Introduction to Astronomy - Quiz with Solutions 4

Quiz 1 Solutions Name Last 1. 2. 3. 4. (please print) First Be sure to attempt all problems. The point values of the problems are indicated at the top of each page. Closed book exam; you may use one page of notes. Wherever possible, try to solve the problems using general analytic expressions. Plug in numbers only as a last step. Problem Grade 1 2 3 4 Total Grader APPROXIMATE VALUES OF USEFUL CONSTANTS Constant c (speed of light) G (gravitation constant) k (Boltzmann’s constant) h (Planck’s constant) mproton eV (electron Volt) M� (solar mass) L� (solar luminosity) R� (solar radius) σ (Stefan­Boltzmann cons) Å(Angstrom) km (kilometer) pc (parsec) kpc (kiloparsec) Mpc (megaparsec) year day AU 1� (arc minute) 1�� (arc second) cgs units 3 × 1010 cm/sec 7 × 10−8 dyne­cm2 /g2 1.4 × 10−16 erg/K 6.6 × 10−27 erg­sec 1.6 × 10−24 g 1.6 × 10−12 erg 2 × 1033 g 4 × 1033 erg/sec 7 × 1010 cm 6 × 10−5 erg/cm2 ­sec­K4 10−8 cm 105 cm 3 × 1018 cm 3 × 1021 cm 3 × 1024 cm 3 × 107 sec 86400 sec 1.5 × 1013 cm 1/3400 rad 1/200, 000 rad mks 3 × 108 7 × 10−11 1.4 × 10−23 6.6 × 10−34 1.6 × 10−27 1.6 × 10−19 2 × 1030 4 × 1026 7 × 108 6 × 10−8 10−10 103 3 × 1016 3 × 1019 3 × 1022 3 × 107 86400 1.5 × 1011 1/3400 1/200, 000 units m/sec N­m2 /kg2 J/K J­sec kg J kg J/sec m J/m2 ­sec­K4 m m m m m sec sec m rad rad Problem 1 (30 points) Short Problems in Gravity (a) Find the gravitational acceleration due to the Sun at the location of the Earth’s orbit (i.e., at a distance of 1 AU). F = ma = GM� m r2 GM� r2 7 × 10−11 N­m2 /kg2 · 2 × 1030 kg a = (1.5 × 1011 m)2 a = 0.006m/s2 a = (b) Show that for a small mass, m, orbiting a much larger mass, M , in an eccentric orbit, that the area swept out per unit time by a line joining the large and small mass is a constant around the orbit (i.e., Kepler’s 2nd law). Hint: use the appropriate conservation law. 1 1 dA � r2 dθ + rdθdr 2 2 1 We note that the 2 rdθdr term is negligible. dA dt dθ dt dA dt dA dt where L 2m 1 2 dθ r 2 dt L = mr2 1 2 L = r 2 mr2 L = 2m = is a constant, given conservation of momentum. (c) The orbits of several stars have been measured orbiting the massive black hole near the center of our Galaxy. One of them has an orbital period of 15 years, and the orbital radius is 0.12 second of arc (as seen from the Earth). Take the distance to the Galactic center to be 8 kpc. Compute the mass of the black hole, starting from F = ma. Express your answer in units of the Sun’s mass (M� ). (Assume that Newton’s law of gravity is applicable for orbits sufficiently far from a black hole, and that the orbiting star satisfies this condition. r � 8000 pc ∗ tan 0.12�� � 1.4 × 1014 m; P � 4.73 × 108 s � 15 yrs � �2 GMbh 2π = P r3 Mbh � 4π 2 (1.44 × 1014 m)3 � 3.7 × 106 M� (7 × 10−11 N­m2 /kg2 )(4.73 × 108 s)2 Problem 2 (25 points) Geometric and Physical Optics (a) Find the theoretical limiting angular resolution of a commercial 8­inch (diameter) optical telescope being used in visible (at λ = 5000Å = 500 nm = 5 × 10−5 cm = 5 × 10−7 m). θ = 1.22 5 × 10−5 cm λ = 1.22 = 2.46 × 10−6 radians = 0.49 arcsecs D 8 × 2.54 cm (b) A large ground­based telescope has an effective focal length of 10 meters. Two astronomi­ cal objects are separated by 1 arc second in the sky. How far apart will the two corresponding images be in the focal plane? s = f θ = 1000 cm × 1 radians = 0.005 cm = 50 µm 2 × 105 (c) A collimated light beam propagating in water is incident on the surface (air/water inter­ face) at an angle θwater with respect to the surface normal. If the index of refraction of water is n = 1.3, find an expression for the angle of the light once it emerges from the water into the air, θair . What is the critical angle, i.e., θwater = θcrit , such that the light will not emerge from the water? θair By Snell’s Law: nwater sin θwater = nair sin θair = sin−1 (nwater sin θwater ) = sin−1 (1.3 sin θwater ) For critical incidence, the angle of refraction, θair , = 90◦ So, using Snell’s Law, 1.3 sin θwater = 1 × sin 90◦ = 1 � � 1 −1 = 50.28◦ Therefore, θwater = θcrit = sin 1.3 Problem 3 (25 points) Short Answer Questions (a) Use the Bohr model of the atom to compute the wavelength of the transition from the n = 100 to n = 99 levels. [Useful relation: the wavelength of Lα (n = 2 to n = 1 transition) is 1216Å.] −13.605eV n2 ΔE = 13.605eV(1/992 − 1/1002 ) = 2.76 × 10−5 eV = hν c = λν hc λ = � 4.5 × 108 Å � 4.5 cm ΔE � � 1 1 1 1 = − 2 2 λ λ0 n f ni En = ­or­ But we are given λ = 1216Å when ni = 2 and nf = 1. This implies that � � 1 1 1 1 , or λ0 = 912 Å = − λ0 12 22 1216 ˚ A Thus, 1 1 = ˚ λ 912 A � 1 1 − 2 99 1002 � or, λ � 4.5 × 108 Å � 4.5 cm. (b) A galaxy moves directly away from us with speed v, and the wavelength of its Hα line ˚ The rest wavelength of Hα is 6565˚ is observed to be 6784A. A. Find v. λ � λ0 (1 + v/c) ˚ and λ0 = 6565˚ where λ = 6784A A. Rearranging, 6784 − 6565 v λ − λ0 � � ⇒ v � 0.033 c c λ0 6565 (c) A star is at a distance from the Earth of 300 pc. Find its parallax angle, π. D = 1pc/π �� π �� = 1pc/300pc π �� = 0.003�� Problem 4 (20 points) Luminosity and Magnitudes (a) A particular star has an absolute magnitude M = ­7. If this star is observed in a galaxy that is at a distance of 3 Mpc, what will its apparent magnitude be? What is the distance modulus to this galaxy? Given: M = −7 and d = 3 Mpc � � � � 3 × 106 d = −7 + 5 log = 20.39 Apparent Magnitude: m = M + 5 log 10pc 10 Distance Modulus: DM = m − M = 20.39 + 7 = 27.39 (b) The differential luminosity from a star, ΔL, with an approximate blackbody spectrum, is given by: 8π 2 c2 R2 ΔL = 5 hc/(λkT ) Δλ − 1] λ [e where R is the radius of the star, T is its effective surface temperature, and λ is the wave­ length. ΔL is the power emitted by the star between wavelengths λ and λ + Δλ (assume Δλ � λ). (i) The star is at distance d. Find the star’s spectral intensity I(λ) at the Earth, where I(λ) is defined as the power per unit area per unit wavelength interval. (ii) Without working out all the arithmetic, show how you could determine the temperature T of the star by measuring the spectral intensity at two different wavelengths, e.g., I(λ1 ) and I(λ2 ). Neither the quantity d nor R should appear in your expression for T . (i) I(λ) = 1 ΔL 2πc2 R2 = 4πd2 Δλ λ5 [ehc/(λkT ) − 1]d2 (ii) The two spectral intensities at two wavelengths, λ1 and λ2 , are given by 2πc2 R2 λ15 [ehc/(λ1 kT ) − 1]d2 2πc2 R2 I2 = I(λ2 ) = 5 hc/(λ2 kT ) − 1]d2 λ2 [e I1 = I(λ1 ) = So, I1 λ5 [ehc/(λ2 kT ) − 1] = 25 hc/(λ1 kT ) I2 − 1] λ1 [e T is the only unknown in this equation and can therefore be determined.