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Telescope mount

2015-08-30T19:44:16Z

Evilscientist: /* Equatorial= */

Very small telescopes, including binoculars, can be hand held. However even very small telescopes and binoculars will show a lot of shake if hand held. Because of this telescopes are often mounted in some fashion in order to steady the image and often to allow the object to be tracked, compensating for the Earth's rotation.

==Types of Mounts==
There are many different types of telescope mounts, though these types do tend to fall into a few broad categories. Fixed, transit, alt-azimuth, equatorial with some other rarer types.

===Fixed===
A telescope on a [[fixed mount]] is immobile and is permanently pointing in one direction (usually up at the [[zenith]]). These telescopes can be made quite large but can only observe a limited part of the sky.

===Transit===
A [[transit mount|transit telescope]] is fixed in [[azimuth]] but not in [[altitude]]. Usually these telescopes are set up to move along the [[meridian]] for [[astrometry|astrometric]] measurements.

===Alt-azimuth===
The [[alt-azimuth mount]] allows the telescope the freedom to move in two axes and thus can look at any part of the sky. With proper computer control these mounts can also track an object.

===Equatorial===
The [[equatorial mount]] has it's rotational axis parallel to the rotational axis of the Earth. Thus equatorial mounts can compensate for the Earth's rotation by simply rotating about this parallel (polar) axis at the rate of the Earth's rotation ([[sidereal rate]]).

===Other types===
Other types of telescope mount include the [[altitude-altitude mount|alt-alt mount]] where both telescope axes are a change in altitude.

[[Category:Telescopes]] [[Category:Observing]]

Flat field

2015-08-22T15:43:33Z

Evilscientist:

In [[CCD]] [[astrophotography|astroimaging]] a flat field (sometimes called a flat image) is an image taken in order to compensate for variances in light sensitivity across the CCD chip. If a CCD sensor is illuminated by a perfectly flat light source, that is that every part of the CCD surface is illuminated by exactly the same amount of light, the number of photons counted by each cell in the CCD will be different. In pretty picture imaging this can cause variations in brightness across the image. In scientific imaging this can introduce significant instrumentation error<ref name="birney1">Birney, D.S., Gonzalez, G., Oesper, D., 2008, Observational Astronomy 2nd ed., Cambridge University Press, Cambridge, p173</ref><ref name="howell1">Howell, S.B., 2010, Handbook of CCD Astronomy, 2nd ed. Cambridge University Press, Cambridge, p67</ref>.
[[Image:Flat_field.jpg|right|thumb|200px|A typical flat field image]]
To compensate for this astronomers take a flat field image. This image is taken by pointing the optical system at a perfectly flat light source and allowing the CCD to be exposed for a period of time. This creates an image that shows not only the variance across the CCD but also any variance caused by the optical system such as dust or vignetting<ref name="birney1"/><ref name="howell1"/>. This image is then saved and used to correct any subsequent application images by dividing the application image by the flat field image<ref name="howell2">Howell, S.B., 2010, Handbook of CCD Astronomy, 2nd ed. Cambridge University Press, Cambridge, p82</ref>.
==Types of Flat Field Images==
There are several methods of creating a flat field image, each one has advantages and disadvantages and astronomers will often use a combination of methods in order to create a good flat field.

===Dome Flats===
As the name suggests, a dome flat is taken in the observatory (which is often covered with a dome). With this type of flat, a screen that is slightly larger than the aperture of the telescope is mounted somewhere in the observatory (usually on the dome) where it is evenly illuminated by artificial lights. The telescope is pointed at the screen and the flat image taken <ref name="birney1"/>.
The advantage of a dome flat is that it can be taken at any time, even during the day. This means that valuable on-sky time isn't used for non-target imaging. The primary disadvantage of the dome flat is that it is often itself not flat due to variances in the lighting used and variances in the screen itself<ref name="birney1"/>. Also without special thought to the paint used on the screen, the screen may also not be truly white and thus not flat in certain filters<ref name="massey">Massey, P. and Jacoby, G.H., 1992, ASPC 23, p240</ref>.

===Twilight Flats===
There is a point in the sky about 20° from the zenith opposite the Sun right at sunset that is basically flat<ref name="birney1"/>. Since this point in the day is called twilight, this makes a twilight flat. The advantage to this type of flat is that it is readily available as all one has to do is wait till the appropriate time, point the telescope at the appropriate point in the sky and take the flat image. The disadvantage of this technique is that the point in time where this point in the sky is flat is limited to periods where the Sun is between 90° and 100° from the zenith<ref name="chromey">Chromey, F.R. and Hasselbacher, D.A., 1996, PASP 108, p 994</ref>. This translates into a period of time of about 40 minutes each night and it is possible to miss this window while setting up for an evening's observing run.
===Sky Flats===
It is also possible to use the night sky itself as a flat field. The difficulty here is that there are things visible in the night sky such as stars that get in the way of making a sky flat. The way a sky flat is made is a point in the sky with few stars is selected. The telescope is pointed at this area, the tracking turned off (so the stars drift) and the flat image exposed. This is done many times. The resulting images are combined using a mode statistic which removes the star trails<ref name="birney1"/>.
The advantage of this type of flat is that the colour is always correct. The night sky is always the colour of the night sky and thus in all filters will show the correct colour. The disadvantages are that even a mode statistic may not remove all the stars, making the image non-flat and that taking a series of sky flats takes time away from on-target imaging.
===References===
<references/>

[[Category:Observing Concepts]]

Celestial sphere

2015-08-12T08:08:23Z

Evilscientist: Add category

[[Image:Celestial_sphere.png|right|thumb|200px|The Celestial Sphere]]
The celestial sphere is an imaginary sphere around the [[Earth]] in which the objects in the sky appear fixed to. If you look at the night sky, it does look somewhat like a bowl or half sphere over the ground. This appearance gave ancient astronomers the idea that a sphere surrounded the Earth and that the [[star|stars]] were fixed to this sphere, which then rotated around the Earth. The [[Sun]], [[Moon]], and [[planet|planets]] were then thought to move against the celestial sphere around the Earth in their own orbits.

Now in modern times we know that there isn't a sphere around the Earth in which the stars are fixed. The stars are all at different distances from the Earth and the ones we can see with the unaided eye are all in our Galaxy and don't rotate around the Earth, but orbit the centre of our Galaxy. That being said it is often useful to think of the sky as a sphere around the Earth for the convenience of mapping and observing.

==Parts of the Celestial Sphere==

There are parts of the celestial sphere that are useful to know when talking about where things are in the night sky or defining the various coordinate systems used to locate objects in the sky.

=== Celestial Equator and Celestial Poles ===
[[Image:Pole_equator.png|right|thumb|200px|The Celestial Equator and Poles]]
If you project the Earth's equator onto the night sky you produce an imaginary line across the sky known as the celestial equator. This means that if you were standing on the Earth's equator looking due east, the celestial equator would start at the horizon, go straight up overhead and then down due west directly behind you. The height above the ground the celestial equator appears depends on how close you are (in [[latitude]]) to the Earth's equator. As with the terrestrial equator on the ground, the celestial equator divides the sky into northern and southern hemispheres.

If you project the Earth's geographic/terrestrial [[Earth's poles|pole]] out onto the celestial sphere, you would then create the north and south celestial poles. Thus if you were to stand on one of the Earth's poles the and looked straight up you would be looking in the direction of one of the celestial poles (north if you were standing on the Earth's north pole, south if you were standing on the south pole). As with the celestial equator the height of the celestial pole depends on your latitude. In fact the height above the northern horizon of the north celestial pole is your latitude (change to the southern pole and horizon for south of the terrestrial equator).

[[Category:Astronomical concept]]

Albedo

2015-08-01T20:44:26Z

Evilscientist:

The albedo of an object is the amount of incident radiation that is reflected back into space by that object. It is the total amount of radiation reflected (<math>F_r</math>) into space by an object divided by the total amount of radiation that hits the object (<math>F_i</math>).

<math>A=\frac{F_r}{F_i}</math>

The more light reflected, the higher the albedo. An object with a higher albedo will appear brighter than an object with low albedo, all other things (size, incident radiation, distance to the object) being equal. This is because the higher albedo object is reflecting more of the light back into space for us to see. A perfect white reflector would have an albedo of 1. A perfect black absorber would have an albedo of 0.

==Albedo of some common objects==
{| border="1"
!Object!!Albedo<ref name="COSMOS_albedo"> http://astronomy.swin.edu.au/cosmos/A/Albedo retrieved on 01 Aug 2015</ref>
|-
|Earth||0.30
|-
|Moon||0.12
|-
|Venus||0.75
|-
|Jupiter||0.34
|}

==References==
<references/>

[[Category:Astronomical concept]]

Mass determination

2015-07-27T05:56:06Z

Evilscientist:

Determining the mass of an object in space is can be done one of two ways. One way if the object is being orbited by another body that is much less massive than the primary, such as a planet orbiting a star. The other way is used when two objects with relatively close masses are orbiting each other. Using these two variations on a theme, we have been able to measure the mass of our Sun, the other planets in our solar system, other stars and even whole galaxies.

Both cases use a form of Kepler's third law of planetary motion, where:

<math>p^{2}=a^{3}</math>

and p is the period of the planet in years and a is the semi-major axis of the planet in astronomical units.

==Mass determination with one object much more massive than the other==

If the object is orbited by a much smaller object, such as a planet about a star, moon or spacecraft about the planet then we can use Newton's form of Kepler's third law.

<math>p^{2}=\frac{4\pi^{2}}{G(m_{1}+m_{2})}a^{3}</math>

With <math>m_1</math> being the mass of one body and <math>m_2</math> being the mass of the other. If one body, <math>m_2</math> is much less massive than the other, it can be safely ignored, dropping it out of the equation thus:

<math>p^{2}=\frac{4\pi^{2}}{Gm_{1}}a^{3}</math>

At this point we can solve for <math>m_1</math> giving us:

<math>m=\frac{4\pi^{2}a^{3}}{Gp^{2}}</math>

Since G is usually given in MKS units as <math>6.67\times10^{-11}N{m^{2}}{kg^{-2}}</math> the semi-major axis (a) and period (p) also need to be in MKS units, metres and seconds respectively.

So an example: Jupiter's moon Io takes 1.769 days to make one orbit. It's semi-major axis is measured to be 421800km. From this we can work out Jupiter's mass. Converting days to seconds gives us a period of <math>1.528\times10^{5}</math>s and a semi-major axis of <math>4.218\times10^{8}</math>m. Substituting in to our equation above gives us a mass of about <math>1.9\times10^{27}</math>kg or about 318 times the mass of the Earth.

==Mass determination with both objects are close in mass==

For objects that are close in mass we need to do something else since the two objects will orbit about a common [[barycentre]]. As it turns out the distance from the centres of the bodies is related to the mass of the two bodies thus:

[[Image:Barycentre_mass.png‎|barycentre]]

<math>\frac{m_{1}}{m_{2}}=\frac{r_{2}}{r_{1}}</math>

So we can, from the ratio and Kepler's Third Law work out the mass of stars that orbit each other based on these two relations:

<math>m_{1}+m_{2}=\frac{a^{3}}{p^{2}}</math>

<math>m_{2}=\frac{(m_{1}+m_{2})}{1+\frac{r_{2}}{r_{1}}}</math>

Where a is the separation distance in astronomical units and p is their orbital period in years. This yields masses in solar mass units (Sun's mass = 1).

So if we observe two stars orbiting each other with a separation of 100 AU and a period of 90 years and we work out that the barycenter is 25 AU from one of the stars we can do the following:

The total mass of the system from <math>m_{1}+m_{2}=\frac{a^{3}}{p^{2}}</math> is:

<math>m_{1}+m_{2}=\frac{(100AU)^{3}}{(90yr)^{2}}=123</math> solar masses.

Using <math>m_{2}=\frac{(m_{1}+m_{2})}{1+\frac{r_{2}}{r_{1}}}</math> nets us the mass of one of the stars:

<math>m_{2}=\frac{123 solar masses}{1+\frac{(100AU-25AU)}{25AU}}=30.75</math> solar masses.

Since we know the total mass of the system we can do some subtraction and find the other star's mass of 92.25 solar masses.

[[Category:Astronomical concept]]