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SCCC 116
Homework 5 - Due 16 March 2006
Prof. Christina Lacey
R&D: 23-5, 23-14, 23-15
P: 23-1, 23-4, 23-12

R&D 23-5
How are Cepheid variables used to determine distances?

Cepheids have the unique property of a relationship between their period of pulsation and their luminosity. By observing the Cepheid and determining its period of variation, its true (absolute) luminosity is known. Comparing the Absolute luminosity with the apparent luminosity allows the distance to be determined.

R&D 23-14
What do the red stars in the Galactic halo tell us about the history of the Milky Way?

The stars in the halo have orbits that form a spherical halo around the Galaxy. This tells us the original shape of the cloud of gas that was the galaxy before stars formed. The entire halo appears to rotate, which further tells us that this cloud was rotating. The formation of a disk is the result of the cloud having some initial rotation.

R&D 23-15
What does the rotation curve of our Galaxy tell us about the Galaxy's total mass?

Studies of the orbits of stars and clouds around the Milky Way allow us to calculate the mass interior to the stars using Kepler's third law. The rotation curve of the Milky Way Galaxy provides strong evidence for dark matter in the Milky Way. The matter we can see is half of the mass that Kepler's law finds.

P 23-1

By definition, an object with a diameter of 1 AU is 1 arcsecond is at a distance of 1 pc: $d (pc) =\frac{1}{\alpha (^{\prime\prime}})$

So using the ratio: an object 200 AU in diameter will be 200 pc in distance and cover 2 $^{\prime\prime}$. If the Andromeda galaxy were solar system-sized it would have an angular size of 2 $^{\prime\prime}$, which is 10,800 times smaller than the 6$^{\odot}$ angular size of the galaxy. Thus, Andromeda must be much larger and/or much farther away than a 200 AU diameter sized object at 100 pc.

P 23-4
Inverse square law: brightness $\propto \frac{Luminosity}{d^2}$. So if a Cepheid is 100 times brighter than an RR Lyra variable, it should be visible at 10 times the distance of the RR Lyra. So if you place the Cepheid 10 times farther away, the Cepheid will be 100 times fainter and equal the RR Lyra's apparent brightness.

P 23-12

The difference between the speed of the star and the density wave is 110 km/s, which is roughly half the speed of the Sun about the Galaxy. The Sun takes 225 million years to orbit the Galaxy once. So the speed of the star catching up to a density wave is half that of the Sun, but there are two arms to cross, so it will take half as much time to cross one density wave. In 4.6 billion years, the Sun has orbited 20.4 times around the center of the galaxy. A star approaching a density wave at half the speed of the Sun around the galaxy will take half as much time, but there are two density waves. So, the star crosses the density waves roughly 20 times in 4.6 billion years.

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Christina Lacey 2006-04-05