1. When you look at an object, your left eye sees the object from a slightly different perspective than your right. Therefore, your left eye will see the object at a slightly different position (relative to distant background objects) than your right eye -- a parallax shift. Your brain translates this parallax shift into an estimate of how far away the object is; the larger the parallax shift (that is, the more discrepancy between your left-eye and right-eye views), the closer the object must be.

The parallax method of finding the distance to a star works the same way, except that the two points-of-view are taken from opposite sides of the Earth's orbit rather than opposite sides of your head. The farther away the star, the less discrepancy there will be between the two perspectives, and the smaller the parallax shift will be. So by measuring the parallax shift of the star, you can deduce how far away it is.

3. The inverse square law says that the apparent brightness of a light source is directly proportional to its luminosity, and inversely proportional to the square of the distance to it. In math-speak:

An ordinary lightbulb has a luminosity of 100 watts, while a star like the Sun has a much higher luminosity of around 10²⁶ watts. If the lightbulb and the star were at the same distance, the star would definitely look much brighter than the bulb. However, since the bulb is millions of trillions of times closer to you than the star, it ends up looking brighter.

10. The fact is, the Sun is mostly hydrogen (75% of its mass -- most of the other 25% is helium). So the Sun's weak hydrogen lines are certainly not due to a lack of hydrogen. But the hydrogen lines in the visible part of the Sun's spectrum (the Balmer lines) will only be strong if many of the hydrogen atoms in the Sun's outer layers have their electrons in the second energy level. At a temperature of only 5,800 K, the hydrogen in the Sun's atmosphere is too cool for this to be the case. Instead, most of the hydrogen atoms in the Sun's outer layers have their electrons in the ground state, and from that position, they cannot absorb the Balmer wavelengths.

12. Since the two stars are the same distance from Earth, the inverse square law of light says that the one with the higher luminosity will look brighter. Which one has more luminosity? Well, this formula

says that the luminosity of a star depends on both its radius and its surface temperature. We are told that both stars have the same radius, but (following the rules of blackbodies) the blue star must have a higher surface temperature than the red. Therefore the blue star will have the higher luminosity, and will consequently look brighter than the red star as seen from Earth.

13.

Technically, the question asks you to sketch the position of "red giants," not "giants." Red giants are the coolest giant stars (the ones that look reddish). They would be found on the right side of the giant region in my sketch.

16. A star's luminosity class can be determined by the width of the hydrogen absorption lines in its spectrum. Main sequence stars (luminosity class V) have the broadest hydrogen lines, giant stars (classes II, III, and IV) have less broad lines, and supergiant stars (classes Ia and Ib) have the narrowest lines.

18. The mass-luminosity relation says that, for main sequence stars, high-mass stars are always more luminous than low-mass stars. There is no real relationship between mass and luminosity for giant stars, supergiant stars, or white dwarfs.

26. Use the parallax formula:

33. We are told that stars A and B look equally bright. So A must have the higher luminosity, to make up for the fact that it's farther away. You can use the inverse square law of light to determine how many times more luminous it is:

A is therefore 16 times more luminous than B.

34. You can use the inverse square law formula to solve this problem, but it's probably easier to just reason out the answer. Star D has the same luminosity as star C, but since it's closer, it must look brighter. In fact, D is 3 times closer than C, so it will look 3² = 9 times brighter.

46. Use the formula that relates luminosity, radius, and temperature:

Rigel is therefore about 99 times larger than the Sun. (The way I solved the problem above is slightly different than how we did it in class. Use whichever way you think is easier.)

50. First of all, I have to say that Zubeneschamali is one of my two favorite star-names. The other one is Zubenelgenubi. Both stars are in the faint constellation of Libra, the Scales. But I digress....

To answer this question, you can use Figures 19-14 and 19-15, on page 438 of your textbook. Here they are:

Locate where the dot representing Zubeneschamali would be plotted on Figure 19-14. With a spectral type of B8 (8/10 of the way from B0 to A0), and a luminosity of 130 L_o (just above 10² L_o), the star's dot should go approximately where the dot marked "Regulus" is. Look up to the top of the graph to read off the star's surface temperature: about 20,000 K.

Now plot Zubeneschamali's dot on Figure 19-15, using 20,000 K for the horizontal coordinate and a smidgen above 10² L_o for the vertical coordinate. The dot ends up a little bit above the diagonal line marked "1 R_o", so Zubeneschamali's radius must be slightly larger than the Sun's.

Alternatively, once you've used Figure 19-14 to determine that Zubeneschamali's surface temperature is approximately 20,000 K, you can use the following formula to calculate its radius:

54. a) Use Newton's version of Kepler's Third Law:

The sum of the stars' masses is therefore 40 M_o.

b) We are told that the mass of one star is 4 times the mass of the other, and we know from the previous part of this problem that the two masses add up to 40 M_o. A bit of playing around with numbers reveals that the stars' masses must be 32 M_o and 8 M_o.

You can also use algebra to find the two masses, as follows:

Last edited 01 Mar 04 M. A. Weinstein.