Solution Manual for To Measure the Sky 2nd Edition by Chromey

CLINTON [COMPANY NAME] [Company address] SOLUTION MANUAL FOR TO MEASURE THE SKY 2ND EDITION BY CHROMEY To Measure the Sky Hints and solutions to the end-of-chapter exercises Chapter 1 2. λ=hcE=1239.85 eV nmE a) λ=124013.6=91.2 nm (Lyman limit) b) 3. fλhas units of brightness/wavelength = Wm3⎡⎣⎢⎤⎦⎥ 4. dF = fλdλ = fυdυ for any amount of flux, so fλfυ=dυdλ=−υ2c=−cλ2 5. The observed flux from either star alone is proportional to its surface area times its surface brightness, so, if we treat both stars as blackbodies, the brightness maximum will be: Fmax=σR2d2T14+a2T24⎡⎣⎤⎦ When star 1 completely occults star 2, the relative brightness of the system is F1Fmax=T14T14+a2T24⎡⎣⎤⎦=11+a2(T24T14)=11+b When star 2 completely overlaps star 1, then the relative brightness is: F2Fmax=T14+a2T24−a2T14T14+a2T24⎡⎣⎤⎦=1+a2T24T14−a21+a2(T24T14)=1−a21+b 6. The energy of 1 photon at frequency 106 Hz is hν=6.63 x 10-34 x 106=6.63 x 10-28 J Flux is fυdυ1,000,0001,000,001∫≅10−26Wm-2Hz-1⋅1Hz=10−26Wm-2=10−26Js−1m-2 N (photons): 10−26Js-1m-26.63×10-28J/photon=15 photons m-2s-1