Solution Manual for To Measure the Sky: An Introduction to Observational Astronomy , 2nd Edition

Solution Manual for To Measure the Sky: An Introduction to Observational Astronomy, 2nd Edition makes solving textbook exercises easier with step-by-step solutions and helpful tips.

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To Measure the SkyHints and solutions to the end-of-chapter exercisesChapter 12.λ=hcE=1239.85 eV nmEa)λ=124013.6=91.2 nm (Lyman limit)b)3.fλhas units of brightness/wavelength =Wm34.dF=fλdλ=fυdυfor any amount of flux, sofλfυ=dυdλ=υ2c=cλ25.The observed flux from either star alone is proportional to its surface area times its surfacebrightness, so, if we treat both stars as blackbodies, the brightness maximum will be:Fmax=σR2d2T14+a2T24When star 1 completely occults star 2, the relative brightness of the system isF1Fmax=T14T14+a2T24=11+a2(T24T14)=11+bWhen star 2 completely overlaps star 1, then the relative brightness is:F2Fmax=T14+a2T24a2T14T14+a2T24=1+a2T24T14a21+a2(T24T14)=1a21+b6.The energy of 1 photon at frequency 106Hz ishν=6.63 x 10-34x 106=6.63 x 10-28JFlux isfυdυ1,000,0001,000,0011026Wm-2Hz-11Hz=1026Wm-2=1026Js1m-2N (photons):1026Js-1m-26.63×10-28J/photon=15 photons m-2s-1

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