1.  (rad) = d/D = d/1AU = 0.53(2  /360  ) = 0.0093, hence d = 0.0093AU = 0.0093 x 1.496 x 10 8 km = 1.392 x 10 6 km or R sun = 6.96 x 10 5 km Can we apply same principles to the stars? e.g.  Cen (like Sun) D = 1.3 pc = 2.7 x 10 5 AU; if R cen = R sun  = 2R sun /D = 0.0093 AU/2.7 x 10 5 AU = 3.3 x 10 -8 rad = 0.007 arcsec (angular diameter of a dime 150 km away!) Can we resolve this small angle with a telescope? Stellar Radii - parts of Chapters 7.3, 6.1 The Sun - resolved as a disk of diameter 0.53   Earth D Sun d Calculation of the Sun’s radius

2. Stellar radii First minimum usually taken as resolution limit of telescope Angular distance from centre to first ring  = 1.22 /D is wavelength of observation, D diameter of telescope,  in radians (see figs 6.8 - 6.10) is wavelength of observation, D diameter of telescope,  in radians (see figs 6.8 - 6.10) Usually say 2 objects resolved if angular separation > /D ResolvedUnresolved  min First Minimum Airy disk - diffraction pattern produced by a circular aperture (telescope mirror).

3. Stellar Radii Example: resolution of human eye Pupil aperture = 1 cm; visible light = 500 nm = 5 x 10 -5 cm (same units)  diffraction = /d = 5 x 10 -5 cm / 1 cm = 5 x 10 -5 rad = 0.003  = 10 arcsec = 5 x 10 -5 rad = 0.003  = 10 arcsec If aperture > 10 cm, then  diffraction = 1 arcsec This is about the limit imposed by the atmosphere To resolve  Cen:  = 0.007 arcsec = 3.3 x 10 -8 rad  >  min = 1.22 /D  D > 1.22 /  In visible light, = 500 nm = 5 x 10 -7 m Therefore need D > 1.22 (5 x 10 -7 m)/(3.3 x 10 -8 ) = 18.6 m Compare with world’s largest telescope, Keck = 10 m

4. Ways to Measure Stellar Radii Method 1: Using Blackbody Laws: Lum. (L) = flux (F - for a BB) x surface area (A - spherical star radius R) L =  T 4 x 4  R 2 L is known if distance is known, T is obtained from spectrum or colour, solve for radius R Examples: There is a tremendous range of stellar radii spanning 7-8 orders of magnitude. +1/2 not -1/2

5. Ways to Measure Stellar Radii Method 2: Using Eclipsing Binaries: Explanation If orbital inclination i ~ 90 , orbital plane is close to line of sight and stars will partially or totally eclipse each other. Duration of eclipses, combined with orbital speeds, gives stellar radii. Light curve of eclipsing binary YY sagittarii Light from both stars visible V~10.0 V~10.7 apparent brightness halved 0.7 2 Primary… Secondary minimum P=2.628 d vBvB vBvB vAvA vAvA v = v A + v B To Sun Orbit seen from above dAdA dBdB v If we could see stellar disks during eclipse brightness time Light curve 1 1 2 2 4 4 3 3 tAtA tBtB v x t B = d B v x t A = d A

6. Ways to Measure Stellar Radii Method 3: Using Eclipsing Binaries: A Real Example Time (days) Radial Velcoty (km sec -1 ) Eclipsing double-line spectroscopic binary AR Lacertae Radial Velocity Curve vAvA v A ~112 km/s vBvB v B ~118 km/s P ~2d Period Difference of magnitude ∆m Light Curve t1t1 t2t2 t3t3 v = v A + v B ~ 230 km/s t A = t 3 - t 1 ~ 0.24 d t B = t 2 - t 1 ~ 0.13 d d A = v x t A ~ 4.8 x 10 6 km R A ~ 3.4 R  d B = v x t B ~ 2.6 x 10 6 km R B ~ 1.9 R 

7. State of Stellar Matter With masses and radii known, the densities and state of matter of stars can be determined. Sun: M sun = 2 x 10 30 kg = 2 x 10 33 g R sun = 7 x 10 5 km = 7 x 10 10 cm Mean density  sun = M sun /(4/3  R sun 3 ) = 1.4 g/cm 3 (a bit more than  water) Giant Stars: M star ~ few x M sun in some cases R star ~ 500 R sun  giant typically < 10 -7 g/cm 3 ! 3000K < T eff < 50,000 K  stars are gaseous