Secondary author: Joanna Robaszewski
Sunday, October 30, 2011
Class Activity: Stellar Properties from Afar Problem 1
Given that the angular width of the sun is .5° and that 1 AU is 1.5 * 1013 cm, it's possible to find the radius of the Sun in centimeters. If we consider the angle subtended by the Sun from a certain point on Earth, we have a right triangle with one leg equal to 1 AU and one angle equal to half of the angular width of the sun, .25° or 15 arcminutes.
By using a trigonometric relation, we can find a value for Rs, the radius of the Sun.
So the radius of the Sun is 6.55 * 1010 cm. Next, we can get 1 AU in solar diameters by simply dividing 1 AU in centimeters by the radius of the Sun in centimeters:
So 1 AU is approximately 100 solar diameters. Now, using Newton's version of Kepler's third law, we can get the mass of the Sun.
This is Newton's version of Kepler's third law, if we assume that the mass of the Sun is significantly greater than the mass of the Earth, which it is. Then P is the period of the Earth, a is 1 AU, G is the gravitational constant, and M is the mass of the sun. Rearranging and plugging in the correct numbers:
We see that the mass of the sun from Kepler's third law is 6.4*1032 g.
Secondary author: Joanna Robaszewski
Secondary author: Joanna Robaszewski
Wednesday, October 19, 2011
Lab Activity 2: The Astronomical Unit
Take a look at this image of Mercury passing in front of the Sun, as taken from the polar Earth-orbiting TRACE satellite.
From this image and a few other facts, it's possible to discover the distance from the Earth to the Sun, which is defined as one AU (astronomical unit). The other information necessary is that the Sun's angular width is 0.5°, Mercury's orbital period is 87 days, and the radius of the Earth is 6.738 * 108 cm.
The slight sinusoidal "wiggle" comes from the fact that the TRACE satellite is orbiting the Earth as its taking images of Mercury, so the image of Mercury projected against the non-moving background of the sun moves up as the satellite moves down to the south pole, and down again as the satellite moves back up to the north pole. By extending the disk of the Sun into a full circle (using a printout of the image and a compass), it's possible to estimate how many "wiggles" would fit into the full height of the Sun, which we know to be 0.5°. From this, we can find the angular width of the "wiggle," which, through some tricky geometry, will give us the distance from the Earth to the Sun. We determined that about 50 "wiggles" would fit into the width of the sun, so the angular width of the sinusoid is .5/50 = 1/100th of a degree.
In the above, ae is the Earth's orbital semi-major axis (1 AU), am is Mercury's orbital semi-major axis, and Re is the radius of the Earth. The angles x, y, and z are also shown on the image above, and the angle y is the angular width of the sinusoid, which we found to be .001°. From considering the triangle formed from the center of the earth to the satellite to mercury and by using the small angle approximation, we find:
We know the radius of the radius of the Earth and the angle y, so we can find ae – am. The small angle approximation requires the angle to be in radians, so y/2 = 8.725 * 10-6 radians. We find that ae – am = 7.723 * 1013 cm. Now we need Kepler's laws to obtain the astronomical unit. Kepler's Third Law states:
This means that the ratio of the periods of Mercury's and Earth's orbits squared is equal to the ratio of the distances from the Sun of Mercury's and Earth's orbits cubed. We don't have ae or am, but we do have ae - am. Plugging this in:
and rearranging for ae:
Solving this, we finally find a value for the astronomical unit! 1 AU = 1.25 * 1014 cm. The actual value for the astronomical unit is 1.49 * 1013 cm, so this is a percent error of:
This is obviously a very high percent error, which is due to the not-very-exact method of estimating the number of "wiggle" widths in the full width of the sun, since that was the only estimated value. Since the calculated value for the astronomical unit is too big, this means that the value for ae - am is too big, which means the estimated width y is too small.
Now we can obtain the distance Mercury is from the Sun from our value for ae - am and our value for ae. We find that am = 4.78* 1013 cm, and the actual distance between Mercury and the Sun is 5.79 * 1012 cm. This percent error is 725%, which is to be expected since our percent error on the astronomical unit is so high.
Finally, we can obtain the mass of the Earth from Kepler's Third Law and the mass of the Sun.
Note that it is also possible to obtain the mass of the sun by neglecting the mass of Mercury in its orbital calculation in Kepler's Third Law,
but since it requires the distance from Mercury to the Sun and that distance has such a large percent error, I will just use the actual value for the mass of the Sun. We find, by rearranging Kepler's Third Law and plugging in the values for Pe, ae, and Ms:
where the mass is measured in grams. The actual value for the mass of the Earth is 5.974 * 1027 g, so my found value has a percent error of 2.5 * 1017 %. This is to be expected, since the value of the astronomical unit is used in the calculation, and if that has a percent error of 742%, then the value for the mass of the Earth will have a percent error of at least that cubed, since the astronomical unit is cubed in the equation for mass of the Earth. The actual percent error is higher because there are other factors in the equation that help in determining the value of the Earth's mass.
From just a few important facts, we've managed to find the astronomical unit, the distance from Mercury to the Sun, and the mass of the Earth. Our values are not very good, but they all stemmed from an estimate using a compass on a piece of paper, so this is to be expected.
From this image and a few other facts, it's possible to discover the distance from the Earth to the Sun, which is defined as one AU (astronomical unit). The other information necessary is that the Sun's angular width is 0.5°, Mercury's orbital period is 87 days, and the radius of the Earth is 6.738 * 108 cm.
The slight sinusoidal "wiggle" comes from the fact that the TRACE satellite is orbiting the Earth as its taking images of Mercury, so the image of Mercury projected against the non-moving background of the sun moves up as the satellite moves down to the south pole, and down again as the satellite moves back up to the north pole. By extending the disk of the Sun into a full circle (using a printout of the image and a compass), it's possible to estimate how many "wiggles" would fit into the full height of the Sun, which we know to be 0.5°. From this, we can find the angular width of the "wiggle," which, through some tricky geometry, will give us the distance from the Earth to the Sun. We determined that about 50 "wiggles" would fit into the width of the sun, so the angular width of the sinusoid is .5/50 = 1/100th of a degree.
In the above, ae is the Earth's orbital semi-major axis (1 AU), am is Mercury's orbital semi-major axis, and Re is the radius of the Earth. The angles x, y, and z are also shown on the image above, and the angle y is the angular width of the sinusoid, which we found to be .001°. From considering the triangle formed from the center of the earth to the satellite to mercury and by using the small angle approximation, we find:
Now we can obtain the distance Mercury is from the Sun from our value for ae - am and our value for ae. We find that am = 4.78* 1013 cm, and the actual distance between Mercury and the Sun is 5.79 * 1012 cm. This percent error is 725%, which is to be expected since our percent error on the astronomical unit is so high.
Finally, we can obtain the mass of the Earth from Kepler's Third Law and the mass of the Sun.
From just a few important facts, we've managed to find the astronomical unit, the distance from Mercury to the Sun, and the mass of the Earth. Our values are not very good, but they all stemmed from an estimate using a compass on a piece of paper, so this is to be expected.
Monday, October 17, 2011
Class Activity: Black Body Radiation Problem 2 b and c
In order to define a relationship between the wavelength corresponding to the peak intensity of radiation emitted from a black body (λmax) and that black body's temperature (T), we must begin with the equation for specific intensity, Bλ(T):
The maximum of this function will be the λ such that
where the derivative is evaluated with T held constant (since we are finding λmax for a given T). If we make a simple variable substitution, the calculation becomes much easier:
Then we need:
This is a difficult equation to solve, but if we approximate ex to second order, then we have:
Plugging in the values for h, c, and k, we find:
where the units of λmaxT are cm · K. Wien's law is:
For a second-order approximation of ex, this is quite close.
Now let's consider very small photon energies, so that hv << kT. Then we can use a second-order approximation for ex again to simplify Bv(T) for small photon energies.
Inserting this into the equation for Bv(T), we obtain:
This is the simplified form of Bv(T) for hv << kT.
Radio astronomers like to talk about "brightness temperature" instead of actual brightness because the brightness is related to the temperature, but the temperature is more helpful in discovering the black body spectrum of an object than its brightness. The black body spectrum, if the temperature is known, gives the most prominent wavelength of a black body's radiation, which is a useful value in classifying an object.
Secondary authors: Joanna Robaszewski, Lauren Gilbert
Now let's consider very small photon energies, so that hv << kT. Then we can use a second-order approximation for ex again to simplify Bv(T) for small photon energies.
Radio astronomers like to talk about "brightness temperature" instead of actual brightness because the brightness is related to the temperature, but the temperature is more helpful in discovering the black body spectrum of an object than its brightness. The black body spectrum, if the temperature is known, gives the most prominent wavelength of a black body's radiation, which is a useful value in classifying an object.
Secondary authors: Joanna Robaszewski, Lauren Gilbert
Class Activity: Optics Problem 3
Telescopes work much the same way as a series of slits in a wall letting light through onto a screen. We know that the pattern of light on the screen is related to the size, spacing, and number of slits in the wall, and if there are several slits next to each other with infinitesimal widths between them, the effect is essentially the same as having one large slit (see image below).
We know that the intensity of the light on the screen is the Fourier transform of the intensity of the light coming through the slits, which, for several slits right next to each other (one big slit), is a top-hat function. The Fourier transform of a top-hat function is a sinc function, or decaying sine function, as shown above. The only difference is that the intensity of light on the screen is the square of the light wave hitting the screen, so the actual sinc function includes negative values. In addition, if the width of the top-hat is W, then in the Fourier transform, the sinc function equals zero at the points 1/W, 2/W, 3/W, and so on.
The width is also related to the wavelength of the light coming through the large slit, since there will be destructive interference where the light from the bottom of the slit and the light from the top of the slit are out of phase by λ/2. Therefore, the width of the center peak of intensity will be:
We know that the intensity of the light on the screen is the Fourier transform of the intensity of the light coming through the slits, which, for several slits right next to each other (one big slit), is a top-hat function. The Fourier transform of a top-hat function is a sinc function, or decaying sine function, as shown above. The only difference is that the intensity of light on the screen is the square of the light wave hitting the screen, so the actual sinc function includes negative values. In addition, if the width of the top-hat is W, then in the Fourier transform, the sinc function equals zero at the points 1/W, 2/W, 3/W, and so on.
The width is also related to the wavelength of the light coming through the large slit, since there will be destructive interference where the light from the bottom of the slit and the light from the top of the slit are out of phase by λ/2. Therefore, the width of the center peak of intensity will be:
This is the angular resolution, and it simplifies to:
Therefore, the angular resolution of CCAT, which is a 25-meter telescope that detects wavelengths of 850 microns = 8.5 * 10-4 m, will have an angular resolution of:
where the angular resolution is measured in radians. The Keck 10-meter telescope observes in the J-band, which is defined as wavelengths between 1.1 and 1.4 micrometers = 1.1 * 10-6 to 1.4 * 10-6 m. Its angular resolution, using the maximum observation wavelength, is:
with D in radians, again. A smaller angular resolution means smaller sources can be resolved, since anything with an angular size larger than the angular resolution can be resolved by the telescope. This means that the Keck telescope, despite being smaller than CCAT, can resolve smaller objects in its observational range.
Secondary author: Joanna Robaszewski
Saturday, October 8, 2011
Halo Stars
I learned something very interesting the other day. I knew that a large portion of matter in spiral galaxies is contained in the galactic halo, the spherical mass (of mostly dark matter) that surrounds the galactic disk. I thought that all the stars were in the galactic disk and the galactic halo was just dark matter. I was surprised to learn that there are a fair number of stars in the galactic halo as well. They orbit the center of the galaxy just like the stars in the disk, but their plane of orbit doesn't line up with the disk, so they spend most of their time in the halo.
Apparently, this is where astronomical feature globular clusters exist. They're large groups of stars held together by their own gravitational force, not large enough to be a galaxy. They orbit the center of a galaxy through its halo, the entire cluster moving as one. Most of the stars in the galactic halo are much older than stars in the disk, because the halo doesn't contain the interstellar dust and gas that the disk has. These areas usually create "star nurseries" where new stars are born, and thus the stars in the disk are much younger than those in the halo.
The difference between old stars in the halo and young stars in the disk is mostly characterized by the metallicity of the stars. Metallicity is the amount, by mass, of elements heaver than hydrogen or helium in the star's make-up. Younger stars tend to have higher metallicities because the dust clouds they were formed out of were at least partly created by supernova explosions of dying stars, which create the violent environments necessary for heavy metals to be created.
I knew of globular clusters in the past, but I always thought they were odd groupings of stars outside of galaxies. I never knew that stars could be part of a galactic gravity system without being part of the disk where most of the stars are.
Apparently, this is where astronomical feature globular clusters exist. They're large groups of stars held together by their own gravitational force, not large enough to be a galaxy. They orbit the center of a galaxy through its halo, the entire cluster moving as one. Most of the stars in the galactic halo are much older than stars in the disk, because the halo doesn't contain the interstellar dust and gas that the disk has. These areas usually create "star nurseries" where new stars are born, and thus the stars in the disk are much younger than those in the halo.
The difference between old stars in the halo and young stars in the disk is mostly characterized by the metallicity of the stars. Metallicity is the amount, by mass, of elements heaver than hydrogen or helium in the star's make-up. Younger stars tend to have higher metallicities because the dust clouds they were formed out of were at least partly created by supernova explosions of dying stars, which create the violent environments necessary for heavy metals to be created.
I knew of globular clusters in the past, but I always thought they were odd groupings of stars outside of galaxies. I never knew that stars could be part of a galactic gravity system without being part of the disk where most of the stars are.
Friday, October 7, 2011
Light Pollution
This time-lapse video of the Milky Way is absolutely stunning...
http://www.ouramazingplanet.com/sorgjerd-video-the-mountain-teide-1405/
...but I can't help noticing the yellow-orange glow on the horizon in nearly every nighttime shot. At first, I thought it was just the sunset, but the glow doesn't change like the stars do as the Earth passes beneath them, and like a sunset does as it gets later in the evening. That glow that obscures everything close to the horizon must be sky glow, or light pollution.
The clouds of the Milky Way are clearly visible, so the light pollution must not be too bad in the Canary Islands, Africa, where the video was taken. But it's still there, and in many places of the world, it's much worse. In the video, it mars the horizon with an ugly yellow glow throughout the night and ruins the peace that comes from watching stars go by overhead. Here in LA, it completely shrouds the view of the cosmos.
One of my least favorite things about living in LA (other than lack of snow) is being unable to see any stars even on a very clear night. Looking up at the stars was always a source of enjoyment and wonder for me, but looking up and seeing nothing but dark gray with an orange tint after a gorgeous clear day is just sad. I can't help wondering how people living without a view of the stars can keep their lives in perspective. I start feeling so lost when I can't look up and see the Earth's place in the universe.
Light pollution is, in my opinion, the worst consequence of large cities. There are all sorts of new streetlights that keep light from escaping upward, but those in charge of such things don't seem to care very much about a view of the stars. The stars bring such an immense inner peace that I really think people of all types could benefit from being able to see them now and then, and know that their hectic lives are not so important in the face of the immovable stars (of course, they're not immovable at all, but to the casual observer, they are).
Thank goodness for astronomy observatories.
http://www.ouramazingplanet.com/sorgjerd-video-the-mountain-teide-1405/
...but I can't help noticing the yellow-orange glow on the horizon in nearly every nighttime shot. At first, I thought it was just the sunset, but the glow doesn't change like the stars do as the Earth passes beneath them, and like a sunset does as it gets later in the evening. That glow that obscures everything close to the horizon must be sky glow, or light pollution.
The clouds of the Milky Way are clearly visible, so the light pollution must not be too bad in the Canary Islands, Africa, where the video was taken. But it's still there, and in many places of the world, it's much worse. In the video, it mars the horizon with an ugly yellow glow throughout the night and ruins the peace that comes from watching stars go by overhead. Here in LA, it completely shrouds the view of the cosmos.
One of my least favorite things about living in LA (other than lack of snow) is being unable to see any stars even on a very clear night. Looking up at the stars was always a source of enjoyment and wonder for me, but looking up and seeing nothing but dark gray with an orange tint after a gorgeous clear day is just sad. I can't help wondering how people living without a view of the stars can keep their lives in perspective. I start feeling so lost when I can't look up and see the Earth's place in the universe.
Light pollution is, in my opinion, the worst consequence of large cities. There are all sorts of new streetlights that keep light from escaping upward, but those in charge of such things don't seem to care very much about a view of the stars. The stars bring such an immense inner peace that I really think people of all types could benefit from being able to see them now and then, and know that their hectic lives are not so important in the face of the immovable stars (of course, they're not immovable at all, but to the casual observer, they are).
Thank goodness for astronomy observatories.
The Celestial Sphere and Observation Planning
If I want to view the star Tau Ceti, I need to know where it will be at what time of the night so I can find it. With just a little bit of knowledge about Tau Ceti's location as well as my own on Earth, I know everything I need to about how to find Tau Ceti.
Let's say I'm observing from Palomar Observatory, which has a latitude of 33º 21' 21'' N and a longitude of 116º 51' 50" W. If I only want to view Tau Ceti when it's on my meridian (the line that runs from north to south directly overhead through the zenith), then it's much easier to figure out where it will be in the sky at what time.
Since an object's right ascension and declination don't change over the course of the year, Tau Ceti's position on the meridian will be the same every night, and the only difference will be the time that it is on the meridian. In order to figure out Tau Ceti's elevation on meridian each night, I need to know its declination and my observation latitude. Tau Ceti's right ascension is 1:4:04.0829 and declination is -15º 56' 14.928". A negative declination means that, from the equator, Tau Ceti will be about 16º below the zenith when it's on the meridian. But I'm not observing from the equator, so I have to subtract my latitude. 33 + 16 = 49, and since I'm 33º north of the equator, it makes Tau Ceti lower in the sky, so its position is -49º (see first image, "Finding Elevation of Tau Ceti").
Now that I've determined the location of Tau Ceti when it's on my meridian, I just need to determine the time of the night that Tau Ceti will be on my meridian so I know when to look for it. The time system LST (local sidereal time) is measured by the stars and thus Tau Ceti will be on my meridian at the same time LST each night all year. But, since the Earth is orbiting the sun as it rotates, LST does not exactly match up with time on earth. One day in LST has an extra four minutes, which means that each month LST has an extra two hours than a month in UT (universal time, measured by days on Earth). LST has its zero point at the vernal equinox, at noon on March 21st, so LST and UT line up at this time. Each month later, LST is an additional two hours later at UT's noon. For example, an LST of 0:00 falls at a UT of 2:00 pm in late April, because April is a month later than March. The graph of Tau Ceti's elevation on the meridian shows the location of Tau Ceti in the sky as well as the time at which it is on the meridian each month of the year (see below).
Now I have everything I need to determine when to observe Tau Ceti from Palomar Observatory!
Primary author: Cassi Lochhaas
Contributors: Joanna Robaszewski, Juliette Becker
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