Thursday, May 28, 2009

Answers to the informal quiz, Part 1: Earth

Okay, one week has passed, and I've been getting lots of feedback from the informal quiz. Since there's quite a few questions, I'll post the answers in parts, starting with the question about Earth. In addition to the correct solutions, I'll also post the most common mistakes. Without further ado, then, here are the answers many have been eagerly awaiting...

  • Why does the Earth experience seasons?
By far the most common mistake here involves distance. Many people believe Earth is closer to the Sun in summer than in winter, thus making it hotter or colder, respectively.

This seems like a reasonable explanation, but breaks down when you consider that the Northern Hemisphere and Southern Hemisphere experience opposite seasons. When it's summer in the USA and Europe, it's winter in Australia, and vice versa. It runs into more trouble when you consider the Earth is actually closest to the Sun in January, and farthest in July...exactly the opposite of what you'd expect for the Northern Hemisphere if this explanation were true.

The real answer here involves the Earth's "axial tilt". The axis about which the Earth rotates is at a constant tilt of 23.5 degrees. This means at certain times of the year, the Northern Hemisphere is more directly facing the Sun, while 6 months later when the Earth is on the other side of the Sun, the Southern Hemisphere is more directly facing the Sun. Note the diagram:

I've marked the equator and the axis of rotation in red. The hemisphere which more directly faces the Sun is the one which experiences summer. The hemisphere angled away from the Sun is the one experiencing winter.

Another common "almost right, but not quite" answer I've heard several times involves the tilt, but in the wrong way. Those folks maintain that because the Earth is tilted, one hemisphere is closer to the Sun than the other. However, this effect is minuscule, since the tilt only accounts for a difference in distance to the Sun of a couple thousand kilometers, while the average distance to the Sun is 150 million kilometers. Again, seasons have nothing to do with distance, it's all about angles.

  • Why is there a 24 hour day-night cycle?
This one is relatively easy. The Earth rotates about its axis every 24 hours, bringing the Sun into view for half of that time period.

(Now, technically, the Earth makes a full 360 degree rotation in only 23 hours, 56 minutes...but because the Earth has also traveled about 1 degree in its orbit around the Sun during that time, the Earth must rotate an extra 1 degree to bring the Sun back to the same relative position. Thus, the 4 extra minutes.)

Most folks get this one, but there are still some who maintain that the Earth doesn't rotate at all, and just orbits once around the Sun per day. Woe is them.

  • Why is the sky blue?
In spite of being a question asked by every 5-year-old in existence, this is without doubt the most frequently incorrect answer.

I've gotten simply wrong responses such as "the sky is simply reflecting the ocean". In the middle of a continent far from any ocean, though, the sky is still blue. I've also gotten the "almost correct" response that the atmosphere is refracting the sunlight. It does involve the atmosphere interacting with sunlight, which is good...but still not quite right.

The correct answer here has to do with scattering, specifically - "Rayleigh scattering". When light has a wavelength close to the size of a gas molecule it's passing near, there's a good chance the light will essentially "bounce" off the molecule and start heading in a different direction.

Moreover, the chance of scattering is also very dependent on wavelength - it scales as 1 over the wavelength to the fourth power. In other words, short wavelengths are much more likely to scatter than long wavelengths. Our eyes interpret the different wavelengths of light as different colors. Red light's wavelength is roughly twice as long as blue light, so blue light is 2 to the 4th power = 16 times more likely to scatter than red light.

So, imagine incoming sunlight coming from the Sun and passing through our atmosphere. Remember, the sun emits every colors of the rainbow, it's just that when the colors are all combined, they appear to us as white light. Now, the red light is more likely to make it through unhindered, while the blue light gets scattered everywhere and appears to an observer on the ground to be originating from a direction other than the Sun. Hopefully this diagram will help explain the concept:

As you can see, most colors on the red side of the spectrum appear to be coming from the direction of the Sun. The blue light (and a bit of the green), however, appear to be coming from elsewhere in the sky. This process happens all over the sky, so it appears that blue light is coming from everywhere.

This also explains why our sun appears slightly more yellow than it would from space. Some of the blue gets scattered out of our line of sight towards the Sun.

  • Why are sunsets red?
The "blue sky" answer above also explains this question. At sunset (and sunrise), sunlight has to pass through far more atmosphere to make it to an observer on the ground, vastly increasing the chances of scattering even the not-so-blue light. Only the very reddest light makes it to the observer without getting scattered...thus, a red sunset.

  • How does the Earth compare to other planets?
So, this is meant as just an open-ended question...I only ask as there seem to be quite a few folks who think that Earth is the largest planet.

Within our own Solar System, planets can be divided into two groups - the inner, terrestrial rocky planets (Mercury, Venus, Earth, and Mars) and the outer gas giant planets (Jupiter, Saturn, Uranus, and Neptune). Earth is the largest of the inner rocky planets, but many times smaller than the outer gas giants. So, it could be said to fit somewhere in the middle, albeit slightly on the smaller side.

On the other hand, if you take the ~300 planets known around other stars, Earth is quite dwarfed. Just about all of these "extrasolar" planets are massive gas giants - many larger than Jupiter - and most of which are found orbiting incredibly close to their parent star. Within that population, Earth is no more than a puny wet rock.

Of course it's quite likely that this known collection of extrasolar planets are not representative of the population of planets as whole. The problem is that our observing techniques for detecting these distant worlds are severely biased towards only detecting very large planets which orbit very close to their parent star. The hope is that with improving technology, we'll start detecting Earth-like planets in a matter of a few years, particularly with space-based missions such as Kepler.

  • Why does a feather fall slower than a bowling ball?
The common mistaken belief here is to think that because the bowling ball is heavier, it falls faster.

It turns out that the only thing which slows down the feather is increased air resistance. In essence, it has a much lower terminal velocity than the bowling ball. (Terminal velocity is the fastest a given object of a given shape can fall - it's the speed at which the Earth's gravitational acceleration is perfectly balanced by the force exerted by air friction.)

Take away the air, and everything - no matter its weight - falls at the same rate. Check out this video from one of the Apollo moon landings - you'll see that in the absence of any air, a hammer and a feather fall at exactly the same rate.

Now, technically the bowling ball, with a larger mass, feels a stronger gravitational force. However, because it has a larger mass, it's inertia is also other words, you need more force to get it going. It turns out that its increased gravitational force perfectly balances its increased inertia, resulting in all things falling at the same rate (in a vacuum).

Thursday, May 21, 2009

An informal quiz...

Based solely on my experiences teaching undergrad students, I've found there's quite a few commonly held misconceptions about astronomy. See if you can answer these seemingly intuitive astronomy questions. (No looking 'em up on the interwebs! That's cheating!)

Answers will be posted in one week...if you really don't want to wait until then, email me and I'll send you the answer key.

  • Why does the Earth experience seasons?
  • Why is there a 24 hour day-night cycle?
  • Why is the sky blue?
  • Why are sunsets red?
  • How does the Earth compare to other planets?
  • Why does a feather fall slower than a bowling ball?
  • Why does the moon go through different phases?
  • Why is the moon bright?
  • Can you see the moon during the day?
  • Does the moon rotate?
  • Why does the moon appear larger on the horizon?
  • Is there gravity on the moon?
  • Compare the Sun to the stars.
  • Why does the Sun shine?
  • What happens to the Sun at night?
  • What is the sun made of?
  • What causes a solar eclipse?
Universe, etc:
  • What's the difference between the Solar System, the Galaxy, and the Universe?
  • What is a star?
  • How are planets different than stars?
  • Where do the stars go during the day?
  • What's the farthest human beings have ever traveled in space?

Saturday, May 16, 2009

Does the universe actually look like that?

Okay, I'll admit it, the last post had a lot of math. For those who are less inclined to slog through equations, let's talk about all those astronomical pretty pictures we see.

Karl asks:
I always hear that photos of cosmological objects (like that photo of the Ring Nebula in your previous post) have been enhanced in some way. What would those things look like if we just saw them exactly as the telescope or camera picked them up? And when the photos are enhanced, are they always enhanced the same way, or does the formula vary?

Finally, does the enhancing serve some scientific purpose, or is it done basically to make the pictures prettier? (I suppose that's a scientific purpose too, in the long view, since pretty pictures make it easier to get funding. Don't worry, I won't tell!)
So, it's totally true, the final images given out for press releases have usually been heavily processed from the original raw images. Just like the pictures you took at a party where you throw them into photoshop and remove your friends' red-eye, there are a set of "standard" processing steps for astronomical images.

That said, there's usually not outright deception the way advertisers airbrush images - no astronomer is going to try and make their planet look skinnier or add lolcat tags. To understand this a little better, let's talk about how modern astronomical images are actually taken.

First off, almost all optical images are taken with a Charged Coupled Device (CCD) mounted to the back of a telescope. This is the same kind of chip that's in the back of your ordinary digital camera, albeit more sensitive and more expensive. Essentially, it's just a thin piece of silicon divided into a narrowly-spaced grid of cells. Each cell in the grid can hold electrons which might get excited when a photon hits them. At the end of an exposure, each cell reports how many energetic electrons it contains. Our image just translates each cell into a pixel, and the brightness of that pixel is just how many electrons it contains.

Now, notice there's absolutely no color information here. The CCD just reports the number of excited electrons, and doesn't know anything about whether it was a red photon or a blue photon which excited this just produces a black & white photo. This means we have to use filters if we want to get any color information. If we put, say, a red filter on our CCD before taking the image, then we know only red photons can get through.

So, first we take an exposure with a red filter, then another with a green filter, and then another with a blue filter. We combine them all at the end to produce our fancy color image.

Okay, you're probably already asking, "then how does my digital camera takes color photos all at once without any color filters?" The answer is that it uses filters all the time - here's a schematic of the filter mosaic used in most digital camera CCDs. By filtering alternating pixels with different colors, in only one exposure the camera can get an image in each filter...albeit at lower resolution than the entire grid. The fancy camera software then interpolates these separate staggered color images to produce a single color image.

So with all this said, let's take a look at an actual single raw image of a galaxy:

You'll want to click on the image above to look at the original with all its glorious artifacts. Let's also take a look at close-up with some artifacts highlighted:

So, there's several issues we have to contend with to make this into a "pretty picture".

In red, I've highlighted a particularly annoying cosmic ray trail (though they're all over the image). Unlike digital camera photos which only open the shutter for a fraction of a second, astronomical images - particularly of faint objects - can be upwards of an hour long. During this time, high-energy particles known as cosmic rays - which are always whizzing around - have a much greater chance of interacting with your CCD and exciting electrons completely independent of any photons coming through the telescope. The annoying ones come in at an oblique angle to the CCD, leaving a trail of excited electrons across the chip. The even more annoying ones do this directly over the CCD cells you're using to capture an image of your object. Thankfully, there are some pretty good cosmic ray removal packages out there which use sophisticated image detection algorithms to remove that's a processing step right there.

In blue, I've highlighted pixel bleed. We're going for a long exposure of a pretty faint galaxy here, so any bright stars in the field will become oversaturated. In essence, the CCD cell containing the image of the bright star begins to overflow with energetic electrons, pouring them out into adjacent cells.

In green, I've highlighted a row of bad pixels. With millions of cells across the entire chip, statistically many are eventually going to fail. For earth-based observatories, it's untenable to keep throwing out CCDs which cost many thousands of dollars whenever some pixels go you work around it. For spacecraft, meanwhile, there's really nothing you can do about bad pixels even if you had the money to replace it.

There's a couple other artifacts noticeable in the original image, as well. Notice the steady gradient of dark-to-light in the background. Unfortunately, not all the pixels have the same sensitivity. Send 100 photons to one cell, and you might get 50 excited electrons...send them to another cell, and you might only get 40.

You have to account for this by taking "flat fields". Essentially, you take images (ideally just before or just after taking your astronomical images) of a uniformly lit surface with each color filter. The idea is that the surface should be sending out a constant number of photons to each cell, so the only signal you'll see will be the change in sensitivity across the CCD. You then divide the astronomical image by the flat field on a pixel-by-pixel basis to remove this sensitivity effect. Finding a truly flat field, though, can be a chore in itself...often times the best flat field you'll get is an image of the twilight sky before the stars come out.

Another artifact you may notice in the original is the weird wavy pattern, particularly noticeable on the left. Ideally you want your CCD chip to be as thin a piece of silicon as possible - this makes it more sensitive. However, particularly for longer wavelengths of light, photons reflecting off the back surface of the CCD can interfere with photons hitting the front surface and produce thin film interference - very similar to the wavy colored patterns you'll see in soap bubbles or with oil on water. Hopefully this too will be removed by flat-fielding.

Finally, as for the purpose of enhancing images, all of the above steps are necessary to get good science. Otherwise, you're just measuring your signal buried in a whole lot of noise. If you're going to take an image this far, though, you might as well go one step further to make a press release photo.

This serves several purposes, but not least of which is to share your own fascination of an astronomical object with the general public. Imagine if the Hubble Space Telescope *never* made press release photos available and only was used for hard science in the journals...public support wouldn't be nearly what it is today. Besides, it's the taxpayer's dollar which goes to fund it - the least we can do is give them some pretty pictures in return.

So, if you want to make a pretty picture, there's one more step you'll have to take - and this is a big one - because the above image was taken through an infrared filter. By definition, the human eye can't see this wavelength of light, so if we were to represent it in "true-color", the entire image should be black.

Creatively mapping various single filtered images to RGB space as well as some tweaking of colors needs to happen to for this to be visible - and aesthetically pleasing - for human vision. This color manipulation doesn't have the same tried-and-true formula as the above sequence of processing steps, and is often just manipulated until one gets something that just "looks good".

Piloting a spaceship through a galaxy cluster.

Blair writes:

In your post, you state

"Now, this gas is actually the material from which stars are created. Its most ubiquitous form throughout a galaxy is as a hot diffuse medium (somewhere around 10,000K)"


"Eventually, a whole cluster of galaxies is created, surrounded by a massive cloud of unused, superheated hydrogen gas (in the incredibly hot 100 million K range) known as the intracluster medium (ICM)."

Could human spaceships fly through these? I guess they may be hot, but the density is very low, so how much heat would the spaceship absorb and could it radiate the heat away?
This is actually a really cool question. My guess is that the intracluster medium is probably sparse enough that one wouldn't need to worry. A little back of the envelope math here:

- Let's say there's 500 galaxies in a cluster, each at around 10^12 solar masses (or about 2 x 10^45 g) so that gives us a total mass of around 10^48 g for the cluster's galaxies. The Intracluster Medium (ICM) should be about 10 times that, or about 10^49 g.

- However, the cluster is huge...if the milky way is roughly 25 kiloparsecs in diameter, a fair estimate for the cluster size is on the order of a megaparsec in radius. That translates to a volume of roughly [(10^6)(3 x10^18 cm)]^3 = 3 x 10^73 cm^3. That leaves us with a pretty low density of about 4 x 10^-25 g/cm^3. We throw in Avogadro's number for good measure, and assume pure hydrogen, and we end up getting about 1 hydrogen atom for every 5 cm^'s actually kinda weird how those huge numbers end up canceling so well.

So, yeah, a given atom will be crazy energetic - temperature scales linearly with kinetic energy - but there's so few of them I don't think it would be a huge issue heat-wise. The radiative constant of the ship should easily compensate for the occasional high-energy atom.

I think the concern would be more of a sputtering problem - at these energies, I'd worry about slow ionization of the ship's outer hull. Each hyper-energetic hydrogen atom the spaceship ran into might strip molecules from the crystal lattice of whatever alloy the spaceship was composed of. Sometimes it'll strip multiple molecules per collision, sometimes none, so let's just do an order of magnitude estimate and say 1 molecule stripped per collision. At 1 hydrogen atom per 5 cm^3, it doesn't seem like a big deal, but I think as it starts to cover spaceship-sized distances it might be an issue. Let's consider this in terms of cross-section:

For each square centimeter of spaceship hull surface hurtling through the cluster, a molecule of the hull will be stripped every 5 cm. If we want to travel from the edge to the center, we're talking about 3 x 10^24 cm, or roughly 6 x 10^23 molecules stripped. Again, weird that we just happen to hit on Avogadro's number again - roughly 1 mole of material per square centimeter will be stripped.

Assuming we're talking about, say, iron here, that's a molecular weight of 56, so 56 g/mole. So, behind each square centimeter of hull, 56 grams will be stripped traveling to the center of the cluster. With iron at a density of 8 g/cm^3, that would mean a 7 cm thickness would be stripped traveling to the center.

So, I guess the answer here is that if you add 7 cm of hull thickness (at least to the front of your spaceship) as an ablative shield, you should be okay. You actually probably want to double that, since presumably you'd like to leave the cluster at some point, too.

Tuesday, May 12, 2009

It's Twitter time.

You can now follow me on Twitter and ask various astronomy question there, too. Yay, technology!

Username: "astronomer_mike"

The Main Asteroid Belt and Future Impacts

Okay, folks, sorry for the languishing blog, but it's been a heck of a semester. Meanwhile, on with the show!

Ben writes:
What's the history of our solar system's asteroid belt? There seems to be a consensus that it's inevitable that another Cretaceous-ending-like asteroid will hit the Earth again someday; do such asteroids typically come from the asteroid belt?
First, a little background: The main asteroid belt is a collection of random rocky debris found in the large gap between Mars and Jupiter. Here's a nice plot from the Minor Planet Center, the organization responsible for tracking and naming the objects. The four inner turquoise circles are the orbits of the four inner planets (Mercury, Venus, Earth, and Mars), and the outer turquoise circle is the orbit of Jupiter. Within the sizable gap between these, each green point is an asteroid in the main belt.

While there's a lot of individual objects known - literally millions - their total mass isn't much. In fact, Earth's moon is roughly 25 times the mass of the entire asteroid belt combined.

There's been a lot of speculation that the asteroid belt comprises the remains of a "failed planet" which couldn't form in our early solar system due to Jupiter's strong gravitational pull constantly rending apart any protoplanets. This may be true, but the jury is still out - it is compelling that there's such a large gap between inner and outer planets, and that there's such a large gravitational force nearby. On the other hand, there obviously wasn't a whole lot of mass to work with in this region (though Jupiter likely stole a good deal of it during its own formation). Moreover, current models suggest that Jupiter may not have even formed in its current location, but actually migrated.

Either way, Jupiter continues to exert a strong gravitational effect on the main asteroid belt today...enter the concept of "Kirkwood Gaps". Now, in the above plot from the Minor Planet Center you'll notice little order in the position of the asteroids locations at a given time.

However, if we plot their *average* distance from the Sun (their so-called "semi-major axis") versus how many are at that distance, something very different happens:

They seem to be grouped into families at specific distances, with no-mans-land in between - our Kirkwood gaps.

More interestingly, an object's average distance from the sun is directly proportional to the time it takes to complete one orbit. Those little numbers at the bottom of each Kirkwood gap (3:1, 5:2, 7:3, and 2:1) correspond with the number of orbits the asteroid makes in a given time versus the number of orbits Jupiter makes. This is what's known as an "orbital resonance". A similar phenomenon occurs with particles in the rings of Saturn in resonance with Saturn's moons, as well as Kuiper Belt objects in resonance with Neptune.

So, what's really going on here? Let's say we place an asteroid in the 3:1 Kirkwood gap. For every 3 times it goes around the Sun, Jupiter goes around once. This will mean the asteroid keeps meeting up with the asteroid in the same part of its orbit over and over...the gravitational force exerted by Jupiter will always be in the same direction each time.

It's a bit like being a little kid on a swing while a big kid keeps pushing you over and over in the same place...eventually, you'll fall out and go flying off. Similarly, asteroids that wander into a Kirkwood Gap due to random interactions won't stay in that gap for long, and Jupiter will send them off on some fairly random orbit. There's good evidence to suggest that many of the asteroids which cross Earth's orbit (i.e. the ones we really need to watch out for), experienced this fate.

So, the answer is a pretty strong yes to this question - a good deal of potentially hazardous impactors probably started out in the main asteroid belt, accidentally wandered into a Kirkwood gap due to mutual asteroid interactions, and were sent hurtling into the inner solar system by Jupiter. Since the nature of this process is inherently chaotic, it's extraordinarily difficult to predict which main belt asteroids this will happen to and which subsequent orbit they'll end up on...but statistically we can say that another major impact is really more a matter of "when" than "if".