Stellar Properties From Afar (Problem 1)

Wednesday, October 26, 2011

Considering the angular diameter of the sun and the astronomical unit, we can estimate the radius of the sun, the AU in solar diameters, and the mass of the sun using Kepler's 3rd law.

Applying basic trigonometric identities and taking the astronomical unit a to be the distances from us to the closest point of the sun to us (i.e., the centre of the circle we see from earth), we can see that:

Multiplying through by a we get a value for the radius of the sun.  It is clear from here that if we divide a by twice the solar radius we can easily determine the answer to the second part of our question.  Finally, we have Kepler's 3rd law:

Where P is the period of the earth and G = 6.7 x 10-8 dyne-cm2/g2.  From here we can solve for the mass of the sun.

Solving the first equation using a = 1.5 x 1013 cm we get the radius of the sun equal to 6.545 x 1010 cm which is very close to the actual value of 6.955 x 10107 s.  Dividing, we get 1 AU = 114.6 solar diameters.  Then, solving for the mass of the sun in Kepler's 3rd law with P = 3.154 x 107 s, we have the mass of the sun equal to 2.007 x 1033 g which is a surprisingly accurate number.

Surface Temperature of Planets

Friday, October 21, 2011

by Eric S. Mukherjee, Nathan Baskin, and I forget who else (sorry).

In this problem we consider the how the temperature of the sun affects the temperature of the earth.  This is possible to estimate by assuming both the sun and the earth to behave like perfect blackbodies.

Assuming the Earth has constant surface temperature and that it behaves like a blackbody, we can estimate the surface temperature using the energy emitted by the sun.  We also assume the sun to be a perfect blackbody.  Under these assumptions we can find the surface temperature of the Earth by knowing the temperature of the sun, the radius of the sun, the mass of the sun, the mass of the earth*, and the radius of the earth.

We start with the equation for flux at the surface of a blackbody (σ is the Stefan-Boltzmann constant):

From this we derive the luminosity of the sun by multiplying through by the surface area:

Then the flux of the sun at the surface of the earth is (where a is the astronomical unit):

If we consider the area of the earth through which the flux passes, it is the circle of area π R2.  Multiplying through by this quantity we get the power input to the earth from the sun.  We then realise that this is necessarily equal to the power output of the earth due to energy conservation which, at the surface of the earth, is equal to σT4π R2.   Thus we have an equation of the form:

Using this equation with R= 695,500 km and T= 5778 K, we get T⊕ = 279 K.

This temperature that we calculate is around 5.5°C which sounds reasonable for an earth without accounting for atmospheric greenhouse effects and allowing for the temperature at the poles.  The true average temperature of the earth is around 16°C but that is measured with the warming effect of the atmosphere.  The sun is not a perfect blackbody which also contributes to the difference between our calculation and the true value.

I'd like to thank the entire Ay 20 class and teachers for collective brainpower due to the fact that I can't remember who exactly worked on this problem and I'm sure we drew from the knowledge of many people in the room.  I'd also like to thank the superior computational power of Wolfram Alpha for bringing to my attention that there exponents matter when calculating ratios and that the temperature of the earth is most definitely not 1270 K.

*Note: It has been brought to my attention by Professor Johnson that the masses of the earth and sun do not actually factor into this calculation at all unless we need them to derive some of our other known constants.

A slight belated correction on AGN

As my readers may remember, a few weeks back I posted about the properties of AGN and how they affect their host galaxies.  One of these ways I listed was star formation rate.  Actually, a bit less than two weeks ago an article was reprinted from UCSD by ScienceDaily that AGN do not stop star formation as previously thought.

Prior research showed a correlation between the presence of AGN and the lack of star formation in galaxies.  This new study claims that this was a function of observational bias.  Older, more massive galaxies are easier to detect, and are also the ones with decreased star formation rate.  This study finds AGN in all types of galaxies including those in which stars are still being formed.

Read the full article here.

Eric Answers

Wednesday, October 19, 2011

Hey all, as you may know I've been collecting astronomy questions from people.  Now I'm going to answer a few of them.

Jessica and choirqueer asked: "Astronomy vs. astrology vs. astrophysics.  What is the difference?"

I've combined the two questions for ease of answering.  Technically, astronomy has more to do with the qualitative or observational study of all objects not contained in the Earth's atmosphere.  Astrophysics is part of astronomy, but is focused on the applications of physics to astronomy and understanding why things are the way they are through physics.  The title of this blog comes from something an old friend of mine used to say despite the fact that he was, in fact, an astrophysicist.
Astrology is completely different in today's world although in antiquity astrology was astronomy.  Astrology is a belief that astronomical phenomena affect our lives as humans on earth and is widely regarded as unscientific.  I personally don't believe constellations and planetary motion have any effect on our lives as constellations are patterns that humans have assigned to stars which aren't even necessarily close together and planets are predictably governed by physics, but belief is very personal and I am not one to tell people they are wrong.

 LilyForest asked: "How noisy is the sun, assuming we could actually here it?"
I actually attended a colloquium at the CfA that dealt with this over the summer.  It was a fascinating topic.  Here's a video from ESA talking about the vibrational modes of the sun way more eloquently than I possibly could.
Basically, the sun produces "noise" due to its surface vibration.  However, this noise is generally not in the human audible range and also cannot reach us on earth.  Here is a clip from Stanford of the audio from 3 modes: Solar Sounds

 NastyNate (Nathaniel) asked: "How many total planets have been discovered and recorded in the universe?"
This is actually a question for my professor, who studies planets outside of our solar system.  However, since this is my blog and not his, according to this website which seems like a credible source run by a Paris Observatory scientist gives the current number a 694 planet candidates found outside of our solar system as of today.

Greg asked: "Given the universe is expanding at an accelerating rate, will the rate of expansion eventually pass the speed of light?
The expansion of the universe is a very tricky subject.  Currently, the recession velocity of galaxies due to the expansion, which is proportion to the speed of light multiplied by the redshift, can be greater than the speed of light for redshifts greater than 1.  However, this is greatly dependant on the coordinate system and reference frame.  Since we can argue that galaxies are moving apart due to expansion of the universe, the short answer is yes: the expansion for distant objects is even currently greater than the speed of light.  The coordinates in which these are moving faster than c, are not the same coordinates used in relativity so this doesn't really contradict relativity.  Presumably this is a result of the odd behaviour of comoving coordinates which are explained best in Ned Wright's tutorial.

Jogirl asked: "Why is Pluto not considered a planet anymore?  How can it be a planet one day and not the next?"
Pluto no longer fits the criteria for a planet.  According to the International Astronomical Union, the current criteria for a planet are the following:
  1. It is in orbit around the sun
  2. It has sufficient mass to have taken on a spherical shape due to self gravity
  3.  It has cleared the neighbourhood of its orbit.
Looking at 1 and 2, Pluto may be a planet.  However, it does not fill the third requirement.  Pluto has very little mass in comparison to the combined mass of numerous objects in its orbit.  In comparison, the Earth is by far the most massive object in its orbit.  Basically, Pluto failed to gravitationally bind or expel the other similarly small objects in its immediate neighbourhood and is therefore one of many similar objects in the area rather than The One Large Thing in its orbit, if that makes any sense.
One answer I can give you for your second question is that science progresses by falsification.  This means that the rules in science are constantly changing and things are being redefined in order to comply with new rules.  As we learn more about the universe we realise that some things we thought before are not the case.  For instance, we now know that the earth orbits the sun.  It's not that one day the sun orbited the earth and then it changed, but that science changed and our theory was then modified to better fit the new model.

Measuring the Astronomical Unit

Tuesday, October 18, 2011

I  apologise in advance for the hurried write up.

We claim that we can measure the astronomical unit using nothing but the following image of Mercury transiting the sun as viewed from a satellite in near earth polar orbit.  From this we can also determine the semimajor axis of Mercury's orbit and the orbital period of the satellite.

Methods and Conclusions
In order to solve this problem, we have to notice the sinusoidal motion of Mercury against the backdrop of the sun.  This is an effect of the parallax using the a distance roughly the diameter of the Earth as a baseline.  We also notice that there is a parallax effect on the sun that has been corrected in the image.  As a result, the angle we measure by comparing the amplitude of the sine wave to the radius of the sun is the angle α between the centre of the sun and the apparent position of mercury against the sun.  We then define β as half the angle from one position of the satellite at the pole to the other pole through the centre of the sun (see figure) and θ/2 as the parallactic angle of mercury compared to the backdrop of the stars.

It is easy to see from this diagram that π = α + β + π - θ/2 or θ/2 = α + β.  Using small angle approximations for tangent, we get β = R/a and θ/2 = R/Δa with R the radius of the earth, a being the astronomical unit and Δa defined as the distance from the Earth to Mercury.

We also know the orbital period of Mercury which is 87 days and the orbital period of the Earth which is 365 days.  Dividing, we get:
Then we use Kepler's 3rd Law we get:
Which gives us Δa = 0.62a.

Therefore, we have:

Drawing a circle on a printout of the satellite image that coincides with the border of the sun, we get the amplitude of the wave to be 1mm and the diameter of the disc to be 218mm.  Using the ratio of these numbers and the fact that the angular diameter of the sun is 0.5 degrees, we get α = 0.00004 radians.  We use R = 6378km and then get a = 9.71e12 cm which is roughly 0.65 of an actual astronomical unit.

We see that the semimajor axis of Mercury is a-Δa = 0.38a.  We also have the orbital period of the satellite using Kepler's 3rd Law:  .  We get P = 1:24:17 hours as the orbital period of TRACE the satellite (using the actual mass of the earth in our calculation).

Our result for the astronomical unit was off by a bit under 50% of the actual value.  This error most likely came from the imprecise measurement of the diameter of the sun on the page and the amplitude of the sine wave across the sun.  This was done by hand with no compass, which left a lot of room for error.

Thank you to Professor Johnson for providing the orbital period of Mercury, Jackie for pointing out the parallax of the sun, and Daniel for measuring the diameter of the circle.

Tau Ceti From Palomar

Sunday, October 9, 2011

I wrote up a mix of 2 and 3 from the work sheet on LST. Hope you don't mind

Palomar is an important observatory for the Caltech astrophysics community. Because of this, we seek to constrain the visibility of the star Tau Ceti from the Palomar Observatory over the course of the year. We determine first whether its declination allows it to be seen and then when it can be observed. We also examine the elevation from month to month. We find that Tau Ceti is observable from Palomar during certain times of the year.

Tau Ceti is a star located in the constellation Cetus at RA = 1:44 (25.0167 degrees) and Dec = -15.9375 degrees in the southern hemisphere of the celestial sphere. The Palomar Observatory is located at 33.358 degrees north and 116.864 degrees west. It is important to understand how and where Tau Ceti can be observed from Palomar in order to study the star. To do this, we determine the elevation of Tau Ceti when it is on the meridian passing through Palomar as well as the times that it is observable optically.


First we determine what the maximum elevation of Tau Ceti is from Palomar. The elevation of a star is the angle it forms with a line connecting the centre of the earth to the observer's local horizon. We calculate this using the equation:

Where a is elevation, delta is declination, phi is the local latitude, and H is the hour angle (the difference in LST from the meridian at the time of viewing). When the star is on the meridian, it is clear that H = 0. We then get:

Since one local sidereal day is a 360 degree revolution of the earth, or the time it takes for the meridian to re-align with the object in question, it is obvious that this is the elevation any time Tau Ceti is on the meridian and also that this is the maximum elevation of Tau Ceti as observed from Palomar.

Next we notice that if we set a = 0, or say the star is at the horizon, we can find the hour angle the time the star rises. This is given by:

Dividing this by 15, we get H = 6.72 hours, which we round down to 6.5 to account for the fact that stars at the horizon are invisible to the ground based observer.

Plotting the elevation as a function of time for the 20th of each month of the year, we get the following graph. We choose the 20th of each month for convenience's sake as the vernal equinox is on 20 March.

This shows, as expected from considering the problem, that at LST = 1:44, or when Tau Ceti is aligned with the meridian, the elevation is always 40.71. The graph also shows the approximate times at which Tau Ceti is aligned with the meridian in UT for planning of observations. Now plotting the time (UT) of Tau Ceti's maximum elevation against the passing months, we get the following graph:

As expected, the change in time is a linear relationship to the time of year. The error bars of this graph mark the total time that Tau Ceti is above the horizon. Regions plotted in red are the times before and after sunrise and sunset, respectively. These are the times that Tau Ceti is actually visible from Palomar because interference from the sun is absent. Predictably, this range is from July to January, with maximum visibility in the October-November region where it is dark the entire time that Tau Ceti is above the horizon. This agrees with observation data that indicates Cetus is best visible in late autumn.

We thank Professor Johnson and Jackie for providing us with this problem to solve. Also, Wikipedia for the image describing horizontal coordinates and Professor Harold Geller for his notes on calculating the altitude of a star. We also thank Microsoft Excel for cooperating long enough to produce these graphs.

Clarifications on AGN

A few days back, Jackie asked:

I'm interested to hear about the ways we can study AGN! I would also be interested in seeing a diagram of one explained. That's a question some of the grad students here have gotten on their qual exam ("Diagram an AGN.") and I still feel a little blurry about the details.

What does it mean that AGN influence the color of their host galaxies?
Now that the horror that is Phys 12, which is by the way the sophomore physics course for astrophysics and physics students is over for the week, I am free to answer.

What Does An AGN Look Like?
To properly understand the parts of an AGN we must consider the two types. The currently accepted "unified" structure of an AGN looks this (personal communication with Dr. Andy Goulding, origin unknown):

The Black Hole and Accretion Disk:
I'd assume you know what a black hole is, but let's talk about the accretion disk. According to NASA, an accretion disk is a flat sheet of gas and dust that surrounds a black hole. The presence of an accretion disk indicates that there is material being incorporated into the black hole and that the black hole is accumulating mass.

The Jets:
It's important to note that jets do not occur in all of AGN, but when they do occur they are characterised by a beam of particles ejected in opposite directions. These are observed in the radio wavelength.

The Broad Line and Narrow Line Regions:
The Broad Line Region (BLR) of an AGN is the region of gas clouds immediately surrounding the black hole. It is characterised by relatively broad emission lines, hence its name. The Narrow Line Region (NLR) of an AGN is an outer layer of gas clouds in the area of the jets that produce strong "forbidden lines" which are not present in denser gases.

The Dusty Torus:
This is simply an obscuring torus around the accretion disk. Current research indicates that it is probably made up of gas and dust, but the distribution of gas and dust in the torus are not yet known (Antonucci, 1993).

Now let's put it all together. The two types refer to the inclination of the AGN with respect to us (the observer).

Type 1 AGN:
A type 1 AGN refers to an AGN in which the BLR is visible. To visualise this situation, a type 1 looks a bit like a donut with an accretion disk at the centre. i.e., a type 1 AGN is face-on as opposed to edge-on. Viewing a type 1 AGN will show a clear view of the BLR as well as the accretion disk and black hole.

Type 2 AGN:
A true type 2 AGN is a bit less straightforward. We call an AGN type 2 when the BLR/accretion disk/black hole area is obscured by the dusty torus. This leaves us only the NLR visible for observation. There are some AGN that are neither entirely edge-on or face-on, since the inclination of galaxies does not change in intervals of 90°. However, a true type 2 AGN is completely edge-on.

AGN and Galaxy Colour
I have to admit, when I wrote that AGNs affect galaxy colour, I did so because my SURF mentor told me to rather than because I truly understood. I still can't claim to fully understand. However, I've spent some time reading through the relevant paper and found an interesting diagram to share (Hickox et al., 2009):

This diagram shows (a) the colour of host galaxies plotted against absolute magnitude with yellow circles, green stars, and red squares showing radio, X-ray, and infrared AGN, respectively and (b) the colour distribution in each of these wavelengths. Notice how in both the X-ray and infrared spectra, the distributions for AGN are shifted to the left in what is called the "green valley". To summarise, AGN do not influence the colour of their host galaxies beyond what is normally seen, but they may influence the proportion of a particular colour in the X-ray and infrared spectra.

CCAT vs. Keck: Angular Resolution and Telescope Size

Saturday, October 8, 2011

by: Eric S. Mukherjee, Nathan Baskin

This is not the actual write up I was going to do for this week. I just thought this problem was a cool look at the many factors that go into designing a telescope and decided to write it up anyway. The other one will follow.

It's very easy to assume that a bigger telescope is always the more precise one. The question is, is this assumption always true? In this post, we examine an order of magnitude solution of problem 3 of the problem set on optics to determine whether this is the case.

The angular resolution of a telescope is the smallest angular separation at which the telescope can resolve two distinct objects. Any angle smaller than this registers as one larger object when viewed by the telescope in question. Caltech is building CCAT, a 25m telescope which observes in the 350-850 micron wavelength. We would like to know how the angular resolution compares to that of Keck, a 10m telescope observing in the near-infrared J-band at 1.25 microns.

We want to find angular resolution, which is given by the small angle approximation:

We then convert all of our known quantities into the same units. In this case, meters. As we can see, the angular resolution has a direct dependence on the wavelength, so we need only convert the minimum wavelength for CCAT:

Immediately we can see that these differ by two orders of magnitude while the diameters, 25 m and 10 m, respectively, are on the same order of magnitude. We can ignore the factor of 1.22 since it is shared by the two equations and is close to 1 and find that

We find that despite the greater diameter of CCAT, there is a difference of roughly a factor of 100 between the angular resolutions of the two telescopes. This was immediately obvious from the differences between the two wavelengths, since the diameters of the two telescopes have the same order of magnitude. Solving the equation for angular resolution we get precisely:


We find in the end that in order to determine a telescope's power we must refer to the wavelength at which it is observing as well as the diameter and that in order to have an equivalent angular resolution at the desired wavelength, CCAT would have to be a 2500 m telescope.