Comments on Homework 4

Part A

A1.	d. A straightforward question about black-body radiation and Wien's law in particular.
A2.	b.
A3.	The stellar wavelength recorded here on Earth is longer than the wavelength at the star (486.3 > 486.1). A radial velocity of recession lengthens the wavelength recorded.
A4.	b.
A5.	d, e, a, c, b. Would the ordering change if these stars were viewed from a different location?
A6.	Betelguese : M I Barnard's star: M V Vega: A V
A7.	To solve this problem we need the luminosities of Vega and Deneb. Consulting the figure, we see LV ~ 100 L LD ~ 100, 000L and LD ~ 1000LV The relation we need is where B is the brightness, L is the luminosity, and d the distance of a star. For a pair of stars such as Vega (V) and Deneb (D), we have Rearranging this, we get and
A8.	O5V, A2V, G2V, G5V, KOII, and M3III. The sequence OBAFGKM is one of decreasing temperature. The numerals 09 order stars within a spectral class by decreasing temperature. For this class, we assume stars of a given spectral class but different luminosity class have the same temperature.
A9.	c.
A10.	c.
A11.	c.
A12.	d.
A13.	Again, B L/d2. Here, the stars have the same L. Therefore, B l/d2. Since p=l/d, where p is the parallax, we may write B p2. Then, X is 25 times brighter than Y.
A14.	Hottest = O Reddest = M Brightest = O Most Massive = O Solar-like = G
A15.	d.
A16.	See Seeds' Table. Read tables such as this with care. He lists the elements by atomic number (the number of protons in the nucleus) not by their concentration in the Sun's atmosphere. The five most common elements are H, He, O, C, N with Ne where I list them by their concentration. Note the dominance of H and He.
A17.	Most answers were poor paraphrases of the text.
A18.	The mass ratio is given by the ratio of the distances of the stars from the center of mass: the more massive star is closer to the center of mass.
A19.	Assume the mass-luminosity relation for main sequence stars: L M4 (or M3.5) For LM4, Then, L=256L.
A20.	c.

Part B

B1.		Note that the two samples - the nearest and the brightest stars - fill up quite different parts of HR-diagram. Why? Red giants are very luminous but rare. Red dwarfs are of low luminosity but common. The luminosity difference between a giant and a dwarf is so great that a quite distant giant is brighter than the nearest dwarf. Same is true for hot stars like Rigel and cool dwarfs like Barnard's star.
B2.	a.	Very few complete answers. Many spoke of observing a shift of a nearby star as the Earth orbited the Sun. But a shift of the star relative to what? Many referred to the parallax angle at the star but few mentioned how this angle is obtained. I repeat here a piece given at the end of Classnotes #2. The angular shift of the nearby star is measured with respect to very distant stars close on the sky to the nearby star. In the diagram, the three arrowed lines point to the same very distant star (hence, the lines may be drawn as parallel). At E1, we measure the angle between this distant star and the nearby star of interest - angle is marked on the figure. Six or so months later, we observe the same pair of stars. Now the nearby star lies to the opposite side of the distant star -- angle is marked . Since the three arrowed lines are parallel, the angles at the star are as marked, and (OK?). The parallax angle is p = ( + )/2. Hence, p is calculable from the measurements of a and b.
	b.	A parsec is the distance at which a star has a parallax angle of one second of arc.
	c.	The angular shifts of nearby stars (relative to distant stars in the same direction) are all less than about 1 second of arc (see Sample one in Classnotes 11). On Earth, the atmosphere blurs star images (see Seeds on seeing). What should be points of light become blurred roughly circular patches of 1 or even more seconds of arc in diameter. The parallax is found by measuring a shift of less than 1 second of arc on photographs taken a few months apart when the image sizes may be different. In space, there is no atmosphere. The stellar images are point-like or very nearly so (remember diffraction!) and, therefore, it is much easier to measure small shifts of the images. Consider, for example, these two images For which image can you accurately define the center?
B3.	a-c.	See Seeds.
	d.	This question refers to the method of spectroscopic parallax. The main sequence shows that there is a tight relation between spectral class (color, surface temperature) and luminosity. Then, a determination of spectral class and a demonstration that the star belongs on the main sequence (luminosity class=V) suffice to infer its luminosity. This with the measured brightness gives the distance ( parallax) from BL/d2.
B4.	a.	Own words? Imagine you are explaining these different kinds of binaries to Mom and Dad.
	b.	The period is about 4 not 2 days. Note there are two locations in an orbit where the radial velocities of the two stars are the same. The period is the time taken for the stars to go from (i) through (ii) back to (i). The top spectrum with the single line may correspond to (i). The next time there is a single line corresponds to (ii). The sequence runs out before the next occurrence of (i) but it is clearly closely approached by the last spectrum in the sequence.
	c.	What drives the stars to orbit each other? The gravittional force between them. Recall that this scales as the product of the two masses divided by the square of a distance between them. In order for a pair of stars to be seen as a visual binary they must be separated by a large distance. Stars comprising a spectroscopic binary will be much closer together than in the case of a visual binary. Then, the gravitational forces and, hence, the accelerations of stars in a visual binary are much weaker than in a spectroscopic binary. This means the stars move more slowly in a visual binary. In addition, the orbits are larger. These two factors mean that stars in a visual binary take a long time to complete an orbit. So long that it's convenient to express the period in years. On the other hand, the stars in a spectroscopic binary driven by larger gravitational forces complete an orbit in the matter of days.
B5.	a.	Atoms and molecules in a flame move into the cooler surrounding air. Collisions with air molecules lead to exchange of energy, cool the hot atoms and molecules, and erase the distinction between them and the air molecules. In the Sun, hot gases are subject to influences: the gravitational pull of the Sun, and collisions with atoms and electrons from all sides. The net effect is that an atom or electron that attempts to disperse (i.e., move outward) is decelerated by gravity, and opposed by collisions with other particles.
	b.	See Classnotes 13. The question called for a discussion of the competetition between the repulsive long-range electrostatic force between nuclei and the attractive short-range nuclear force between nuclei.
B6.	a.	Classnotes 13
B7.	b.	Classnotes 13.