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(rad) = d/D = d/1AU = 0.53(2/360) = 0.0093, hence d = 0.0093AU = 0.0093 x 1.496 x 108 km = 1.392 x 106 km or Rsun = 6.96 x 105 km
Can we apply same principles to the stars?
e.g. Cen (like Sun) D = 1.3 pc = 2.7 x 105 AU; if Rcen = Rsun = 2Rsun/D = 0.0093 AU/2.7 x 105 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 Stellar Radii - parts of Chapters - parts of Chapters 7.3, 6.17.3, 6.1
Stellar Radii Stellar Radii - parts of Chapters - parts of Chapters 7.3, 6.17.3, 6.1
The Sun - resolved as a disk of diameter 0.53
Earth
D
Sund
Calculation of the Sun’s radius
Stellar radiiStellar radiiStellar radiiStellar radii
First minimum usually taken as resolution First minimum usually taken as resolution limit of telescope limit of telescope
Angular distance from centre to first ring Angular distance from centre to first ring = 1.22 = 1.22 /D /D
is wavelength of observation, D diameter of is wavelength of observation, D diameter of telescope, telescope, in radians in radians (see figs 6.8 - 6.10)(see figs 6.8 - 6.10)
Usually say 2 objects resolved if angular Usually say 2 objects resolved if angular separation > separation > /D /D Resolved Unresolve
d
min
First Minimum
Airy disk - diffraction pattern Airy disk - diffraction pattern produced by a circular aperture produced by a circular aperture (telescope mirror).(telescope mirror).
Stellar RadiiStellar RadiiStellar RadiiStellar RadiiExample: resolution of human eyeExample: resolution of human eye
Pupil aperture = 1 cm; Pupil aperture = 1 cm; visible light visible light = 500 nm = 5 x 10 = 500 nm = 5 x 10-5-5 cm (same units) cm (same units)diffractiondiffraction = = /d = 5 x 10/d = 5 x 10-5-5 cm / 1 cm cm / 1 cm = 5 x 10= 5 x 10-5-5 rad = 0.003 rad = 0.003 = 10 arcsec = 10 arcsec
If aperture > 10 cm, then If aperture > 10 cm, then diffraction diffraction = 1 arcsec= 1 arcsecThis is about the limit imposed by the atmosphereThis is about the limit imposed by the atmosphere
To resolve To resolve Cen: Cen: = 0.007 arcsec = 3.3 x 10= 0.007 arcsec = 3.3 x 10-8-8 rad rad > > minmin = 1.22 = 1.22/D /D D > 1.22 D > 1.22//In visible light, In visible light, = 500 nm = 5 x 10 = 500 nm = 5 x 10-7-7 m mTherefore need D > 1.22 (5 x 10Therefore need D > 1.22 (5 x 10-7-7 m)/(3.3 x 10 m)/(3.3 x 10-8-8) = 18.6 m ) = 18.6 m
Compare with world’s largest telescope, Keck = 10 mCompare with world’s largest telescope, Keck = 10 m
Ways to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar RadiiMethod 1: Using Method 1: Using Blackbody Laws:Blackbody Laws:
Lum. (L) = flux (F - Lum. (L) = flux (F - for a BB) x surface for a BB) x surface area (A - spherical area (A - spherical star radius R)star radius R)
L = L = TT44 x 4 x 4RR22 L is known if distance L is known if distance is known, T is is known, T is obtained from obtained from spectrum or colour, spectrum or colour, solve for radius Rsolve for radius R
Examples:Examples:There is a tremendous There is a tremendous range of stellar range of stellar radii spanning 7-8 radii spanning 7-8 orders of magnitude.orders of magnitude.
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L = σT4 4πR2 → L
L§
=T
T§
⎛
⎝ ⎜
⎞
⎠ ⎟
4R
R§
⎛
⎝ ⎜
⎞
⎠ ⎟
2
→ R
R§
=L
L§
⎛
⎝ ⎜
⎞
⎠ ⎟
1
2 TT§
⎛
⎝ ⎜
⎞
⎠ ⎟
−2
MV( )§= +4.8 Teff( )§
= 5800K
Betelgeuse : MV = −5.5 Teff ≅ 3500K
MV( )Betel− MV( )§
= −2.5 logLBetel
L§
⎛
⎝ ⎜
⎞
⎠ ⎟
− 5.5 − 4.8( ) = −1.5 logLBetel
L§
⎛
⎝ ⎜
⎞
⎠ ⎟
LBetel
L§
≈ 13,200
→RBetel
R§
= 13,200( )−1
23500K5800K
⎛ ⎝ ⎜
⎞ ⎠ ⎟−2
≅ 320
Sirius B : LLß
≅ 0.03 Teff ≅ 27,000K
faint companion to Sirius A( )
RSirius B
R§
= 0.03( )12
27,000K5800K
⎛ ⎝ ⎜
⎞ ⎠ ⎟−2
≅ 0.008approx . sizeof Earth
1 2 3
+1/2 not -1/2
Ways to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar RadiiMethod 2: Using Eclipsing Binaries:Method 2: Using Eclipsing Binaries:ExplanationExplanationIf orbital inclination i ~ 90If orbital inclination i ~ 90 , ,
orbital plane is close to line of orbital plane is close to line of sight and stars will partially or sight and stars will partially or totally eclipse each other. totally eclipse each other. Duration of eclipses, combined Duration of eclipses, combined with orbital speeds, gives with orbital speeds, gives stellar radii.stellar radii.
Light curve of eclipsing binary YY sagittarii
Light from both stars visible
V~10.0
V~10.7
apparent
brightness
halved
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m1 − m2 = 2.5 logb2
b1
⎛
⎝ ⎜
⎞
⎠ ⎟
0.7 2
Primary… Secondary minimum
P=2.628 d
vB
vB
vA
vAv = vA + vB
To Sun
Orbit seen from above
dA
dB
v
If we could see stellar disks during eclipse
brightnes
s
time
Light curve
1
1
2
2
4
4
3
3
tA
tB
v x tB = dB
v x tA = dA
Ways to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar RadiiWays to Measure Stellar Radii
Method 3: Using Eclipsing Binaries:Method 3: Using Eclipsing Binaries:A Real ExampleA Real Example
Time (days)
Radial Velcoty (km sec-1)
Eclipsing double-line spectroscopic binary AR Lacertae
Radial Velocity Curve
vA
vA~112 km/s
vB
vB~118 km/s
P ~2d
Period
Difference of magnitude ∆m
Light Curve
t1 t2t3
v = vA + vB ~ 230 km/stA = t3 - t1 ~ 0.24 d
tB = t2 - t1 ~ 0.13 d
dA = v x tA ~ 4.8 x 106 km
RA ~ 3.4 R
dB = v x tB ~ 2.6 x 106 km
RB ~ 1.9 R
State of Stellar MatterState of Stellar MatterState of Stellar MatterState of Stellar MatterWith masses and radii known, the densities and state With masses and radii known, the densities and state of matter of stars can be determined. of matter of stars can be determined.
Sun: Sun: MMsunsun = 2 x 10 = 2 x 103030 kg = 2 x 10 kg = 2 x 103333 g gRRsunsun = 7 x 10 = 7 x 1055 km = 7 x 10 km = 7 x 1010 10 cmcm
Mean density Mean density sunsun = M = Msunsun/(4/3/(4/3RRsunsun33) = 1.4 g/cm) = 1.4 g/cm33 (a bit (a bit
more than more than water) water)
Giant Stars: Giant Stars: MMstarstar ~ few x M ~ few x Msunsun in some cases in some casesRRstarstar ~ 500 R ~ 500 Rsunsun
giantgiant typically < 10 typically < 10-7-7 g/cm g/cm33!!
3000K < T3000K < Teffeff < 50,000 K < 50,000 K stars are gaseous stars are gaseous