy : Principles and Practices" is to measure the azimuth from North eastwards 0° to 360°. For a southern hemisphere observer, this system changes to from South ...


6

Stars and gases at a wide range of distances from the Galactic center orbit at approximately 220 kilometers per second. From Wikipedia That's much faster than the Voyager probes relative to the Sun (Voyager 1: 17 km/s). Hence the probes will orbit the galactic center roughly the same way as our Solar System, even after occasional hyperbolic encounters with ...


5

Azimuth is conventionally calculated as a degree value between 0 (inclusive) and 360 (exclusive). North is assigned 0 (also by convention), which puts East at 90, South at 180, and West at 270. Sometimes 0 is set to South intead of North. This is the traditional means of measuring solar azimuth, even though North is the most accepted convention. Positive ...


4

You need to define an agreed standard for orientation of axis and origin. What co-ordinate system you choose as a standard depends on your purpose. Here are some examples and links : Celestial coordinate systems. Typical reference points are defined using very distant objects whose position can be measured accurately ( typically quasars ).


4

"The instrument will be tested in Earth orbit for one year, ..." What would it then orbit? It will continue to orbit the Earth, at least for a while. It will eventually end up not orbiting the Earth because it will be placed in low Earth orbit, where orbits decay due to atmospheric drag. The project is a technology demonstrator, which means a ...


4

Comets can look like stars. Sometimes the outside of the coma and tail of the comet are too faint to see with the naked eye, and only the central part of the coma is bright enough to be seen. In this case the comet will appear like a (slightly fuzzy) star. Look at the Great Nebula in Orion. It appears star-like, perhaps a little fuzzy. Other clusters and ...


3

The DSAC to be launched in June 2019 is a Technology Demonstration Mission, in which they only test whether it works as well in space as it does on the ground. If it passes the tests, similar devices could simplify navigation for future missions elsewhere in the solar system. It is one of many payloads sharing space on an Orbital Test Bed satellite, which ...


3

Your formula is a crude approximation. It's ok if you just want a rough idea of the number of orbits the Moon makes over a long period of time, but it's not useful for calculating the daily motion of the Moon. The Moon's orbit is fairly eccentric (about 0.0549), so its daily speed varies quite a bit. The Babylonians discovered that the mean daily motion is ...


3

To expand a little on James's answer: The pole angle method doesn't care what time of year it is, the celestial pole isn't going anywhere. ;) However, if you're in the tropics, the altitude of the celestial pole is rather low, which can make accurate observation difficult. Of course, if you're doing this on another planet, (or even in Earth's southern ...


3

Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.


3

Navigating the solar system, you'd probably track your probe using Earth-based radio telescopes in the equatorial ICRF, though you might describe the overall trajectory in terms of the ecliptic. Navigating at sea, you might measure the Sun's position in horizontal coordinates at a time in GMT to determine your own geographic coordinates. Practical uses in ...


3

I've been trying to figure out how to calculate the latitude of my location with just the shadow of a stick cast by the sun. I am doing this under the assumption that I am stranded in some place with no almanac or any other data. (not even day or month). You really cannot find your latitude with only the sun and sticks at your disposal and no ...


2

Yes, it can be used, but it is not easy. For the latitude, you usually need to know the Alt of Polaris. You can derive that if you know the Alt of two of the three, but the (spherical geometry) formulas for that are not simple. For the longitude, you need Greenwich time and the RA of any star. Any star includes Vega, Altair and Deneb ;)


2

Usually when the direction to an object is given relative to a star, the directions are related to equatorial coordinates (the coordinates on the sky). They are not related to the compass or cardinal points. It is best to avoid directions relative to the horizon/compass/cardinal points since those depend on the time and location. In your example, 1 degree ...


2

Pulsar, or X-ray navigation is being actively tested right now. Currently an X-ray telescope mounted on the International Space Station completed a test where a small navigation computer was able to solve for it's own position and orbit in space relying only timing information from X-rays produced by a group of pulsars. I've written more about it in the ...


2

One's latitude is easy to establish. You just need to find the angle of elevation of the pole, the point about which the stars appear to revolve each night. A stick and a plumb line can do this. One's Longitude is nearly impossible to find unless you have an accurate clock. To find Longitude you need to measure the position of the stars at a known absolute ...


2

Picture was taken somewhere near 30°22'02.0"N 75°00'35.7"E Bathinda, Punjab, India. Answering your question about time - how it enters into the sight reduction: RA, all by itself, isn't useful. It's the angle between the plane of the meridian of a celestial location and the plane of the meridian of the first point of Aries. That angle is more or ...


2

I do not have Stellarium, so I cannot confirm what is causing it with 100% certainty. But by using JPL's Horizons website, I can confirm that the position on the Earth's surface changes the apparent Right Ascension (R.A.) and Declination (DEC) of Venus. The following output is for an observer at 80 degrees W longitude and the latitude listed. The time was ...


1

I wrote a test program to compute the deviation between the wristwatch estimation and the real sun position, based on a program by dabasler. The results quality depends on latitude, month and time of day. Between latitudes 45 south and 45 north, the results are barely usable (can be off by 45 degrees or more). Typically, you get the worst results around 09:...


1

The method is generally reliable between 9 am and 3 pm. The other consideration is when one is using the method in either of the tropics: Cancer or Capricorn. It is generally accepted that in the northern hemisphere, the sun is located in the southern part of the sky and the opposite in the southern hemisphere. This applies to latitudes greater than the ...


1

Suppose we could look at Earth from Venus at that time: Simulated image courtesy NASA/JPL-Caltech The central meridian is 101°56' W. If you observe Venus from that longitude, its hour angle is 0h 0m 0s whether your latitude is 89° N, 27° N, or 63° S; there is no east-west parallax. The 0° meridian is just beyond the right limb in this view. Relative to a ...


1

I remember answering a question on the origin of declination last year (Why is declination positive in the northern hemisphere?) which led me to discover this paper on the origin of celestial co-ordinate systems. In it they seem to indicate that the ancient Greeks were the first to use Right Ascension to describe locations in the sky around 300 B.C. Whereas ...


1

The best inertial frame is that which is comoving with the cosmic microwave background. There is a notable dipole in the microwave background, the result of the Solar system's motion relative to it. The invariance of this dipole provides evidence that the frame of the barycentre of the solar system is also a very good inertial frame. This is of no ...


Only top voted, non community-wiki answers of a minimum length are eligible