# Tag Info

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This is an integral, the number of stars of mass between $XM_{SUN}$ and $YM_{SUN}$ is : $N(XM_{SUN} \to YM_{SUN}) = \int_{XM_{SUN}}^{YM_{SUN}} \xi_0 (\frac{m}{M_{SUN}})^{-\alpha} \frac{dm}{M_{SUN}} = \int_X^Y \xi_0 (m')^{-\alpha} dm'$ So, it is clear that, with the additive properties of the integral, we have : $N(XM_{SUN} \to ZM_{SUN}) = N(XM_{SUN} \to ... 3 You would expect type Ia supernovas to be standard candles in neutrinos for the same reason they work that way in light. Alas, with the current detector packages they are only good in our local group (this would be considerably improved with the implementation of a GADZOOKS!-like scheme to gadolinium-doped superK or by several proposed but very expensive ... 3 Kepler's 3rd Law: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. If we take Jupiter's semimajor axis of 5.2 AU (avg value), then objects at 2.5 AU, 2.82 AU, 2.95 AU, 3.25 AU have orbital periods shorter than Jupiter by a factor of 3:1, 5:2, 7/3 etc. The fact that the ratios for the deepest ... 1 There seems a lot of conjecture in any event. Venus could have been a meteor, with an innate spin, that swung by the Sun and have been captured into our Solar systems anticlockwise orbital arrangement. Retaining her original spin momentum, clockwise relative to the others. 0 If I understand your question right, you want to know what is so special about the red and infrared bands / filters in studying the spatial distribution of galaxies. It is true that the red and IR bands are used inreasingly much in the mapping of galaxies at high$z$. This is because at these redshifts, we often use the Lyman-Alpha line to both detect and ... 0 They should be visible any time at night (note that longitude has little to do with observing meteors), but there are many things that could thwart your observing. Bright lights: In a big city you will see far fewer meteors. In a big enough city, you won't see any. On that note, if you keep checking a mobile device with a backlit screen, you probably won't ... 0 If we are assuming that we are restricted to observing them via light only, then we can use the angular resolution relation, $$\theta\approx1.22\frac{\lambda}{D}$$ where$\theta$is the angular resolution,$\lambda$the wavelength observed, and$D$the diameter of the aperture. Note that this only applies to optical and radio telescopes. In order to ... 0 The sum of two angles in different planes is known as a dogleg angle. A dogleg angle is not the same as an angle. The operation of adding two angles in different planes to get a dogleg angle is well-defined mathematically. The reasons astronomers use the the longitude of the perihelion instead of the argument of the perihelion are circular orbits and ... 0 To give an indication, comet ISON was seen first by two amateur astronomers when it was at magnitude 18.8, so a large body approaching from a distance might be found at about the same magnitude. Depending on albido, a moon size object could reach that magnitude at about twice the distance of pluto. The actual distance would also depend on how fast it was ... 2 Since Pluto is about the size of our moon, clearly we can detect an object of that size at least at the distance of Pluto. We can probably spot something like that some distance into the Kuiper belt with current technology. Of course the ability to detect something and actually detecting it are two different things. We have to be looking in the right ... 4 There is indeed an optimal angle where Venus is brightest. Have a look at the following figure: The distance between Earth and Sun is$\Delta$, between Sun and Venus is$r$, and between Earth and Venus is$\rho$. The amount of light that Venus receives from the Sun is $$f \sim \frac{\pi a^2}{r^2},$$ where$a$is the radius of Venus. When viewed from ... 4 The Earth receives approximately$6.8\text{mW/m}^2$of reflected sunlight from the moon (see below for details of how I calculated that). However, the sunlight is also absorbed by the moon and this raise the surface temperature. So the moon also emits thermal radiation towards the Earth (assuming the highest day time temperature of 400K, see comments below ... 2 Assume the moon is at roughly the same distance from the Sun as the Earth and so receives the same amount of solar energy / area. Find the area of the moon facing the Earth (hint, it's roughly the area of the moon's disk). Multiply by the reflectivity of the moon (about 12%). But that power is reflected from the moon in all directions, so you need to ... 1 Think about this : People that live on the equator have 12 hours of day and twelve hours of night all year ! If the earth's axis was perpendicular to the plane of its orbit (ecliptic) everyone would have twelve hour days and nights all year. But that ain't the model. The axis is ~ 23.5 degrees to the perpendicular. The plane containing the Earth axis and ... 3 There are two causes of this and other effects which are basically that noon as measured by the Sun, i.e. when it is due south, is usually not the same as noon by local clock time. The two phenomena are: the speeding and slowing of the Earth in its elliptical orbit round the sun, and the inclination of the Earth's axis to the plane of the ecliptic. If we ... 1 Both ecliptic and galactic coordinates are spherical coordinate systems that involve measuring angles on the celestial sphere. There are two equivalent ways to convert between such two coordinate systems: A transformation by deriving a general rotation matrix, for example using Euler angles; Finding an appropriate spherical triangle and calculating its ... 9 It is happening because of the acceleration of the Earth orbital speed around the Sun (Earth is near the perihelion). Between December 13 and December 31 the Earth is speeding up and also it is normally rotating around its axis. These 2 movements (constant rotation and increasing orbital speed) add up to create the observed apparent movement of the Sun on ... 0 Perhaps you're looking for something you can just punch into a spreadsheet instead of a generic matrix transformation? I've found it's easier to go from equatorial to other systems. So you can move it from ecliptic to equatorial (in degrees):$\alpha=tan^{-1}(\frac{sin(\lambda) *cos(\epsilon)-tan(\beta)*sin(\epsilon)}{cos(\lambda)})\$ ...

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Elliptic coordinates are a general coordinate system while galactic coordinates are a set of 2-dimensional spherical coordinates with a heliocentric origin. Thus the conversion between the two would just be the conversion between elliptical and 2-dimensional spherical (i.e. polar) coordinates. This is easy to derive by writing each system in terms of ...

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So if that is the case where does all the other colors (which are not in the absorption spectra) of the sunlight come from? It is important to realize that the radiation of atoms, molecules, and any pieces of matter, although can have pronounced lines in the spectrum, always contains some small amount of radiation at all frequencies. That is, the ...

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A stars emission comes from thermal radiation, which will approximately be equal to black body radiation. However part of this spectrum will be absorbed by the outer layers of this star. It is true that absorption lines and emission lies coincide, so these frequencies will be re-emitted. But they are scattered in all directions, so also back into the star. ...

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It's all because of the wavelength of light. In most bands the radio telescope is about the size of the wavelength it's observing - so it can only see a single point at once anyway. It would be like having an optical telescope that was a tiny microscopic pinhole - there wouldn't be much point in having a megapixel camera behind it. Radio telescopes that ...

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The Iridium satellites can be easily seen naked eyes. They also have flares that permit to see them even easier. This is when the flares will happen and daily predictions for all brighter satellites.

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I think it is possible if you look at Ganymede it is bigger than Mercury but however if you know about other planetary systems there are gas giants much bigger than Jupiter. It is very possible for a gas giant to have a moon similar in size of earth however not in terms of mass. We can speculate that if the gas giant is a brown dwarf than it may be possible ...

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I got the Starblast 4.5 with the Dob mount. Mine came with 17mm and 6mm Explorer II lenses. I upgraded the 6mm lens to an Orion Long Eye Relief 6mm eyepiece. It made a big difference in eye relief and field of view. Jupiter and it's moons look good and I can almost make out 2 bands using a Meade #8 light yellow filter at 75x. I thought about using a ...

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