AST111 PROBLEM SET 4 SOLUTIONS Homework problems 1. On Astronomical Magnitudes You observe a binary star. Ordinarily the binary has a magnitude of 10 and this is due to the brightness of both stars. The smaller star passes behind the larger one and you observe a magnitude of 11. What are the magnitudes of both stars (individually)? Answer: We call the flux of the larger one A and the flux of the fainter one B. The larger one has a magnitude of 11, as it is seen during secondary eclipse. The magnitude of both together m A = 11 m both =.5 log 10 (A + B) + constant = 10 The magnitude of only the larger one Subtract m A =.5 log 10 (A) + constant = 11 m both m A =.5 log 10 (A + B) +.5 log 10 (A) = 10 11 ( ) A + B.5 log 10 = 1 A The ratio depends on the difference in magnitudes ( ) 11 10 A + B = 0.4 = log.5 10 A 10 0.4 =.51 = A + B A = 1 + B A and this means that B is 1.51 times brighter than A, or B/A = 1.51. 1
AST111 PROBLEM SET 4 SOLUTIONS The magnitude of the fainter one m B =.5 log 10 (B) + constant Subtract the two magnitudes m B m A =.5 log 10 A +.5 log 10 B =.5 log 10 (B/A).5 log 10 (1.51) = 0.45 We add this to the magnitude of A to get the magnitude of B m B = 11 0.45 = 10.55. On signal to noise and sky background You plan to observe an object and you predict that you will record about 100 photons per second at your telescope from the object. However at the same time you will also receive 10000 photons per second in sky background. So you expect to detect 10100 photons per second from object + background and 10000 photons per second from the background alone. To detect the object you subtract the background signal from the total of background + source. (a) Using Poisson statistics, how long do you need to integrate to detect the object with a signal to noise of 4? The noise here is approximately that from the sky or N = 10000t = 100t 1/ where t is the integration time in seconds. The signal is only S = 100t. Setting S/N = 4 we find 100t = 4 10000t = 400t 1/ or t 1/ = 4 or t = 16 seconds. A better calculation: We recall that if you have measurements of a, b with errors σ a, σ b the error of c = a + b is σ c = σa + σb Likewise the error of d = a b is σ d = σa + σb The noise of the sky plus source is 10100t. The noise of the sky alone is 10000t. The noise of the sky + source subtracted by the sky is estimated by adding the two in quadrature or N = 10100t + 10000t t 1 100
AST111 PROBLEM SET 4 SOLUTIONS 3 The signal is still S = 100t. The signal to noise S/N = 100t 100t 1 = t 1 4 = t 1 t = 3s (b) How long would you need to integrate to detect the source at the same signal to noise of 4 if there was no background contamination? In this case S = 100t and N = 100t. The ratio S/N = 100t = 4 giving us 100t = 16 or t = 16/100 = 0.16 second. 3. Black bodies Consider two stars that have the same radius. One is 16 times more luminous than the other. (a) What is the ratio of their temperatures? (b) The fainter star s spectrum peaks at λ 1. At what wavelength does the brighter star peak? Answer: a) L = 4πR σt 4. The ratio of luminosities is 16 so the ratio of their temperatures is 1/4-th power of this or. The brighter star has twice the temperature of the fainter star. b) The Wien law states that the peak wavelength is inversely proportional to the temperature. The brighter star has a temperature that is twice the temperature of the fainter star. So it s peak wavelength is 1/ of this or λ = λ 1 /. Workshop problems 1. On Magnitudes A star like the Sun has a radius of R = 7 10 10 cm. A Jupiter-like planet has a radius of R p = 7 10 9 cm. During mid-transit the planet blocks star light. (a) Approximately what fraction of light is blocked?
4 AST111 PROBLEM SET 4 SOLUTIONS (b) Show using a Taylor series for small x and to first order that x log 10 (1 + x) x.30 It may help to know that for f(y) = log 10 y, the derivative f (y) = 1 y. with =.30. (c) In magnitudes, how much is the stellar flux deficit mid-transit? Answer: a) Depends on the ratio of areas or f = πr /(πr p) = 0.1 = 0.01. b) f(x) = log 10 (1 + x) Evaluate at x = 0 f(x = 0) = 0 Take formula for f raise to power of 10 Write 10 as e to a power Take derivative w.r.t x Solve for f (x) Evaluate at x = 0 10 f(x) = (1 + x) e log(10)f(x) = 1 + x log(10)f (x)e log(10)f(x) = 1 f (x) = 1 log(10)f(x) e f 1 (x = 0) = Now expand using Taylor series near x = 0 f(x) f(x = 0) + xf (x = 0) +... Use our formulas for f(x) and f (x) evaluated at x = 0 x f(x) c).5 log 10 (1 0.01) = 0 0.01.5 = 0.01.5.30 0.01
AST111 PROBLEM SET 4 SOLUTIONS 5 The object is 0.01 magnitudes fainter during transit.. Numbers of photons and Light buckets (a) About (to order of magnitude) how many optical wavelength photons does the Sun emit per second? The luminosity of the Sun is L = 3.8 10 33 erg/s and its effective temperature is T eff = 5780K. (b) Suppose you have a 1 cm diameter telescope in space at 1AU. How long do you need to integrate to measure the solar radiance to a signal to noise of 10 9? Again to order of magnitude. Answer a) Most of the light is coming out in the optical. Taking λ = 0.5µm = 0.5 10 4 cm as the typical wavelength, the energy of this photon is E = hν = hc/λ = 6.67 10 7 3 10 10 /0.5 10 4 = 4 10 1 erg Approximately the number of photons emitted per second is n = L/E = 10 45 s 1 This estimate could be done more accurately by taking the black body spectrum B ν and dividing by hν and integrating over all wavelengths. You need to take into account some factors of solid angle to do this correctly as you need to have the answer consistent with the stellar luminosity. b) The light collecting area is πd /4. The fraction of light collected is f = πd /4/(4πr ) = d /(16r ). At 1 AU this is f =.8 10 8. The number of photons collected per second is r = fn = 3 10 8 10 45 s 1 = 3 10 17 s 1 The signal is S = rt the noise is N = rt, the signal to noise is S/N = rt. We set this to be S/N = rt = 10 9 giving rt = 10 18. Now solve for t 3. On the diffraction limit t = 10 18 = 3s 3 10 17 s 1 On Earth the geosynchronous orbit is has a radius of 4,164 km (from the center of the Earth).
6 AST111 PROBLEM SET 4 SOLUTIONS (a) For a satellite at the radius of this orbit, what sized mirror is needed to resolve a road (width a few meters) at a wavelength of 1 µm? (b) What wavelength do you need to work at this diffraction limit for a telescope that is only 3 m in diameter? Answers: a) 1µm = 10 6 m. The equatorial radius of the earth is 6,371 km. The distance to the geosynch orbit is 4164-6371 =35793 km. Take a road width of 4m. The angle is 4m/ (35793 1000) = 1.117 10 7 radians. Using the diffraction limit D = λ/θ = 10 6 m/1.117 10 7 = 8.9m. b) About 3 times smaller or 0.3 µm = 3000Å. 4. Black bodies (a) The cosmic microwave background has a spectrum with temperature.75 K. At what wavelength (in cm) does the spectrum peak? (b) The cosmic microwave background temperature, seen in the past by a local observer, depends on redshift. At redshift z a person at that time would have seen locally a higher temperature, 1 + z times higher. At a redshift of 0.3 what temperature would have been seen by a local observer at that time and what would have been the peak wavelength of the spectrum? Answers: a) Using the Wein displacement law λ = 0.9cmK/.75K = 0.106cm b) The temperature would be 1.3 higher or 3.54 K and the peak wavelength 1.3 smaller or 0.081 cm. Notes for those interested: The photons are free (not scattering or being absorbed as the universe is optically thin at long wavelengths). The numbers of photons per unit volume is approximately σt 4 /E max where E max is the energy of the peak (more accurately integrate the black body spectrum divided by hν). The volume increases as the universe expands as (1 + z) 3. So taking into account the
AST111 PROBLEM SET 4 SOLUTIONS 7 increase in temperature and decrease in volume we can reconcile that the numbers of photons in a comoving volume is staying fixed. Via Friedman s equation radiation evolves as a 4 and radiation energy density goes as T 4.