1	HW #8 Solutions Question 1 (1 point) Which of the below sequences shows objects with increasing densities? a.Earth -- Sun -- White Dwarf -- Black Hole -- Neutron Star b.Red Giant -- White Dwarf -- Neutron Star -- Black Hole c.Earth -- Neutron Star -- White Dwarf -- Black Hole d.Sun -- White Dwarf -- Earth -- Black Hole -- Neutron Star e.White Dwarf -- Red Giant -- Sun -- Earth -- Neutron Star This should be something you know from the book and lectures. Density, recall, is a mass divided by a volume. If you know the masses and the sizes of these objects, even roughly, it should let you easily put them in density order.
2	HW #8 Solutions Question 2 (1 point) We believe that we receive hundreds of pulses per second from so-called "millisecond pulsars" because the neutron star emitting them: a.Is spinning hundreds of times per second. b.Is actually pulsing (due to thermal instabilities) and changing its size hundreds of times per second. c.Is in a binary system orbiting another star hundreds of times per second. d.Is seen through a certain kind of nebula that causes "scintillation". e.Is surrounded by an accretion disk which has material orbiting the star hundreds of times per second. According to the “lighthouse” model of pulsars, we’re seeing the neutron star’s magnetic north (or south) pole once per spin, and the rotation axis and the magnetic axis are offset.
3	HW #8 Solutions Question 3 (1 point) Assume that relatively young white dwarfs and neutron stars have similar temperatures (both very hot, although cooling at different rates). White dwarfs are hard to spot because the radiating surface is small (about the size of Earth). Neutron stars are even smaller (for this questions assume 10 km radius). How much fainter are they? a.About 4 million times fainter. b.About 400000 times fainter. c.About 40000 times fainter. d.About 4000 times fainter. e.about 400 times fainter. We know that the luminosity of a star depends on its temperature to the 4^th power times a constant times its surface area. If temperature is the same, this question really boils down to the relative surface areas. Ballpark, a neutron star has a radius = 10km, while a white dwarf is about the size of the Earth (radius = 6378 km). Surface area depends on the radius squared. Squaring 6378/10 gives about 400,000.
4	HW #8 Solutions Question 4 (1 point) The use of Cepheid variable stars to find distances relies on the fact that a. their luminosities are related to their ages. b. their luminosities are all the same. c. their radial velocities are related to their periods. d. their luminosities are related to their periods. Question 5 (1 point) You observe two Cepheid variable stars, A and B, which have the same average apparent magnitude. Star A brightens and dims with a period of 5 days, while star B has a period of 18 days. Which star is closer to Earth? a.Star A b.Star B c.Star A and B are at the same distance d.You cannot tell from this information Make the following argument to solve this one easily. The period-luminosity relationship is precise, but precision is not needed here. The star with the longer period will be more luminous, so that is star B. But we want to know the distance. They both appear to have the same average apparent magnitude. If one is more luminous than the other, the more luminous one must be farther way – if it wasn’t, it would appear brighter. Therefore the less luminous star is closer, and that is star A.
5	HW #8 Solutions Question 6 (1 point) Which of the following is not a characteristic of the stars of the disk component of our galaxy? a. circular orbits b. random highly inclined orbits c.higher metal abundance d. young stars e. star formation regions Question 7 (1 point) If the sun is 5 billion years old, how many times has it orbited the galaxy? Assume a circular orbit for the sun around the center of the galaxy, and look up in the text its orbital speed and distance from the galactic center. a. 5 times b. 5 billion times c. 200 times d. 20 times e. 5000 times Approach this problem like a time, rate, distance problem to figure out an orbital period, then see how many go into 5 billion years. If the orbital speed is 220 km/s, and the oribital radius is 8.5 kpc, we can get the period (time): time = speed/distance and we must convert. 8.5 kpc = 8500 pc x 3.1 x 10 ¹³ meters/pc = 2.6x10¹⁷ km. So time = 2πx2.6x10¹⁷ km/220 km/s = 7.5x10¹⁵s. What is that in years? There are about 31 million seconds per year, so this period is 7.5x10¹⁵/3.1x10⁷ = 2.4x10⁸ years, or about 240 million years. 5 billion divided by 240 million is 5x10⁹/2.4x10⁸ = 20.8.
6	HW #8 Solutions Question 8 (1 point) If all the mass in our galaxy were centrally concentrated, we'd expect velocities to fall with increasing distance according to Kepler's laws. This is not seen in the disks of spiral galaxies. Galactic rotation curves appear "flat" with increasing distance. This must be due to a. The gravitational influence of massive globular clusters in the halo. b. The difficulty in measuring velocities of stars in the galactic disk because of all the gas and dust. c. The fact that Kepler's laws do not apply over the 25 kpc size of the Milky Way galaxy because the effect of gravity travels only at the speed of light. d. The Spiral density wave traveling at a different speed than the stars in the disk. e. The gravitaional influence of "dark matter" in the halo. Question 9 (1 point) The location of the sun relative to the center of our galaxy was FIRST CORRECTLY determined by a. the relative brightness of the visible Milky Way in different directions. b. the positions of the globular clusters and the properties of the variable stars in them. c. the position of supernova. d. radio maps of the galaxy. Question 10 (1 pt) If you find a metal-rich star it is more likely to be a member of which Population? A. Population I (the disk)
7	HW 8 Solutions Question 11 (1 point) One fundamental requirement for an object to be a good tracer (i.e. marker) of the galaxy's spiral arms is that it be a.moving with small radial velocity (relative to earth). b.moving with large radial velocity (relative to earth). c. a short lived and therefore young object. d. a very long lived and therefore probably old object.
8	HW #8 Solutions Question 12 (1 point) The black hole in the center of our Milky Way galaxy seems to be about 2.6 million times the mass of the sun according to the text. (Recently updated to 3.6 million solar masses, so use the new value here.) What is the radius of its event horizon? Think about how big this is compared to other astronomical objects (e.g., Earth, the sun, the solar system, the distances between stars). a.About a billion km. b.About 5.5 thousand km. c.About 5.5 million km. d.About 11 thousand km. e.About 11 million km. Use the equation for the even horizon radius, R = 2GM/C² and plug in the right numbers. Or remember that the horizon for a solar mass black hole is 3 km, and multiply by 3.6 million.
9	HW 8 Solutions Question 13 (2 points, extra credit). The answer is tidal forces which you can get easily if you read the story “Neutron Star.”