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

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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 =Wm3⎡⎣⎢⎤⎦⎥4.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+a2T24⎡⎣⎤⎦When 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+a2T24−a2T14T14+a2T24⎡⎣⎤⎦=1+a2T24T14−a21+a2(T24T14)=1−a21+b6.The energy of 1 photon at frequency 106Hz ishν=6.63 x 10-34x 106=6.63 x 10-28JFlux isfυdυ1,000,0001,000,001∫≅10−26Wm-2Hz-1⋅1Hz=10−26Wm-2=10−26Js−1m-2N (photons):10−26Js-1m-26.63×10-28J/photon=15 photons m-2s-1

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7.A magnitude of zero means:−2.5 log(2.65×10−8)+K= 0⇒K=−18.94.8.Assume that the band-pass magnitude is directly proportional to the monochromatic flux at thecenter of the band, and use the magnitude difference formula to compute the monochromaticflux at the center of the B band in star X:fx=f1⋅10−0.4(mx−m1)=375⋅10−0.4(17.79)=2.87 x 10−5Jy(b) The total energy collected isE= (monochromatic flux) x (bandwidth) x (area) x (exposure time)Or, converting units:E= ([2.87 x 10-5Jy] x [10-26J s-1m-2Hz-1/1 Jy) x (2.5 x 1014Hz) x 100 s x 5 m2= 3.59 x 10-14JThe corresponding number of photons isN=Eλhc=(3.59×10−14)(4.4×10−7)(6.63×10−34)(3×108)=79429.The difference between the magnitude of two stars together and the magnitude of one staralone is:mboth−mone=−2.5 log(F+FF)=−2.5 log 2=−0.753, somboth=mone−0.753=7.5910.(a) Apply the reasoning of the previous problem: The magnitude difference between the totalnebula and a one arcsec square is:Δm=−2.5 log144=−5.40m144=m1−5.4=17.77−5.40=12.37(b) If distance to the nebula doubles, the surface brightness stays the same, the angular areadecreases by a factor of 4, and the total apparent magnitude increases by 2.5 log(4) = 1.51and becomes 13.88.11.As in the previous problem, if distance to the nebula doubles, theintrinsicsurface brightnessstays the same, the angular area decreases by a factor of 104, and the total apparentmagnitude increases by 2.5 log(104) = 10 and becomes 22.37. However, in this case one

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cannot observe the intrinsic surface area or surface brightness because of the limitedtelescopic resolution. Note that the intrinsic diameter of the nebula would beDintrinsic=4πAintrinsic⎡⎣⎢⎤⎦⎥12=4π144104⎡⎣⎢⎤⎦⎥12=0.141 arcsecThis is considerably smaller that the actual observed image diameter of 1.2 arcsec, so theobserved image will be spread over an area ofAobs=π4(1.2)2=1.13 arcsec2The observed surface brightness will therefore be reduced by a factor ofAintrinsicAobs=0.01441.13=1.27×10−2over the intrinsic value. Or in magnitudes, the surface brightness isSobs=Sintrinsic−2.5 log(1.27×10−2)=17.77+4.74=22.51 mag arcsec-212.(a) The distance modulus formula, m – M= 5 log(r)-5, gives:13.25 – (-19.6) + 5 = 37.8 = 5log(r)r=3.72×107pc=37.2 Mpc(b) To correct for the absorption, repeat the computation above but with the apparentmagnitude brightened by 1.5 magnitudes:13.25 – (-19.6) – 1.5 + 5 = 36.3 = 5log(r)r=1.82×107pc=18.2 Mpc13.Begin with the definition of flux:F=L4πr2, then letF=flux observed from an arbitrary distance , rm= apparent magnitude observed from a distance rF10=Flux observed from a distance of 10 pcM= apparent magnitude observed from a distance 10 pc (the absolute magnitude)Applying the formula for magnitude difference:Δm=m−M=−2.5 logFF10=−2.5 log 102r2=5 logr−514.Note that the definition of logarithms meansF=eln Fand thereforelog10F=ln F⋅log10e

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So, how much does the magnitude change if the flux changes by amountΔF? Take thederivative of the magnitude equation:m=−2.5 log10F+K, that is:dm=−2.5d(lnF⋅log10e)=−2.5dFFlog10e=1.086dFF, soΔm!ΔFF15.It is important that the aperture used for each star be identical in size; i.e. use the same totalnumber of pixels for each star image. One of many reasonable measurements might go likethis:341626333722252529192825222044342226143030201917317098663725353639392320349922910738284610215993372233671036736326924039324869302233342936246524136324468242822171632244685157844222182527261718302935243027322316292524302820352223282828242626171930353026Background:the 18 pixels 3 X 6 box in the lower left corner have a mean value of 24.11, amedian of 25, and a mode 25. Either 24.11 or 25 is a reasonable choice for the background.Assume B=24.11.Brighter star:This star is very symmetric around a point midway between the pixels withvalues 393 and 363. Add the values of the 16 shaded pixels: Sum of star and background =2680, so the total number of counts for the brighter star alone is 2680 – (16 x 24.11) = 2294.2Fainter star:Also very symmetric, but around the center of the pixel with value 229. Takethe 21 pixels indicated, but give only 13 pixels full weight (dark shading) and give the 8outer pixels a weight of 3/8. Thus the effective number of pixels in both star apertures are thesame. For the fainter star alone the total counts are:1056 + ((3/8) x 253) - (16 x 24.11) = 765.1Magnitude: If the magnitude of the brighter star is 9.000, the magnitude of the fainter ismf=−2.5 log(765.1 / 2294.2)+9.0=10.192

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Chapter 21.(a) The tenth datum is unknown, so the population mean can’t be computed. The nine knownvalues are not a random sample. Since they are the nine smallest values, the sample mean isbelow the population mean by an unknown amount.(b) The population mediancanbe computed because the rank of the missing value is known.The median (the expected survival time) isμ12=(2.6 + 2.9)/2 = 2.75..2.(a)x=x12=5.This is a sample, sos2=1102 25+16+9+4+1(){}=11, ands=3.31.(b)x=x12=5.s2=224 16+2⋅9+3⋅4+4⋅1{}=4.167, ands=2.04.3.(a) is a uniform distribution, withP(i)=111,for integers 0≤i≤10. (elseP(i)=0)(b) is a “triangular” distribution withP(i)=1255−x−5⎡⎣⎤⎦,for integers 0≤i≤10. (elseP(i)=0)4.(a) In the absence of further information, this appears to be a Poisson process, with a meanrate of 1.7 events per year per square km orr= 1.7 x 10-4events per yr per 100 m2roof.(b) The probability of zero events per year on the roof is given by the Poisson distribution:P=PP(0,r)=11exp(−1.7×10−4)=.99983(c) The probability of more than one penetration in 40 years will be:P=1−PP(0, 40r)−P(1, 40r)=1−exp(−6.8×10−3)−6.8×10−31exp(−6.8×10−3)=1−0.99322(1+.0068)=2.6×10−55.Because the integral of this function between the limits of zero and infinity diverges, it cannotbe interpreted as a valid probability function. (The integral of a normalized continuousprobability function equals unity.) This distribution withγ> 0 becomes arbitrarily large asaapproaches zero. A student might argue that the mean will be driven this limit, and so in asense this represents a “typical” member of the population. But the function also permits

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indefinitely large valuesa,so in another sense a mean of zero does not represent thepopulation at all. Imposing upper and lower limits on the permitted values of the randomvariable will allow a computation of a valid mean and standard deviation. (If one investigatesthe mathematical definitions of the mean and standard deviations of continuous distributionseven slightly, it is easy to show that imposing only a lower limit ona—and no upper limit —leads to a well-defined mean ifγ> 2, and a well-defined variance ifγ> 3.)6.Follow the reasoning in the example on page 48. Transform the variableqinto the standardnormal variablez:z=(q−0.8) / 0.6,zlow=−1.50,zhigh=−0.833Now compute the probability that a single trial will result in the discovery of an earth-likeplanet. Use the tabulation of the standard normal distributionPSN(z)=G(z)and itsintegralgiven inAppendix C:Prob(earthlike)=G(z)dz−1.5−0.833∫=P(1.5)−P(.833)=.933−.796=.137So in 500 trials, on should expect 68.5 earthlike planets discovered.7.Counting photons is a Poisson process. The fractional uncertainty in countingNevents(equation 2.15) is(a)0.05=1N,soN=400. (b)N=40,000.9.The standard deviation of the sample of four measurements is 13.6 km/s, which implies anuncertainty in the mean of13.6 /4=6.8km/s. If the astronomer measures N additionalstars, to reach an uncertainty of 2.0 km/s, then:2.0=13.6N+4and therefore N=43.8.Assume she spends equal amounts of time measuring the target and the background. Thecounts are N*=Nb=Nmeas/2. The uncertainty in the star brightness can be computed from thevariance:σ∗2=σmeas2+σb2=3N∗So the relative uncertainty isσ∗N∗=3N∗=6Nmeasand (a) for 5% uncertaintyNmeas= 2400, (b)Nmeas= 240,000.

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9.The uncertainty in the mean is related to the scatter in the population and the number ofsamples, N, by :σmean=σNWe estimate the scatter in the population from the standard deviation (N-1 weighting) of thefour measurements as 13.6 km/s. The above equation then implies a sample size of N = 46would yield an uncertainty in the mean of 2 km/s.10.Because we know that the fluctuations in the value ofrin 10 second exposures are normallydistributed with scatter of .05 mV, we can use the Central Limit Theorem to conclude thatuncertainty ofrin 100-s exposures will be reduced by a factor of110. We will assumethat this detector accumulates voltage in a linear fashion as exposure time increases, so thevalues for each of the Ns in equation 2.39 on the longer exposures will each increase by afactor of 10. Making these substitutions in equation 2.39:σ*2=1660+850+.055⎛⎝⎜⎞⎠⎟2110⎛⎝⎜⎞⎠⎟2.7556+.7225()×106⎡⎣⎤⎦=2510+3.5=2513N*σ*=81050.1=16.211.(a) A straight-forward average of the five trials and a treatment of the five results as randomvariables gives:x=21.2,σmean=37.25=16.6(b) An alternative approach is to use that fact that these are (presumably) counts ofphotons, so an arrival 106 photons in five seconds suggests (Poisson distribution) anuncertainty in the average of106 5=2.06The divergence of these two methods suggests that something may be amiss, as does the factthat the mean and median values are quite different. The fact that the result of trial 2 differsfrom the mean by more than almost 2σin method (a) is a little suspect. However, if we reallybelieve that we are counting photons (as in method b) then the second trial is 29σlarger thanthe mean, and it is highly unlikely that we are correct in including it as a valid measurementof the same process as the other trials. Often a single deviant event like trial number 2 willalert us to a systematic error.12.As in the Chapter 1 problem, it is important that the aperture used for each star be identicalin size; i.e. use the same total number of pixels for each star image, and centering of theapertures will require the use of fractional pixels. Repeating the measurements made in thesolutions for Chapter 1:341626333722252529192825

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222044342226143030201917317098663725353639392320349922910738284610215993372233671036736326924039324869302233342936246524136324468242822171632244685157844222182527261718302935243027322316292524302820352223282828242626171930353026Background:the 18 pixels 3 X 6 box in the lower left corner has a total count of 440 and a meanvalue of 24.44. Uncertainty in the total count is440=20.98and uncertainty in the averagebackground is440 / 18=1.165.Note that the standard deviation of the sample of 18 pixels is s=4.49, which implies an uncertainty in their mean value of 1.06. This is consistent with theuncertainty computed under the assumption of Poisson statistics.Brighter star:This star is very symmetric around a point midway between the pixels with values393 and 363. Add the values of the 16 shaded pixels: Sum of star and background = 2680, so thetotal number of counts for the brighter star alone isFSTD=2680 – (16 x 24.44) = 2288.9 with anuncertainty given by:σstd2=2680+1618⎛⎝⎜⎞⎠⎟2440=3027.7soσstd= 55.0.Fainter star:Also very symmetric, but around the center of the pixel with value 229. Take the 21pixels indicated, but give 13 pixels full weight (dark shading) and give the 8 outer pixels aweight of 3/8. Thus the effective number of pixels in both star apertures are the same. For thefainter star alone the total counts are:Ff= 1056 + ((3/8) x 253) - (16 x 24.44) = 759.8 and theuncertainty is:σ2f=1056+38⎛⎝⎜⎞⎠⎟2253+1618⎛⎝⎜⎞⎠⎟440=1439.25soσf= 37.9Magnitude: If the magnitude of the brighter star is 9.000, the magnitude of the follows from theratioR=Ff/FSTD= 759.8/2288.9 = 0.332:mf=−2.5 log(759.8 / 2288.9)+9.0=10.197To get the uncertainty in the magnitude of the fainter star, first note that for the uncertainty in theratio,R(for a product or ratio, the relative variances add), we have

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σ2FfFSTDFfFSTD()2=σ2FfFSTD0.332()2=37.9759.8⎛⎝⎜⎞⎠⎟2+552288.9⎛⎝⎜⎞⎠⎟2=(0.055)2Then either use the result from Problem 1.13, or note, sincemf=−2.5 log(Ff/FSTD)+9.0=−2.5 log(R)+9.0σm2=(−2.5)2σRlogeR⎛⎝⎜⎞⎠⎟2=(1.086)2σRR⎛⎝⎜⎞⎠⎟2=(0.060)2This assumes that the cataloged standard magnitude of the brighter star is perfectly known. (Notalways the case!). Somf= 10.197 ± 0.060.

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Chapter 3Nαδδφ∆BA12Nδ+∆δθα∆BArδBAr∆δCProblem 1Problem 41.Draw the great circle connecting the two points, and the hour circle of each point. Theresult is a spherical triangle whose vertices are the celestial pole and the two points:triangleNABin the diagram. The sides of this triangle are, respectively, the angle ofintertestφ, and the complements of the declinations of the objects, (90°-δ1) and (90° -δ2). The angle opposite sideφisΔα.Apply the law of cosines and simplify:cos(⇤) =cos(90⇥1)cos(90⇥2) +sin(90⇥1)sin(90⇥2)cos()cos(⇤) =sin(⇥1)sin(⇥2) +cos(⇥1)cos(⇥2)cos()2.Apply the result of the previous problem, letting:1= latitude of NYC2= latitude of Mexico, or LA= longitude difference, NYC and other, so for example:cosφNY-LA=sin41°sin34°+cos41°cos34°cos(118° – 74°) = 0.818The NY-LA distance (35.2°) is longer than the NY-Mexico City distance (30.6°).3.The distance the Earth travels in a sidereal year (the orbit circumference) is 2πau. Butthe amplitude of the radial velocity variation is the orbital velocity of the Earth. Ifaisthe length of the au in kilometers, then its estimated value isa=12π(29.167±.057 km/s)(31557940 s)=(1.488±.003)×108km4.Again, from the spherical triangle with points A, B and the north celestial pole (N).Then apply the law of sines, noting thatr,Δδ,andΔαare all small:

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sinΔαsinr=Δαr=sinθsin(90−δ − Δδ)=sinθsin(90−δ)thus:Δα=rsinθcosδFor triangle ABC, all sides are very small, so plane trig results apply, i.e.:Δδ=rcosθ(Youcando this problem less directly by first applying the law of cosines on△NCB, and making use of the double-angle formulae and the series expansion ofcos for small angles)5.This is identical to problem 4 withθ= 90° and r = 1000 arcsecs.Thus:Δα=1000arcseccosδ⋅1 sec time15 arcsec=66.7scosδe.g.atδ=85°:Δα=66.7scos(85)=764s=12m: 44s6.Mean ST= Apparent ST –EOT.During December, EOT decreases from about+11minutes on the 1stto -2 minutes on the 30th. On the apparent clock, earliest sunset isat the solstice. However, the declination of the sun changes very little close to thesolstice, and consequently the apparent time of sunset changes very little near thesolstice. Because of more positive values in EOT before the solstice, the mean solarclock will read earlier at sunset on the days before the solstice than on the solsticeitself.Zone time will differ from local mean solar time with longitude within a time zone, sothe zone time of sunset will likewise differ with location by some constant offset, butthe date of earliest sunset will be the same at any longitude. (This date does dependon latitude)7.At the solstice the sun is at RA = 18 hrs and a midnight the sun has HA = - 12 hrs, soat that time RA = 6 hrs is on the meridian. So at midnight the sidereal time is 6 hrs.At sunset at latitude = + 40, the celestial sphere indicates the sun is at about HA+4:40 when on the horizon. Therefore the local solar time at sunset is 4:40 PM).8.d=14.17×10-3=240pc,θ=ad⇒a=θd=11 arcsec⋅240 pc = 2640 au

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9.Rate of increase ofradiusisvr=12Δλλc=0.08656⋅3×105=36.6km/s.(b) The time needed to reach this present radius at this speed is the age of theremnant:T=avr=2640au37km/s⋅1.49×108kmau=1.06×1010s3.15×107s/year=338 years(c) Assume the proper motion of the edge of the remnant is due to a tangentialvelocity equal to the measured radial velocity. Using the relationship for propermotion and distance, the new estimate for the distance isdμ=υT4.74Δθ60= 257 pc10.(a) from the parallax equation, the distance is:d=ap206265=(1.535×108)(206265) / (0.32)=9.984×1013kmThe relative uncertainty in d isσdd=σpp⎛⎝⎜⎞⎠⎟2+σaa⎛⎝⎜⎞⎠⎟2⎧⎨⎪⎩⎪⎫⎬⎪⎭⎪12=(.04 / .32)2+.05)2(){}12=0.135or 1.34x1013km. (b) The modern estimate is (10.748 ±0.019 )x 1013km.11.Refraction does not change the apparent angular diameter of the sun in the horizontaldirection. In the vertical direction, the apparent angular diameter is reduced by thedifference between the refractive effect on the top of the solar image and therefraction at bottom of the solar image If R(z) is the refraction at zenith angle z, thenthe reduction in the angular size of the minor axis when the center of the sun is atzenith angle isΔb=R(z– 0.267°) –R(z + 0.267°)12.(a)There is no month 6 frame because the object is on the ecliptic, so it will be behindthe sun at that time.(b) The conservative approach is to set up a reference frame based on the three“background” stars. Since the scales and orientations of the four images are identical,we only need to set the zero point of each frame to be the MEAN POSITION (centerof “mass”) of the three background stars on each frame. Then measure x,ycoordinates for each object relative to this center on each frame. The combined resultsare illustrated graphically in the included figure. Each of the three background starshas roughly the same position in each frame, and the moving object does not. From

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the deviations of individual measurements, it looks like these three stars (opencircles) probably have very small but detectablerelative systematic motions in latitude (y-direction),but because these are small and not systematic we willignore them for purposes of the problem. The figureshows the proper motion over one year (dashed line)of the one rapidly moving object. The proper motionmeasures, in arcsec/yr:μx=1.75,μy=3.2,The displacement of the moving object from thedotted line at month 3 measures the parallax (length oflinep3) as does, independently, the length ofp9. Theresults, again in arc sec, are about:p3=1.12,p9=0.8,Averaging these, the parallaxp= 0.96 arcsec, implies a distance ofd=1.04 pc, andμ=3.65 arcsec/yr. Uncertainties, estimated from the scatter in four measurements of thethree background stars, as well as from the disagreement of the two parallaxestimates, are about 0.15 arcsec.(c) The distance of one parsec means that this object, although very nearby, is not amember of the solar system. An object bound to the sun in a circular orbit would at adistance of 1 pc, by Kepler’s third law, have a period and proper motion of aboutP=d32≈(2×105)32=9×107yrμ= 360⋅60⋅609×107=0.013 arcsec/yrTherefore, unless some very peculiar effects have mimicked the heliocentric parallax,the object is moving TOO FAST to be an asteroid. (d) The tangential velocity is about17.6 km/s.14.(no problem 13).If the uncertainty in the wavelength of an absorption line ofwavelengthλisδλ, then the uncertainty in the radial velocity determined from thismeasurement is just:δv=cδλλ=cRHereRis the spectrograph resolution and the extreme right-hand side assumesthatδλis completely determined by spectrograph resolution. If N lines are measured,then the uncertainty in the mean radial velocity should be (assuming constant R):δλmean=1NcR=0.67 km s−103912pp39

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15.Statistical parallax cannot be used to find the distance of this cluster, since there is noreason to suppose that the magnitude of the tangential velocity of the cluster isroughly equal to its radial velocity. If one assumes that the two are in fact equal, thenthe tangential velocity equation suggests a distance ofd=514.74×0.145=74 pcthe propagation of errors givesσ2=d21Nσv2v2+σμ2μ2⎧⎨⎩⎪⎫⎬⎭⎪=8.6 pcThis, however, is a meaningless result, since the distance is completely unknown.Chapter 42. Beta Cen has a parallax of 6.21 mas, as measured by HIPPARCOS.3. (a) Only seven stars are currently listed: four confirmed and three suspected variablesin the GCVS. The Cepheids are CE Cas A, CE Cas B and CF Cas . (b) CF Cas hasV=10.80, (c) P=4.87522 days, Julian date epoch of max light = 2437022.191, (e) Themost recent treatment is An D.; Terndrup D.M.; Pinsonneault M.H.:Astrophys. J.,671, 1640-1668 (2007).5. Between 2015 and 2020, Eris will be within two degrees of RA 1:43, DEC -3°,Makemake within three degrees of 12:52 +25°.

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