# Tag Info

25

The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. Bovy & Tremaine (2012)). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 ...

24

The sun is within a few parsec (15-25 pc) of the galactic plane, slightly above. The thin disk of the milky way (containing ~85% of the stars and gas) has a density going roughly like $\rho_0 \exp(-|z|/300 pc)$, while the thick disk (older stars, a few percent) has a scale height of 1000 pc instead. So if we want to move up (galactic north) above 90% of the ...

16

The Galactic Coordinate system is a longitude-latitude coordinate system that is used to define the positions of objects in space, most commonly objects within our own galaxy. It uses the center of our galaxy as the focal point (i.e., where we consider $(0^{\circ},\:0^{\circ})$), much like we use the position just off Africa as the focal point of the Earth's ...

13

The article is referring to the black hole known as V4641 Sagittarii. It was originally thought to be the closest known black hole to Earth, at 1,600 light-years (0.5 kiloparsecs). It's a binary system which emits X-rays. Original distance measurements were made based on observations during and following an X-ray outburst in 1999. Analysis of the jets ...

8

Salem is at 42.5° north. At the equator, the position of the sun varies by 23.5 degrees on either side of the vertical over the course of each year, due to earth's 23.5° axial tilt with respect to the ecliptic. Saturn, however is not on the ecliptic. It departs by about 2.5° over the course of its 29.45 year, 9.5 AU orbit. So once every 29.5 years Saturn's ...

8

It is impossible to have the planets line up like in your image. This is done for illustration purposes only. If such an image was to scale, either the bodies would be extremely small and impossible to see, or the image would be extremely large and impractical to use. For example, the Sun is about 1,392,000 km in diameter. The Earth is about 12,756 km in ...

7

To add to the excellent answer by barrycarter, there are 2 planetarium-like codes, that I know of, that run on a mac and would make excellent tools for viewing certain astronomical objects. The codes are Stellarium and Celestia. Both turn your computer into your own planetarium where you can search and view objects in space.

7

I don't know whether this picture is a fake, but you could get this sort of image without cheating, if it were taken a long way away from Rio, and the magnification cranked up. In other words, get far enough away from Rio that its angular diameter is about 1 minute, and take the picture from there.

7

Assuming you had access to the relevant astronomical catalogues and data, then yes it would. Looking at constellations would of course be hopeless and most stellar catalogues only contain star positions for a relatively small volume in our Galaxy. However, there are classes of object inside and outside the Galaxy that would do the job. A displacement in ...

6

You use Spherical Trigonometry Given $A_1$ and $A_2$ are the respective azimuthal coordinates of the two objects, and $a_1$, $a_2$ their respective altitudes, the angular seperation $\theta$ is given by $$\cos \theta = \sin a_1 \sin a_2 + \cos a_1 \cos a_2 \cos (A_1-A_2)$$

6

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which ...

6

I'm assuming you're talking about physical distances (as opposed to any of the other distance measures in cosmology). The comoving distance to a galaxy at redshift $z$ is $$d_C(z) = \frac{c}{H_0}\int_0^z \frac{dz}{\sqrt{ \Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k(1+z)^2 + \Omega_\Lambda }},$$ ...

5

They do move - just far too slowly for you to detect by eye even over several human lifetimes. Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space Douglas Adams, "The Hitch-hikers Guide to the Galaxy" Even the ...

5

I like @DrChuck's answer and this Astronomy Picture of the Day shows how this has some plausibility: ...but you could get this sort of image without cheating, if it were taken a long way away... and the magnification cranked up. Screen shot from YouTube video Moon Setting Behind Teide Volcano From Astronomy Picture of the Day; 2018 June 4 Moon ...

5

Not much has changed. Each update is consistent with its predecessor. The original ICRF is defined by 212 extragalactic sources. The ICRF2 is defined by 295 sources, 97 of which are shared by ICRF1 The reason for the change was greater accuracy. The 295 sources in ICRF2 allow for locations accurate to less than 40 micro-arcseconds (µas) compared with 250 ...

5

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 ...

5

It turns out that 0, 0 on ecliptic coordinates is not the "winter" direction, nor the "summer" direction. And it turns out to be a good thing. Zero longitude is defined using the "spring" direction, and this means that converting these points to another coordinate system is really easy. The earth's axis is tipped 23 degrees, but imagine the axis on which ...

5

The pulsar map used by Pioneer and Voyager is a diagram of the position of the Sun relative to 14 galactic pulsars. It encodes their positions, distances, and pulse periods, which in theory makes it a decent tool for identifying where the Solar System is (although there are some issues with it). To make a similar map for an arbitrary point in the galaxy, you'...

5

You can equally well just type in a negative year and Stellarium will navigate you into the corresponding year B.C. Don't forget that the year 0 didn't exist. 1 A.D. follows directly after 1 B.C. A while ago I've confirmed that Stellarium is very accurate down to 1500 B.C. using a babylonian list of lunar occultations of stars. Thus, it should be reasonably ...

5

In the linked algorithm, the "zenith" is the position of the sun at sunrise. This is the number you need to change. If you want to calculate the time when the top of the sun reaches the horizon you use the "official" zenith of 90⁰50' This is intended to account for refraction and should estimate the time when the sun first appears over ...

5

Wikipedia has a nice planetary distribution picture, but it is in 1-dimension and based off aphelion (furthest position to the sun) and perihelion (closest position to the sun) only. Unfortunately, this picture doesn't take into account the planets' orbital inclinations to the Sun's equatorial plane. Nor does it take into account that the perihelion and ...

5

Jacob's staff is probably all you need. It is a pole with a cross arm. By positioning the arm at a certain point on the pole and sighting down the pole one can get a position on the sky. A skilled user can achieve sub-degree accuracy with such a tool. The position of the cross beam is easy enough to calculate by trigonometry, or by careful construction of ...

4

To add to the other answers, if you were in a place where Saturn could be directly above, make sure you get there just over an hour and a half early. Now; do you need to be there when Saturn is directly overhead or when the light from Saturn is directly overhead? The closest Saturn gets to earth is approximately 1,200,000,000 km (750,000,000 miles), ...

4

Heres more python than you can shake a telescope at. I just used @RobJeffries' algorithm. This is just a python script, the real answer to the question is @RobJeffries' answer and I've just scripted it. The mathematics behind generating statistically uniform distributions is explained very nicely there as well. Python is below the plots. You can see on an X-...

4

Just to be clear, the "height from sea level" (mentioned in the title) does not change what star is circumpolar. Whether the ground is at 0 units above sea level or 1500 units above sea level does not make a difference. I think what you intended to write is the height above the ground makes a difference ("angle of dip from the height" mentioned in the ...

4

There is no simple formula. First, it's not true that the angular speed of a planet relative to the sun can be found from its orbital period. This is because planets don't move in circles, but in eclipses and the speed of the planet changes, as described by Kepler's second Law. When describing the motion of a planet in the sky you need to consider the ...

4

As I do not want to reinvent the wheel, is there any star recognition framework that you've heard of? Preferably open source, the language is of marginal importance as long as I can get the relevant data in and out. I'm thinking of something that I can feed with an image file and get back an array with the recognized objects with their x/y ...

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