telescope - Alt-Az or Polar?

With the wedge in place, you will only do polar alignments. If you want to go back to az/el, you must remove the wedge.

For me polar alignments are much faster and easier than az/el alignments.
For polar:

1. level the mount


2. Set your latitude on the wedge


3. Let the auto alignment point the scope at dec 90 and manually spin the mount until you align on Polaris. You use the manual controls on the wedge for this, not the hand controller.


4. Do a one star alignment on an overhead star. I always use Vega in the summertime. You use the hand controller for this.

If you haven't moved the mount since last time (or if it is permanently mounted) all you have to do is power up and do a one star alignment. Easy!

earth - Where exactly does the Moon flip, given that it appears the other way up in the other hemisphere?

This doesn't really answer your question, but the orientation of the moon is changing constantly, because we tend to compare the moon to the horizon, and think of the horizon as fixed. Here's a video demonstrating the effect (simulated) for the full moon over London on August 28-29 (all times GMT):

https://www.youtube.com/watch?v=QXVpRfaJXmY

If you were changing location, you would see a similar gradual effect.

The only real place you would see a true "flip" is near where the moon is at zenith (overhead), because, as you passed the zenith point, the moon may go from being 89.999 degrees high in the western sky to 89.999 degrees high in the eastern sky (and flip directions because you're now comparing it to a different horizon).

Realistically, when the moon is that high, you probably wouldn't be comparing it to the nearest horizon.

black hole - Why do X-ray binaries such as GRS-1915 have active and quiet states?

Such objects are surrounded by an accretion disk, that is a disk of gas, the gas being accreted onto the central object. However, at this stage of its life, the accretion disk around the black hole of a binary system like GRS-1915+105 is strongly depleted. Therefore, the accretion rate is not sufficient to support a coutinuous burst. Matter fills the outer disk until a critical density is reached, that triggered an outburst. Mass from the companion star is lost by a Roche lobe overflow mechanism.

References:

Does naked eye miss a lot between local stars and distant milkiness?

According to Wikipedia,

Theoretically, in a typical dark sky, the dark adapted human eye would see the about 5,600 stars brighter than +6m while in perfect dark sky conditions about 45,000 stars brighter than +8m might be visible.

(By the way, "+6m" means stars of apparent magnitude 6 or brighter.)

Under the best conditions, a person could see 45,000 stars. That's quite a lot. However, the Milky Way alone has over 200 billion stars (and that's a low estimate!). So we can only see a small fraction of the total stars in the Milky Way. Earth-based optical telescopes can do a lot better, though.

What are the units of distance in this subtended angle calculation?

The units of radians are 'dropped' because unlike most units, they are dimensionless. Recall that the definition of radian angle measure is the ratio of the length of a circular arc to its radius. Thus, radian angle measures have units of $[mathrm{Length}]/[mathrm{Length}] equiv 1$, i.e. dimensionless.

Radians are units in the sense that they give information about what standard the angle quantity is measured by. Thus, the convention of writing $mathrm{rad}$ is useful in that it distinguishes it from other ways of measuring angles (e.g., in your question, arc-seconds), but in terms of dimensional analysis, $mathrm{rad} equiv 1$, so $mathrm{m}/mathrm{rad} = mathrm{m}$.

amateur observing - Massive Nearby Stars

Rob Jeffries nailed it, but I'll add a few points.

25-30 stellar mass stars are quite rare. O-type stars begin at a mass of about 16 suns and they're about 0.00003% of the main sequence (that's 1 in 3 million). They're also short lived, a few million to maybe 10 million years in main sequence. That's part of the reason there are so few of them.

If we look at what stars are near us - lets just go 20 light years, there's about 150 stars and star like objects - source: http://www.solstation.com/stars/s20ly.htm

Each time you double the distance, you have 8 times the volume, so within 100 light years (and this is a very rough estimate), 5 to the 3rd power * 150, roughly 20,000 stars within 100 light years of earth. So if a star of the size you're looking for is 1 in 3 million (or less common than that at 25-30 solar masses), and there's 20,000 stars within 100 light-years, it's not surprising that there's none that close to earth. That doesn't mean there's never been one that close, cause star orbits in the Milky way move around in relation to each other quite a bit over time, but mostly stars of that size don't get that close cause they're very rare.

Loosely related footnote, but SN 1054, which is now the Crab Nebula was so bright when it went super nova that it could be seen during the day (for about 23 days) and that's 6,500 light years away. We can see stars that large well over 100 light years. Betelgeuse is 640 light years away and it's the 8th brightest star in the sky. They don't have to be that close to be impressive.

light - What is the average color of the universe?

According to astronomers at Johns Hopkins University, it's beige. Note, they previously announced it was turquoise, but they identified a bug in their calculations, redid the numbers, and now it was determined the average color is beige. Kind of boring, yes, but kind of what you'd expect.

In short, they collected "detailed light measurements from more than 200,000 galaxies. They then constructed a "cosmic spectrum," which represents all the energy in the local universe emitted at different optical wavelengths of light" and averaged it. More details at link below.

NPR : The Color of the Universe Is...

solar system - Why do the rings around the Gas Giants get their 'ring' shape?

There are different theories as to how planetary rings form, and so there are different answers to your question.

One theory of planetary ring formation is that a small (or large, depending on the ring size) moon wandered into the gas giant's Roche limit (the radius from the celestial body inside which no second celestial body can stay intact due to tidal disruptions by the larger body). The moon would have been orbiting the planet, and would thus have been moving in a flat plane. The resulting debris would have continued to move in a plane, because there would be no forces inducing them to move "up" or "down".

Another theory is that the debris is left over from the early solar system. A protoplanetary disk formed around the Sun; it contained all the material that makes up the planets. Over time (and through a lot of collisions!), bits of the material slammed together and formed planets. The trouble is, not all of that material became planets. Some became asteroids, some became comets, and some became moons (although the theory of the formation of the Earth's moon is a bit different). Some of this material could also have coalesced around planets. Now, the protoplanetary disk would already have been in a disk shape. Why? Think of a rotating object - like a planet. You might have read that the Earth is not a sphere. Aside from the obvious deviations of mountains and valleys, some of the deformation comes from the Earth's rotation. Any rotating body will flatten a little around its equator; a spherical body will become an oblate spheroid. The same thing happened for the protoplanetary disk - it became a disk because it was rotating! Any material that coalesced around planets would already be in a plane - that's why the planets' orbits aren't at odd angles to each other. So this material would be in a flat shape if it was captured by the planets, and would become a circular ring because of rotation. And if, at that point, there was material that wasn't in a flat plane, it would become flat by the same process that made the protoplanetary disk flat.

I hope this helps. That was a cool question.

Edit

There are stellar systems where the protoplanetary disk isn't in the same plane as the star's rotation. This could affect planetary orbits; however, such systems have not been thoroughly studied because of their rarity.

Second Edit

Just realized something. The gas giants are the most massive objects in the solar system, besides the Sun. Is it a coincidence that they all have rings? Probably not. I can think of two answers to this, which I suppose relate to your question: 1) the gas giants have rings because they were massive enough to attract moons, and thus more likely to have a moon enter into their Roche lobes, or 2) they gathered up more material (here's the "material from the early solar system" hypothesis) in the early solar system because they were so massive. The other planets might not have had enough gravity to attract this material.

Third Edit

Wow, these edits are out of control. I'll end them sometime. Anyway, you could probably test the moon-in-Roche-lobe theory and the early-solar-system-debris theory by analyzing the ring content and looking for similarities among the material in the ring structure.

Gravitational tidal force problem - Astronomy

if we have spherical mass held by its self-gravity, there will be a limit to distance - let's say it is $r$ - that the center of this body may approach a much more massive body (lets we call $M$ body).

Our spherical mass here would be held - as I understand - by gravitational tidal force. But the question is what is the distance $r$ between to masses his effect will hold?

$$r=left(frac{2M}{m} right)^{1/3} times R$$

$m$ - Spherical body mass

$R$ - Spherical body radius

$M$ - Massive body

I guess this is the equation for this limit, but I don't understand the conditions of this limitation.

observation - How can the orbit of a Kuiper Belt Object be differentiated from the transit of a rogue planet?

I'm not speaking from an informed position here, but two things come to mind.

First is that a rogue planet is likely to be traveling very fast, relative to our solar system. An object from within our solar system (a Kuiper/Oort object) is going to have an orbital velocity. Something that isn't a part of the system at all could be traveling much, much faster.

Secondly, if a rogue planet somehow slowly drifted into our system at almost no speed, and then got pulled into a sort of orbital speed by the sun, the direction of its orbit might give it away. For example, it's orbital inclination could be way off--even perpendicular to the normal orbital plane. It also might be orbiting in the wrong direction--that is--clockwise, instead of ante-clockwise.

That being said, there's no reason that some disturbed Kuiper belt object might not have an eccentric orbital inclination or even an anti-clockwise orbit. But the larger the object, the less likely this is, because it would need a larger and larger disturbance to throw it so out of whack.

orbit - How can I predict space directions?

If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in the 2D view), it's actually well above or well below Neptune's orbit. Pluto and Neptune never come closer than about 17AU.

At least for now. Minor perturbations over tens of millions of years may change that. Or might not. Pluto and Neptune currently have a 3:2 resonance, and such resonances tend to be stable.

You might want to look at http://en.wikipedia.org/wiki/Pluto, especially the section "Relationship with Neptune".

celestial mechanics - Roll, Pitch and Yaw of Orbital Planes

You're using the wrong terms. Engineers use yaw, pitch, and roll to describe the orientation of a vehicle. Some erroneously call these rotations Euler angles. Astronomers and physicists use true Euler angles, a rotation about the Z axis of some reference plane, followed by a second rotation about the once-rotated X axis, followed by a third rotation about the twice-rotated Z axis. Note that the Tait-Bryan angles (aka Cardano angles) use a sequence of rotations about three distinct axes. Euler angles use only two axes.

The first rotation is the planet's axial precession angle. The second rotation is the planet's axial tilt, or obliquity. The third rotation represents the planet's daily rotation. The rates at which the precession and obliquity change are much smaller than the quickly-changing third angle.

solar system - Does the Moon Have Enough Water for Robert Zubrin's Mars settlement Plan To Work?

Self-sufficiency is an incredibly broad term. We could argue that yes, there is water on the Moon, and that yes, there are viable ways to produce required electricity in self-sustainable ways, but the real question is, are there areas on the Moon that would be viable for both at the same time.

You see, the most likely place where surface or near subsurface water could exist on the Moon and be suitable for mass extraction are its polar, permanently dark regions. Indeed, the ISRO (Indian Space Research Organisation) Chandrayaan-1 spacecraft has detected evidence for water locked in surface lunar regolith minerals in lunar south polar region, water that likely originates from asteroid and comet impacts embedding it deep within the lunar core and released as magmatic water closer to the surface. Any free-form water in other regions of the Moon that are exposed to sunlight and Solar radiation would sublimate to its gas form directly and with ionisation lose hydrogen atoms, so while hydrogen and oxygen atoms might still be present to some extent embedded into the surface layer minerals, extraction would likely be too elaborate there.

But, wherever you'd find your water source, you would still require a great deal of electricity to power your extraction plant, later use electrolysis to separate molecular water into its constituent atoms, and compress it in cryogenic conditions to their diatomic liquids that are suitable as propellant components, diatomic liquid oxygen (or LOX) as your oxidizer, and double as much in molecular quantity of diatomic liquid hydrogen (or LH2) as your rocket fuel. Problem with electricity is, unless you brought your own and a great deal of it on you to power your plants, you will likely want to use be it solar power, or tap into the in lunar regolith embedded helium-3 (or ³He) and power your third-generation Helium-3 fusion reactor. See for example this answer of mine on Space Exploration on how that could be done.

So the main conundrum to exploiting Lunar resources, for the time being, remains finding sufficient and viably mineable resources of water where there is also self-sustainable ways of generating required electricity. One option that I can think of is staying on the most exposed to the Sun lunar equator and extracting deuterium and tritium hydrogen isotopes, as well as helium-3 from lunar regolith, all of them embedded there from Coronal Mass Ejections (CME). Required oxygen could be produced by crushing oxidized minerals and letting them sweat with the presence of hydrogen isotopes into ionized water, and helium-3 could be used as previously mentioned to sustain a fusion reaction producing required electricity to later break water molecules into its constituent atoms of hydrogen and oxygen by electrolysis.

How much of these hydrogen and helium isotopes are actually embedded in the lunar regolith, and how long these deposits persist in it, possibly staying there for at least some time due to the static charge of the regolith as it is bombarded by the Solar radiation, this is however a whole different question and one we can't currently yet answer. The study of the Lunar exosphere and dust environment is the sole purpose of the LADEE (Lunar Atmosphere and Dust Environment Explorer), that we barely just launched there. We will know in roughly one year, if it will be able to provide conclusive scientific evidence for these theories I've just mentioned.

How do we know Milky Way is a 'barred' spiral galaxy?

There are several different lines of evidence which together form a coherent picture: that of a barred galaxy. Moreover, as most disc galaxies are barred, we should expect the same from the Milky Way. The various evidences are:

The observed light distribution (2MASS) shows a left-right assymmetry in brightness and the vertical height. This is explained by the near end of the bar being located on that side.

The observed gas velocities show velocities which are "forbidden" in an axisymmetric or near-axisymmetric (spiral arms only) galaxy. These velocities occur naturally from the orbits of gas in a barred potential

The velocity distribution of stars in the Solar neighbourhood shows some asymmetries and clumping which is most naturally explained by orbital resonance with the bar rotation.

The extent, pattern speed, and orientation of the bar is consistent between all three of these.

Calculate Distance To Stars - Astronomy

The currently accepted answer is not relevant for finding the distance to a star like Proxima Centauri.

Here's how parallax works. You measure the position of a star in a field of stars that are (presumably) much further way. You do this twice, separated by 6 months. You then calculate the angle that the star has moved against its background stars. This angle forms part of a large triangle, with a base that is equal to the diameter of the Earth's orbit around the Sun. Trigonometry then tells you what the distance is as a multiple of the distance from the Earth to the Sun. [In practice you perform many measurements with any separation in time and combine them all.]

The "parallax angle" is actually half this angular displacement, and a star is said to be 1 parsec away if the parallax angle is 1 second of arc. So 1pc is 1 AU/$tan (theta) = 3.08times10^{16}$ m. The larger the parallax, the closer the star.

The Gaia satellite is currently mapping the entire sky and will estimate tiny parallaxes
with precisions of $10^{-5}$ to $10^{-4}$ arcseconds (depending on target brightness) for about a billion stars.

Parallax - as illustrated at http://www.bbc.co.uk/schools/gcsebitesize/science/21c/earth_universe/earth_stars_galaxiesrev4.shtml

Now in reality, it is a bit more difficult than this because stars also have a "proper motion" across the sky due to their motion in our Galaxy relative to the Sun. This means you have to do more than two measurements to separate out this component of motion on the sky. In the case of Proxima Centauri the motion against the background stars due to proper motion is larger than the parallax. But the two components can clearly be seen and separated (see below). It is (half) the amplitude of the curved motion in the picture below that corresponds to the parallax. The proper motion is just the constant linear trend with respect to the background stars.

HST images of the path of Proxima Centauri against background stars. The green curve shows the measured and predicted path of the star against the background field over the next few years.

Parallax measurements work best for nearby stars, because the parallax angle is larger. For more distant stars or those without a parallax measurement, there are a battery of techniques. For isolated stars, the most common is to attempt to establish what type of star it is, either from its colour(s) or preferably from a spectrum that can reveal its temperature and gravity. From this one can estimate what the absolute luminosity of the object is and then from its observed brightness one can calculate the distance. This is known as a photometric parallax or spectroscopic parallax.

asteroids - What is the connection between moons, rings, and gaps?

It's mainly by the clear definition of the rings, and their mere existence.

Without a moon the rings would be short-lived, hence unlikely to be detected just in time shortly after they've formed. And they would tend to wash out to broader rings. The space between the rings seems to be empty. That's easier to explain by one or more moons keeping the gap clean, and keeping the rings on their track (shepherd moons).

A way of thinking of the shepherd moons:
Our Earth's moon slows down Earth's rotation by tidal forces and friction.
Think of a ring a little inside the orbit of a small moon as an analogon of Earth inside the orbit of Moon: It's slowed down by tidal forces and friction (non-elastic collisions between ring particles). Slowing down ring particles results in an orbit closer to the planet/asteroid. This way a small moon can act repulsive to a nearby ring, confining it.

More details about the dynamics of planetary rings, including confinement by shepherd moons, see this paper, p. 491 ff.

orbit - Semimajor axis variations in co-orbital moons

I've been playing with simulations of co-orbital bodies similar to Saturn's moons Janus & Epimetheus- horseshoe orbits where the two bodies are of comparable mass- and I'm seeing some very odd patterns that I can't explain.

What "should" happen is that the body with the slightly smaller orbit will eventually overtake the outer body; when their mutual interaction becomes significant compared to their interactions with the primary, the lower body is accelerated forward into a higher orbit, while the higher body is accelerated backward into a lower orbit, they switch positions around their average orbital radius, and then separate again.

What I'm seeing in simulation, however, is a little more complicated. When I simulate the Saturn+Janus+Epimetheus system, the moons' semimajor axis very slowly migrate farther away from their average up until they swap; right around the swap, the difference in semimajor axes is maximized, and they drift closer together for half a cycle and then farther apart again before the next swap. This creates "u" and "n" shapes on the plot of semimajor axis over time, with the plots for each moon in opposite phases.

For Janus and Epimetheus, the effect is very small, but increasing the masses of the moons amplifies it, to the point that the maximum separation in orbital radii that occurs during the moons' closest approach to each other is ten or more times larger than the "normal" separation during the middle of a cycle.

There is of course the possibility that this an artifact of errors in my simulation software, but I think that is unlikely given the following:

1. All other aspects of the Janus+Epimetheus system are reproduced perfectly- individual orbital periods are correct, average semimajor axes are correct, the frequency of the swaps is correct, and my simulation conserves energy to within less than 1 part in 1 trillion over a petasecond of simulation time (approx. 31,689 years).


2. The effect is perfectly regular, and simulated systems remain stable. If the effect was a result of simulation errors, I wouldn't expect them to behave so nicely.

My first thought was that perhaps the semimajor axis of each moon was being affected by the added gravity of the other moon when they are on opposite sides of the primary from each other, and when they approach more closely they won't be pulling each towards the primary as strongly. That, however, should result in a uniform expansion and contraction of both moons' orbits as they get respectively closer and farther away from each other. What I actually see is that the inner moon's orbit gets smaller while the outer moons orbit gets larger, and vice-versa.

So, is this likely to just be a simulation error after all (if it helps, I am using the 4th-order Hermite integration algorithm)? If not, what is the explanation for it? Has such an effect been documented before, and if so, where?

Vehicle liquids freeze on Moon

We often think of the freezing point as being the temperature where water freezes (273.15 K). There are several scales for measuring temperature, for simplicity lets use Kelvin here. If you have watched a pot of water boil away you are aware the water has 3 states, vapor (steam), Liquid (water) & solid (ice). The point where a solid turns to a liquid as often called the melting point.

According to Wikipedia temperatures on the moon range from 26 K to 390 K. If we look at a list of melting points we see that element like Hydrogen & Neon have melting points below 26 K but they also boil around 20 - 27 K, so they would not be a good liquid for using on the moon. As they would vaporize in all but the coldest lunar winter day.

The lunar equator has temperatures ranging from 100 k to 390 K. As we look at the list of melting points there are some that are liquid around 100 k but they all have fairly short liquid ranges Oxygen has one of the bigger ranges of about 45 degrees but that is between 54 K and 90 K. It seem our biggest risk then is not in finding an element that will flow during lunar cold spells, but rather one that will not boil during the warm spells. Even water which boils at 373.16 K can boil on the moon.

There are also considerations for pressure, as it can modify the melting & boiling points. In short there is no single element that will work as liquid on the moon, so any fluid would either need to be compound or solution (antifreeze) to keep from freezing, some heat would likely be required for the coldest periods, and pressures maintained to prevent boiling from extremes of temperature or exposure to vacuum.

How are the compositional components of exoplanet atmospheres differentiated?

How are exoplanetary atmosphere compositional spectra distinguished from those of the parent star(s), from the composition of the planetary surface or any other factor? Is it actually possible to determine the atmospheric composition precisely using this method?

Are there any specific examples of such an analysis having been performed on an exoplanet?

galaxy - How do we calculate the escape velocity of galaxies?

In short, the escape velocity from the region we find ourselves in the milky way is 544 Km/s +- 10%.

This number, and an extensive explanation of how it was calculated can be found on a paper called The RAVE Survey: Constraining the Local Galactic Escape Speed

Also important to consider is the fact that we already are at aproximately 220 Km/s because of the sun's orbital velocity, so the actual value needed to get there from earth is much lower.

Solar Eclipses for dummies: Step 1 - moon equatorial orbit around a planet without axial tilt

1. Every new moon would be a total solar eclipse.

Incorrect. Even if the Earth had no axial tilt and the Moon had an equatorial orbit, that doesn't mean that every new moon would be a total solar eclipse. Yes, the moon would always generate some kind of an eclipse at every new moon, but it's orbit shape hasn't changed -- it's still an ellipse. It's further at some points in its orbit and closer at others. When it's further away at new moon, it appears smaller than the sun in the sky, making an annular (ring) eclipse, not a total eclipse.

2. The path of totality would always cover the same area: a 250km wide corridor with its centre at the equator (so 125km over the northern hemisphere and 125km over the southern hemisphere).

For the same reason as (1), the path of totality would vary from a maximum amount down to nothing. For annular eclipses there would be a similar path of annularity instead. But whichever path it is would always be centered on the equator.

3. I think the path of totality wouldn't always cover the same area, e.g. always over Africa, so Q1: Is there a simple formula to predict which areas would be in the path of totality?

I'm sure that there are formulas that would predict which areas would be in the path of totality/annularity, but they would not be simple. They would depend on exactly when new moon occurs and the exact distance of the moon from the earth. That would control which strip along the equator would see the eclipse and how wide it would be. Because the moon slows down in its orbit when it's further away from the earth, and it speeds up in its orbit when it's closer to the earth, the time between new moons is not constant. The fact that Earth's own orbit around the sun is also elliptical, and it also slows down/speeds up when it is far/close from/to the Sun complicates this as well.

4. In real life, the moment of totality can vary from seconds to seven minutes. But in my study example, Q2: the period of totality should always be the same, right? Because the crossing path of moon and sun is always the same (while in real life the Moon can overlap the Sun at different angles and going in different directions) Q3: How can I calculate this period of time?

Again, because of the elliptical orbit of the Moon, the length of totality or annularity still can vary. Even during a single eclipse, the length of totality depends on your exact location on Earth. Your exact location determines your exact distance from the moon during the eclipse, which will vary depending on whether the Moon is getting closer/farther and how fast. There are formulas for calculating this, but I'm sure they are quite complicated even with the moon orbiting over the equator and with Earth having no axial tilt.

5. What about the latitudes for whom the solar eclipse would only be seen as partial? I understand that the nearest to the path of totality, the more covered the sun will be, and vice-versa. Q4: Is there a set value that says e.g. for each km, another degree is visible?

Generally, the closer you are to the equator, the more the sun would be covered. But this would vary slightly depending on the distance of the Moon from the Earth, and it wouldn't depend solely on your latitude. As Gerald points out, the moon's shadow, in which you would see a total solar eclipse, is cone-shaped. Different points on the Earth close to the equator would experience a different cross-section of the cone, which may be narrower or wider depending on your exact location. Where the eclipse is partial, the apparent size of the moon would control how much of the Sun is covered. A further-away moon would appear smaller in the sky and thus cover less of the Sun. Close to the borderline where the Moon appears to graze the Sun, the Moon may miss the Sun entirely if it's far enough and it appears small enough.

astrophotography - Why is there a black stripe in the Hubble images of Pluto?

That photograph is a composite of two images taken with different exposure times.

To be correct we'd have to say that the exposure of the two photographs is different, i.e. the outer photo was created by absorbing more light. In this case we can assume that the focal ratio (derived from Hubble's lens aperture) and the luminance of the scene (how much light is travelling in the lens' direction) are identical for both photographs, which leaves only the exposure time as a free variable when it comes to determining exposure.

This is necessary because we're photographing objects with very different brightness. For Pluto to show up a relatively short exposure time is required, but its moons reflect much less light and would need a longer exposure time to visible. As long as the sensor is exposed, Pluto would continue to increase in brightness to the point that it becomes washed out. Objects that are significantly brighter become over-exposed resulting in a loss of detail and fidelity, known as blown-out highlights in photography. In our case Pluto would turn into a solid white dot compared to the more detailed picture that is now possible. You can draw a parallel with false color images rendered from infrared: this composite is not what the human eye would see if it was capable of picking up this level of light and detail.

In another Hubble image, NASA explained the reason why composite imaging is used:

This is a composite image because a single exposure of the stellar background, comet Siding Spring, and Mars would be problematic. Mars actually is 10,000 times brighter than the comet, so it could not be properly exposed to show detail in the Red Planet. The comet and Mars also were moving with respect to each other and could not be imaged simultaneously in one exposure without one of the objects being motion blurred. Hubble had to be programmed to track on the comet and Mars separately in two different observations.

Source: Hubble Sees Comet Next to Mars

Very long exposure times are often necessary since relatively little light is coming our way from distant planets and stars. As the Hubble website explains for its Deep Fields images:

Hubble has made a series of very deep observations taken in very dark parts of the sky. Like using a long exposure on a digital camera, these long exposure shots (up to several weeks) reveal very faint details that are not normally visible in shorter exposures.

Source: "What are the Hubble Deep Fields?", Spacetelescope.org FAQ.

Wikipedia summarises a paper by Robert E. Williams and the HDF team, "The Hubble Deep Field: Observations, Data Reduction, and Galaxy Photometry" as follows:

Between December 18 and December 28, 1995—during which time Hubble orbited the Earth about 150 times—342 images of the target area in the chosen filters were taken. The total exposure times at each wavelength were 42.7 hours (300 nm), 33.5 hours (450 nm), 30.3 hours (606 nm) and 34.3 hours (814 nm), divided into 342 individual exposures to prevent significant damage to individual images by cosmic rays, which cause bright streaks to appear when they strike CCD detectors. A further 10 Hubble orbits were used to make short exposures of flanking fields to aid follow-up observations by other instruments.

Source: Hubble Deep Field, Wikipedia, retrieved 2014-12-09