Happy Halloween

It's a ghost!

This is actually known as the Ghost of the Cepheus Flare. The Cepheus Flare is a molecular cloud about 1200 lightyears away. This "ghost" is actually a 2-lightyear-wide dust cloud at the very edge of the flare, seen by reflected starlight. The bulges in the cloud's shape are nebula regions; the section of dust on the far right of the image is collapsing in on itself, and is likely in the first stages of forming a binary star.

Can You See the Seahorse?

Above is a picture of a seahorse in the Large Magellanic Cloud, a nearby satellite galaxy of the Milky Way. What looks to us like a seahorse (an example of pareidolia, the same phenomenon that lets us see constellations in groups of stars and the man in the moon) is actually a pillar of dust 20 lightyears long.

The seahorse is very close to another well-known object in the LMC, the Tarantula Nebula:

The Tarantula Nebula is very energetic, and is in the process of forming a star cluster, the center of which is just above the cutoff of the Seahorse image. As stars form in the cluster, the resulting solar wind will slowly erode the Seahorse dust pillars over the next million years or so. 

I think it is often pictures like this that first get people interested in astronomy. There are several reasons for this. The first is rather obvious: they look really cool. Pictures of astronomical objects can be really artistic, and sometimes breathtaking in their patterns and complexity. This can be enhanced when one is able to see something familiar in the picture, like the Magellanic Cloud's Seahorse. 

But then there's another dimension. It's fun to imagine a seahorse in the picture, but it's even more interesting to learn what the image really is. A towering pillar of smoky dust, billowing and reforming as the nearby nebula grows and grasps it ever more tightly. It's a monument taller than the highest skyscraper we will ever build, and it's over 150,000 lightyears away, a distance greater than we could possibly imagine. Yet through the technology of the Hubble Telescope, we are able to gaze into the deep, dark night, and find a seahorse sculpted by gravity and eons.  

Nuclear Winter

So, in Battlestar Galactica, the Cylons have just launched a nuclear attack on the Twelve Colonies, including the Earth-like world of Caprica. This is, of course, terribly sad, but we will assuage our grief with science.

Each bomb the Cylons use kicks up a lot of dust into the planet's atmosphere. And increasing the amount of optically thick dust in the atmosphere reduces the sunlight that reaches the planet's surface, which results in lower surface temperatures. Can we figure out the relationship between the amount of dust in the atmosphere and the cooling of the surface?

OK, so suppose the Cylons drop N bombs, and each bomb moves a mass M of dust into the atmosphere. Each dust particle has a a mass density rho and a radius r (assume a spherical dust particle).

First, we should find the total number of particles moved to the atmosphere. Note that



where mp is the mass per particle and Vp is the volume per particle. So we can write



where the units of mp are kg/particle. NM is the total mass of particles. So if we take



we have units of particles, so Np is the total number of particles.

Next we need to find the fraction of flux that gets through the dust and to the planet's surface. We'll assume that



is the surface area of a particle of dust, and



is the surface area of the planet, with R being comparable to the Earth's radius. So then we have



for the ratio of the flux that reaches the planet's surface to the flux that reaches the planet. Since temperature is related to flux by



we can find the factor relating the original temperature to the new with



The only problem with this formulation is with the flux ratio. When the second term is greater than one, which occurs for large numbers of bombs or masses of dust, the result is a negative ratio, which doesn't make sense.

Worked out with David and Mee.

Also, on a semi-related note, if you are a fan of nuclear winter and/or post-apocalyptic fiction, I recommend The Road by Cormac McCarthy.

The Big Bang

If you watch the TV show "The Big Bang Theory," you may be somewhat familiar with this song. 

One quibble, though. The members of the band Barenaked Ladies appear to be proponents of the Big Crunch hypothesis for the Ultimate Fate of the Universe, but all current evidence shows that the universe is not only expanding, but accelerating in its expansion. So, if things keep going the way they are, the universe won't reconstitute itself into a singularity, but will keep expanding outward. Most of the universe will move so far away so fast the it will pass beyond our local event horizon (its light won't be able to reach us because it is expanding away too quickly).

The most commonly held hypothesis for the fate of the universe is the Big Freeze,  or heat death. This occurs when the universe achieves maximum entropy, leaving the universe bereft of free energy capable of doing work. Sadly, this would put an end to most of the interesting parts of the universe.

That could be a bit depressing, so let's end with something more optimistic:

The Distance to the Sun

The above image is a series of pictures of Mercury passing between the Earth and the Sun. It was taken from a satellite in polar orbit around the Earth. Is there any way we can use information from this picture, and some other basic astronomy knowledge, to find the distance from the Earth to the Sun?

The above diagram can help us find everything we need to know. We want to find a, a variable that stands for the astronomical unit, or the distance from the Earth to the Sun. Supposing we have only the orbital periods of the Earth and Mercury, and the two pictures above, can we find it? Kepler's Laws and trigonometry say yes.

First, we can use the picture of Mercury's motion to find the angle alpha in the upper left of the diagram. If you look at the picture, you may notice that Mercury's "height" relative to the Sun oscillates up and down. This is due to the movement of the satellite around the Earth: when it's near the North Pole, Mercury's position relative to the Sun appears different compared to when viewed from the South Pole. The angle between Mercury's high and low points is alpha.

We know that the height of the Sun (the diameter) is 0.5 degrees relative to the total sky. By comparing Mercury to the Sun in the above picture, one can estimate Mercury's oscillation in degrees. It's not very big: about 0.0005 degrees. This is alpha. In radians, alpha is 0.00000876. 

OK, what next? Take a look at the top triangle, whose angles are alpha, beta, and pi minus theta. Since the sum of these angles is equal to pi (using radians), we have

Using trigonometry and small-angle approximations, we have



Solving for delta-a, we get



Since the total distance a is the sum of delta-a and the distance from Mercury to the Sun, we can write



So, how does this help us? Fortunately, Kepler's Third Law of Planetary Motion says that the square of a planet's orbital period is proportional to the cube of the planet's semimajor axis, which we call a. So we have



We know the radius of the Earth (using 6300 km), and that the Earth's orbital period is 365 days, so if we also happen to know that Mercury's orbital period is 87 days, we can solve for a. Though I must warn you, it involves some pretty hideous algebra. Do not attempt at home.

But, if you do, you get around 200 million kilometers. How much is it really? About 150 million kilometers. Cool.

Given this, we can use a refined version of Kepler's Third law to find the mass of the Sun:



Assuming we're using Earth, we know everything in this equation now except for the mass of the Sun (We can neglect the Earth's mass, as it is ridiculously tiny compared to that of the Sun). Using the values we've calculated, I get



for the mass of the Sun. What is it really?



So we end up a bit under on our estimate, coming up with only about 30% of the mass of the Sun.

Worked on with Mee and David.

The Sun

For your edification and enjoyment. And now, just for your enjoyment:

More on Blackbodies

Previously on Looking Up, we explored the Blackbody properties of a Y Dwarf orbiting a Sun-like star, and we found the peak wavelength emitted by the Y Dwarf given its temperature. From this, we were able to calculate the photons per second per area from the Y Dwarf that reached an observer 30 lightyears away. But many questions were left unanswered: how many photons came from the Sun-like star? What is the flux ratio of the Y Dwarf to this star? Tonight, the thrilling conclusion: