Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Measurements of Star PropertiesFor a refresher on Trigonometry, please consult the Tutorial on Measurement BasicsCommon Name Scientific Name Distance (light years) Apparent Magnitude Absolute Magnitude Spectral TypeSun - -26.72 4.8 G2VProxima Centauri V645 Cen 4.2 11.05 (var.) 15.5 M5.5VcRigil Kentaurus Alpha Cen A 4.3 -0.01 4.4 G2VAlpha Cen B 4.3 1.33 5.7 K1VBarnard's Star 6.0 9.54 13.2 M3.8VWolf 359 CN Leo 7.7 13.53 (var.) 16.7 M5.8VcBD +36 2147 8.2 7.50 10.5 M2.1VcLuyten 726-8A UV Cet A 8.4 12.52 (var.) 15.5 M5.6VcLuyten 726-8B UV Cet B 8.4 13.02 (var.) 16.0 M5.6VcSirius A Alpha CMa A 8.6 -1.46 1.4 A1VmSirius B Alpha CMa B 8.6 8.3 11.2 DARoss 154 9.4 10.45 13.1 M3.6VcRoss 248 10.4 12.29 14.8 M4.9VcEpsilon Eri 10.8 3.73 6.1 K2VcRoss 128 10.9 11.10 13.5 M4.1V61 Cyg A (V1803 Cyg) 11.1 5.2 (var.) 7.6 K3.5Vc61 Cyg B 11.1 6.03 8.4 K4.7VcEpsilon Ind 11.2 4.68 7.0 K3VcBD +43 44 A 11.2 8.08 10.4 M1.3VcBD +43 44 B 11.2 11.06 13.4 M3.8VcLuyten 789-6 11.2 12.18 14.5Procyon A Alpha CMi A 11.4 0.38 2.6 F5IV-VProcyon B Alpha CMi B 11.4 10.7 13.0 DFBD +59 1915 A 11.6 8.90 11.2 M3.0VBD +59 1915 B 11.6 9.69 11.9 M3.5VCoD -36 15693 11.7 7.35 9.6 M1.3VcBegin with the right triangle expressionTan θ = R / dAnd for small angles, θ (radians) = R / dMeasurements of Star PropertiesShow the steps required to develop the expressiond (pc) = 1 / θ (arc sec)RParallax AngledUsing distance measurements in AU, θ (radians) = R / d → d (AU) = R (AU) / θ (radians)To make the units the same as the above expression, we need two conversions:Conversion of the distance to parsecsd(AU) = d(pc) (? AU / ? pc)Where the conversion factor will be left unstated at this point, andConversion of the angle in radians to arc secθ (radians) = θ (arc sec) (1 radian / 206265 arc sec)Measurements of Star PropertiesShow the steps required to develop the expressiond (pc) = 1 / θ (arc sec)Sod (AU) = R (AU) / θ (radians)d(pc) ( ? AU/ ? pc) = R (AU) / { θ (arc sec) (1 radian / 206265 arc sec) }Sod(pc) = { R (AU) / θ (arc sec) } (206265 arc sec / radian) (? pc / ? AU)If we DEFINE 1 pc = 206265 AU, thend(pc) = { R (AU) / θ (arc sec) } (206265 arc sec / radian) (1 pc / 206265 AU)d(pc) = R (AU) / θ (arc sec) (arc sec/radian) (pc/AU)If the measurements use the orbit of the earth for the baseline, R = 1 AU andd(pc) = 1 AU / θ (arc sec) (arc sec - pc/radian)Measurements of Star PropertiesShow the steps required to develop the expressiond (pc) = 1 / θ (arc sec)d(pc) = 1 / θ (arc sec) (arc sec - pc/radian)For convenience, the units at the end are dropped, andd(pc) = 1 / θ (arc sec)Measurements of Star PropertiesShow the steps required to develop the expressiond (pc) = 1 / θ (arc sec)Measurements of Star PropertiesBarnard’s star is observed from the earth. Observations are made of the location of the star from opposite extremes of the earth’s orbit around the sun. What parallax would be observed for Barnard’s star?The right triangle information is as shown below:RParallax Angled (radians) = RdWe know from the previous table that the distance to Barnard’s star is 6 light-years and the radius of the earth’s orbit (from class) about the sun is 1 AU Barnard’s star is observed from the earth. Observations are made of the location of the star from opposite extremes of the earth’s orbit around the sun. What parallax would be observed for Barnard’s star?Using the small angle approximation described in class,Measurements of Star PropertiesUsing Appendix 2 from the book,1 light-year = 63,200 AUTherefore, the distance to Barnard’s star isd = 6 (63200 AU) = 379200 AUAs a result, the parallax (that is, the parallax angle) for Barnard’s star isθ = R/d = 1 / 379200 = 2.6 x 10-6 radiansBarnard’s star is observed from the earth. Observations are made of the location of the star from opposite extremes of the earth’s orbit around the sun. What parallax would be observed for Barnard’s star?Measurements of Star PropertiesReview:It is common to provide the small angles in arc sec, soθ (degrees) = (2.6 x 10-6 radians ) (360 degrees / 2π radians)= (149 x 10-6 degrees)θ (arc sec) = (149 x 10-6 degrees) (3600 arc sec / 1 degree)= 0.54 arc secBarnard’s star is observed from the earth. Observations are made of the location of the star from opposite extremes of the earth’s orbit around the sun. What parallax would be observed for Barnard’s star?Measurements of Star PropertiesUsingθ(arc sec) = 1 / d (pc)And1 light-year = 63,200 AUFinallyd(pc) = 6 (63200 AU ) ( 1 pc / 206265 AU) = 1.836 pcθ(arc sec) = 1 / d (pc) = 1 / 1.836 = 0.54 arc secIT WORKS !!Barnard’s star is observed from the earth. Observations are made of the location of the star from opposite extremes of the earth’s orbit around the sun. What parallax would be observed for Barnard’s star according to the “shorthand” formula d (pc) = 1 / θ (arc sec)?Measurements of Star Properties (radians) = wdMeasurement made same time during the yearwdd x (radians) w =If the time interval between measurements is measured, then v = w/ tMeasurements of Star PropertiesA star 15 pc from the sun has a proper motion of 0.1”/year. What is its transverse speed? If spectral lines are red shifted by 0.001 %, what is the magnitude of its true speed? Refer to the following figure to see the situationA star 15 pc from the sun has a proper motion of 0.1”/year. What is its transverse speed? If spectral lines are red shifted by 0.001 %, what is the magnitude of its true speed? Using small angles θ << 1, w = d θ Where θ is measured in radians. Since 1 radian = 206265 arc sec, the proper motion of this star is { (0.1 arc sec ) (1 rad / 206265 arc sec) } / year = 4 x 10-7 rad / yearAnd, since the proper speed isW / year = d (θ / year) = ( 15 pc ) (4 x 10-7 rad / year) = 6 x 10-6 pc / yearMeasurements of Star PropertiesA star 15 pc from the sun has a proper motion of 0.1”/year. What is its transverse speed? If spectral lines are red shifted by 0.0001 %, what is the magnitude of its true speed? The star moves 4 x 10-7 rad / year. This can be converted to a transverse speed by determining the distance associated with the 4 x 10-7 rad traveled by the star in one year. Using the triangle from the previous slides, the distance traveled in one year is given by w = d θ = (15 pc)(4 x 10-7 rad) =
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