ASTR 1040 Spring, 2005
1. How much less would you weigh on the summit of Mt. Everest (altitude 8.84 km) than at sea level?
Use the law of universal gravitation, noting that in this problem all of the terms except d, the distance between you and the center of the Earth, are constant. Therefore the ratio of your weight at the summit of Mt. Everest to your sea-level weight is given by the inverse ratio of the squares of the distances, which are the Earth's radius of 6367 km when you are at sea level, and the earth's radius plus 8.84 km when you are on Mt. Everest. The ratio is (6367/6375.84)2 = 0.99723.
2. A binary star system is observed. The orbital period is P = 17.28 years, and the semimajor axis, in angular units, is 0.44". The distance to the system is 22 pc (recall that 1 pc = 206,265 AU). The distance from star 1 to the center of mass of the system is 0.273 of the total distance between the two stars. What are the masses of the individual stars?
You need to use Kepler's third law for this, which is simplest to do if you keep things in solar units: period in years, semimajor axis in AU, and masses in solar mass units. Then the law is (m1 + m2)P^2 = a^3. You are given values for P and a, but must convert units in order to solve for the sum of the masses. To convert a from arcseconds to AU, first convert the angular value of a to radians: theta = 0.44"/[(60"/')(60'/°)(57.30°/rad)] = 2.13 X 10^-6 rad; then convert the distance (22 parsecs) to AU (22 X 206,265 = 4.54 X 10^6. Now use the small-angle approximation: a = (theta)d = (2.13 X 10^-6)(4.54 X 10^6) = 9.67 AU. The period is 17.28 years, so now we can find the sum of the masses: (m1 + m2) = a^3/P^2 = (9.67)^3/(17.28)^2 = 3.03 solar masses. The ratio of the masses is known from the relative distances of the two stars from the center of mass: m1/m2 = r2/r1 = (1 - 0.273)/(0.273) = 2.66. Hence we have the sum of the masses (3.03) and the ratio (2.66), and we can solve for the individual masses: m1 = 2.20 solar masses and and m2 = 0.83 solar masses.
3. Suppose a star has a luminosity of 780 times the solar luminosity, and its spectrum shows a wavelength of maximum emission of 1074Å. What is the radius of the star, in both basic units (meters) and in solar units?
Write the Stefan-Boltzmann law for the Sun and for the star, and then divide and solve for the ratio of radii: R*/RS = (L*/Ls)^0.5 (Ts/T*)^2. The ratio of luminosities is given, but we must use Wien's law to find the temperature of the star: T = 0.0029/(1074 X 10^-10 m) = 27,000 K = 4.67 times the Sun's temperature of 5780 K. Now we can substitute into the relation above: R*/Rs = (780)^0.5(1/4.67)^2 = 1.28. The star has a radius 1.28 times that of the Sun, or 8.91 X 10^8 m.
4. The energy needed to ionize a hydrogen atom (from its ground state) is 2.18 X 10-18 J. What wavelength of photon does this energy correspond to? What kind of telescope is needed to observe at that wavelength? How fast would an electron have to be moving to have this much kinetic energy?
To find the photon wavelength, use the relation E = hc/lambda, or lambda = hc/E. Substituting, we get lambda = (6.63 X 10^-34)(3 X 10^8)/(2.18 X 10^-18) = 9.12 X 10^-8 m = 912Å. This is in the far-ultraviolet part of the spectrum, so an ultraviolet telescope would be needed. To find the electron speed corresponding to this energy, recall that kinetic energy KE is given by mv^2/2. Solving for the velocity v, we find v = (2KE/m)^0.5 = [2(2.18 X 10^-18)/(9.11 X 10^-31)]^0.5 = 2.19 X 10^6 m/sec = 2190 km/sec.
5. The Rydberg formula for the spectral lines of hydrogen is
1/lambda = R(1/n^2 - 1/m^2), n = 1, 2, 3, ...; m = n+1, n+2, ...∞.
First three lines and ionization limit for the Paschen series:
n = 3, m = 4: 1/lambda = R(1/9 - 1/16) = 5332.64 cm^-1; lambda = 18,752Å
n = 3, m = 5: 1/lambda = R(1/9 - 1/25) = 7800.89 cm^-1; lambda = 12,819Å
n = 3, m = 6: 1/lambda = R(1/9 - 1/36) = 9141.67 cm^-1; lambda = 10,939Å
n = 3, m = ∞: 1/lambda = R(1/9 - 0) = 12,189.2 cm^-1; lambda = 8204Å
First three lines and the ionization limit for the Brackett series:
n = 4, m = 5: 1/lambda = R(1/16 - 1/25) = 2468.25 cm^-1; lambda = 40,515Å
n = 4, m = 6: 1/lambda = R(1/16 - 1/36) = 3809.03 cm^-1; lambda = 26,253Å
n = 4, m = 7: 1/lambda = R(1/16 - 1/49) = 4617.47 cm^-1; lambda = 21,657Å
n = 4, m = ∞: 1/lambda = R(1/16 - 0) = 6856.25 cm^-1; lambda = 14,585Å
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