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was grade ' W ' Closed book, closed notes.
You are expected to know the general formulae
A sheet of constants vyill be provided. . ' There. will be an extra credit assignment dealing with this exam
—that will be discussed next class.. T 5‘3: NASA, NOAO, ESAand The Hubble Heritage Team (STScl/ HertzprungRussell
Diagram
Horizontal axis: spectral class, B — VI or T 1:n:u:u:i
(increasing to left). EIIIIIiEil. r: '. mm. 300le :Em lEIIII DEM] Vertical axis: Luminosity or
absolute magnitude. main sequence: diagonal
line running through all the
spectral classes. Some TL combinations
not realized in nature. [I Lilli Wide range in L for stars of
the same [mum Low L population: white [Minimum dwarfs. AST 203 (Spring 2011) if: E [biogKEEN: Binary Stars (Kulner, Ch. 5) AST 203 (Spring 2011) Binary Stars ~ 1/2 of all stars are in binary systems. A
:3
(I)
5
.1
(I)
E Period + separation tell us about the
gravitational force. Binary systems can tell us about
stellar masses. visual binaries: both stars seen independently—details of the
orbits of both stars in the system can be measured. optical doubles: two stars that appear close in the sky, but do not
orbit one another, and may in fact be quite a distance apart. AST 203 (Spring 2011) Binary Stars (HewholooksMnkipedia) Albireo: a double star. It is not known whether these stars are
gravitationally bound—if so, the period is > 100,000 years. AST 203 (Spring 2011) Binary Stars 7 iii:7 ii "if ' Ci ‘5': =1" iiiﬁ‘f} mass retro: 2.00
path of primary star ECCEHtFICItYT 000
path of unseen companion W astrometric binary: we only see one of the stars in the system—
wobble in its proper motion indicates an unseen companion. AST 203 (Spring 2011) Binary Stars “ rpf eclipsing binary: magnitude __
variation with a regular period. Nearly edgeon orbit
showing eclipses. n e 9: . . . . J .
ﬂ.‘ ‘0.“ '0.“ us [LIB 0.2 v.5 $6 “.5 0.5 Ms
UIBITEFI‘ESE
a ( d Small hot star and larger
cool star eslipsing: V HHGNI TIDE 10.5 “1.4 is NJ MM 8 1a.. Flux drops when
either star blocked. Stellar radii can be
found from eclipse
ft: m * shapes.
:1. ll. :1 i. l l, 4;.
HP flux Binary Stars composite spectrum binary: spectrum does not correspond to
known spectral type but rather is a blend of two—one from each
star in the system. spectroscopic binary: shows shifts in the position of the spectral
lines about some average wavelength (Doppler shift). AST 203 (Spring 2011) Doppler Shift Source moving toward us:
wavelength of emission is
compressed—blueshift. Source moving away from us:
wavelength stretched out—redshift. Only the relative speed along the
line of sight matters. As long as this velocity is not relativistic, shift in wavelength is: d°PP'era“
AA i A — A0 12—7. A—O A0 C
Here, A0 , is the rest wavelength of the emission. AsT 203 (Spring 2011) Doppler Shift We can also write an expression for the Doppler shift relating to
frequency. Recall that A : C/l/ SO dA : —ch//I/2 Therefore, AsT 203 (Spring 2011) Doppler Shift Example
Ha is one of the strongest lines in stellar spectra. If we have a star with a radial velocity of 10 km/s, what is the
observed wavelength shift? Recall that Ha has a wavelength of 656.28 nm.
106 cm 8—1 U
A Z _: ,2 —= .22
A A00 656 8nm3xlolocm871 00 nm This is a small, but measurable shift. AST 203 (Spring 2011) Spectroscopic Binary AST 203 (Spring 2011) Orbital Motion Our knowledge of orbital motion started with the planets. Kepler described three laws of planetary motion that were later
derived by Newton. ’lSt law: The orbit of each planet is an ellipse with the Sun at
1 foci. 2nd law: The radius vector to a planet sweeps out equal
areas in equal time 3rd law: The square of the period of the planet is
proportional to the cube of the semimajor axis of
the orbit. In reality the planets and the Sun orbit the common center of
mass of the system. AST 203 (Spring 2011) Circular Motion Starting point: circular orbits Consider two stars of masses m1
and m2 orbiting the center of mass
atdistances 7‘1 and 7‘2. The center of mass lies on a line
connecting the two masses, so mlrl : m2r2 Note: 71 + T2 is the distance between the centers of the stars,
not their surfaces. Newton showed that for spherical objects,
the force of gravity is equivalent is all the mass of the object
were placed at the center. AST 203 (Spring 2011) Circular Motion The stars are always opposite one another—period is the same: 27W"
P : — : const.
1) Each star experiences a gravitational force from the other:
77217722 F:G—
(7”1+7“2)2 Gravitational force balanced by centripetal acceleration. For star
1, we have: 77111]? AST 203 (Spring 2011) Circular Motion Rewriting, vi : G m2 r1 (M + r2)2
we know that 271'7‘1
’U :
1 P so 47T2T1 Gmg P2 (M + r2)2
Defining R = 71 + T2 , we have 7" m 7'
R=r1+r2=r1<1+—2) :r1(1+—1)=—1(m1+m2)
7'1 m2 m2 AST 203 (Spring 2011) Circular Motion Note: your text Putting it all together, we have misses the power of
2 3 2 on Tr—be sure to
477 R 2 (m1 + m2)P2 look at the errata.
G Solar system: Kepler's third law, also known as the harmonic law. P is an observable. distance to the system + resolve components —> separation.
Eclipsing binary: get the period by studying dips in lightcurve. This allows us to determine the total mass of the system. Can
we do any better? AST 203 (Spring 2011) Circular Motion Ratio of the masses: m1 /m2 : rz/r1
Together, we can find the mass of each component!
Nature isn't this simple however, the orbital plane may be inclined to our line of sight
the orbits may be elliptical, instead of circular AST 203 (Spring 2011) Mass of the Sun The easiest star that we can find the mass of is our Sun. Let's write down Kepler's law for the Sun and the earth: 4 2
gift 2 (MG + M@)P2 What are R and P ? R is just 1 astronomical unit, 1.5 X 1013 cm P is just 1 year We take M@ + M69 % MG SO 4W2R3 471'2 1.5 x 1013 cm 3 A.
MG: 2 Z is ( 2 i2 ) 7 2:2X1033g
GP (6.67 X 10 dyn cm g )(3.16 x 10 s) AsT 203 (Spring 2011) Mass of the Sun
Solar system: MG + Mg m MG
P2 = 1:553
for P in years and a in AU, k = 1 Kepler's Third Law Nauru?
102 — Uranus+1 
Kepler discovered this Jup.:r::n’+
. . l _
observationally In 1609. 10 ,v
T: Mars+’l
I I 3‘ 100 _ Earthﬂ;
Newton used this to derive M “9””?
I ercury
the Inverse square Q _] . F"
. 10 ' / /
dependence of graVIty. {Calm
‘ Ganymede
10'2 — ’ Europa
I. ID
1073 3 ‘ 2 Il I0 ‘1
10' 10' 10 10 10 AST 203 (Spring 2011) semiemajor axis (AU) Doppler Shift Eliminate the separation in favor of velocity. First, we use 271'?"
v : —
P
now, the total separation is
P
RZTllT'Q : gﬁﬂ +112) so we can completely eliminate the separation: P ‘
—(’01 + 1J2)d : m1 + m2 27rG AST 203 (Spring 2011) Inclined Orbits Complication: inclined orbits. Doppler shifts only measure the velocity along
the line of sight, UT. . If the orbit is inclined by some angle 1’, then we
measure 1),. = v sini and our period/velocity relation becomes P (“m + 02,1)3 . . 2 m1 + m2
2770 $1113 2 AST 203 (Spring 2011) Orbital
wane Plane
of sky Inclined Orbits Eclipsing binary: inclination close to 90°. With enough observations, we can trace out the orbit in the sky,
we can determine the inclination from the orbital characteristics. When we don't know i, the best we can do is assume that it is
90°, and get a lower limit to the sum of the masses. AsT 203 (Spring 2011 :1 Example Observations of a binary system show that it has a period of 10
years. The radial velocities of the two stars are determined to be 10 km/s and 20 km/s. Assuming circular orbits, with an inclination angle of 90°, what
are the masses? We begin by finding the sum of the masses via: P (1)177“ + U2,r>3 m1 + m2 = — . .
2770 Slug 2 _ 10 yr  3.16 x 107 s/yr (106 cm/s+ 2 x 106 cm/s)3
_ 27r  6.67 X 10—8 dyn cm2/g2 singz' : 2.04 X 1034 g/sin3z' = 10.2 MQ/sin3i AST 203 (Spring 2011) Example Since the periods of the stars are the same, then
r1/v1 : 72/222
and from the center of mass relation, we have
m2/m1 : rl/r2 : v1/v2 (Note this is also just conservation of momentum: mlvl : 777,202 ) Thus the ratio of masses is just ml/mg : U2/U1 : 2 If we take 1' = 90° , then we have
m1 + m2 = MG ml/m2:2 AsT 203 (Spring 2011) Example
SO m1 = MG and m2 = M@ More massive star has lower velocity (momentum conservation) If we instead take 1' = 45° , then the sum of the masses is
m1 + m2 = 10.2 MQ/sin3z‘ = 28.8 Me
and we have or m1 2 192 and 7712 = 9.6 doesnot change. AsT 203 (Spring 2011) ...
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This note was uploaded on 05/04/2011 for the course AST 203 taught by Professor Simon,m during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Simon,M
 Astronomy

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