Download The Properties of Stars

The Properties of Stars
Masses
Using Newton’s Law of Gravity to Determine the Mass of a
Celestial Body
Any two particles in the universe attract each other with a force that is directly proportional
to the product of their masses and inversely proportional to the distance between them.
F
GMm
r2
The force on a particle outside an object with spherical symmetry is the same
as if all of its mass were concentrated at its center.
Newton’s law of gravity, combined with his laws of motion enable us to determine the mass
of a celestial body by observing its effect on a second celestial body.
For example, we can find Jupiter’s mass by measuring the orbital radius and period for each
of its Galilean satellites and using Newton’s form of Kepler’s third law of planetary motion.
M Jupiter 
9
2 3
a Callisto  1.883  10 m
4 a
GP 2
M Jupiter  1.900  10
M Jupiter
6
P Callisto  1.442  10 s
27
kg M Jupiter 
1.900  1027
5.977  10
4 1.883 10 


 6.673 10 1.442 10 
2
24
9
11
3
6
2
Earth Masses M Jupiter  318 Earth Masses
We’ll use a similar method to find the masses of stars in binary star systems.
Using Newton’s Form of Kepler’s Third Law to
Measure the Masses of Stars
In using M Jupiter 
4 2a 3
GP
2
to determine the mass of Jupiter, we assumed that the mass
of Jupiter is much greater than the mass of any of its moons.
However, in binary star systems, the two bodies have comparable masses and the relevant
form of Kepler’s third law is
M1  M 2 
4 2a 3
GP
2
where M1 and M2 are the masses of the two stars.
P is the orbital period of the stars and a is the average distance between them.
Because the masses of stars are very large, but a relatively small multiple of the mass of the Sun,
it is convenient to use solar mass units. In that case, Kepler’s third law is
M1  M 2 
a3
P
2
M1 and M2 are multiples of the Sun’s mass if a is in AU’s and P is in years.
Center of Mass
M1 r2 v 2
 
M 2 r1 v1
The black dot is the center of mass, and the colored disks are two stars at distances
r1 and r2 from the center of mass. M1 and M2 are the masses of the two stars and v1
and v2 their orbital speeds.
3 Types of Binary Star System
•
•
•
Visual Binaries
Both stars are visible, so their orbits can be plotted.
Spectroscopic Binaries
The stars appear as a single star but, because of the Doppler effect, the
spectral lines can be seen to shift as the stars move in their orbits.
Eclipsing Binaries
The stars appear as a single star, but we see the orbits edge-on, so the stars
periodically eclipse each other.
In order to use M1  M2
a
3
2
to find M1 and M2, we have to (1) measure a
P
and P and (2) determine what fraction of the total mass belongs to each star.
In order to accomplish (2), we need the concept of center of mass defined
below.
When two bodies move through space and are acted on only by their
mutual gravitational forces, there is a point between them that moves in a
straight line. That point is called the center of mass of the two bodies.
The center of mass of a pair of bodies satisfies the equation M1r1 = M2r2
where r1 is the distance from M1 and r2 is the distance from M2.
Visual Binaries
The figures below show the observed positions of the stars in two different visual binary
systems). Poor visual binary observations result from (a) the period being so long that few
observations have been made or (b) the stars being so close together and/or so far from
Earth that their angular separations cannot be measured accurately. The solid lines in the
figures represent the ellipse that best fits the data (colored dots and plus signs).
Visual binaries have periods between 1 year and thousands of years.
Plenty of observations of stars
separated by several arcseconds.
Few observations
of stars separated
by a fraction of a
arcsecond.
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Example 1
Consider a visual binary star system in which star 2 is 5 times farther from the center of
mass than star 1, the period is 200 years and the semi-major axis is 100 AU. Calculate (a)
the total mass and (b) the mass of each star.
The equations to be used are
We are given
P  200 years and
3
So, M  M  100
1
2
200 2
We are also told that
M 2
M1 r2

M 2 r1
25
 4.17 M
6
and
M1  M 2 
a3
P2
a  100 AU
M1  M 2  25
M1
 5 , so
M2
5M 2  M 2  25
M1  5M 2  20.8 M
6M 2  25 M
Spectroscopic Binaries
In spectroscopic binary systems, the two stars are too close together to be resolved by a
telescope. Because their separations are less than about 1 AU, their periods are as short as
a few hours or as long as a few months.
In the animation, the black dot represents the center of mass. Actually, the center of
mass moves, but we are looking at the system from the viewpoint of someone at rest
relative to the center of mass.
Finding the Masses of Spectroscopic Binaries
Finding the mass of the stars in a binary star system requires observations that give (a) the
sum of the masses and (b) the ratio of the masses. This can easily be done if the system is
a well-observed visual binary. In that case, we can plot the orbit and measure a and P. r1
and r2 can be determined by observing the motion of the system long enough to locate the
center of mass.
For spectroscopic binaries, it isn’t so easy. In that case, we must extract information from
the combined spectra of the stars. Since the Doppler effect only gives the star’s radial
velocity (the component along the observer’s line of sight) and most orbits are tilted, we
are usually able to only determine a lower limit to the total mass.
We’ll just consider the simplest case: the angle between the line of sight and the orbit
is 0º, and the orbit is circular. The figure shows the orbits from above the orbital plane.
The red circle is the orbit of the red star and the blue circle is the orbit of the blue star.
Earth is to the right.
vA
Toward
P
Q
R
S
Earth
vB
The blue star (A) is moving away from us, so its
spectrum is red-shifted while that of the red star (B) is
blue-shifted.
When the stars arrive at the points P, Q, R, and S they are
moving across our line of sight so we see no redshift.
The Spectrum
Usually, the spectrum will show two sets of lines that change positions as the stars move
along their orbits. In the following figures, wavelength increases toward the right and only
the hydrogen Balmer lines are shown. In each case, the Balmer lines observed in the
laboratory are displayed on the bottom for comparison with the binary’s spectrum on the top.
The first figure shows the spectrum at a time when the stars are moving across our line of
sight so there is no wavelength shift, and spectra of the two stars are superimposed.
In practice, the photographs of the stars’ spectra would be black and white, but I’ve used
blue for star A and red for star B.
Star
Lab
t = 0 (spectra
of A and B
are superimposed)

Star
Lab
t = ¼ of the
period later.
Which star is
moving toward
us?
The Radial Velocity Graphs
The following graph shows the radial velocities of the two stars as a function of time. The
time scale isn’t the same as that used in the previous slide.
Radial Velocity vs. Time
150
Stars moving across our line of sight
Radial Velocity (km/s)
100
50
0
0
1
2
3
4
5
6
-50
-100
Star A moving away from Earth.
-150
Time (days)
Star B moving toward Earth.
Note: the radial velocity of the center of
mass has been subtracted before the
graph was drawn.
Calculations
From the radial velocity graph, we can read the orbital velocities of the two stars vA and vB
as well as the orbital period P.
The radii of the orbits are
The semi-major axis is
The total mass is
v P
rA  A
2
v P
rB  B
2
a = rA + rB
MA  MB 
The ratio of the masses is
and
a3
P2
MA vB

MB vA
When the orbits are circular and the angle between them and the line of sight is zero (i.e.,
we are seeing them edge-on), the last two equations permit us to calculate the masses of
the two stars. In that case, the stars will also periodically eclipse each other.
Example 2
A spectroscopic-eclipsing binary star system has a period of 3.00 years. The maximum
radial velocities of the stars relative to the center of mass are 20 km/s (for star A) and
10 km/s (for star B). Calculate (a) the ratio of their masses and (b) the individual
masses, assuming that the orbit is observed edge-on from Earth,
(a) vA = 20 km/s, vB = 10 km/s
(b)
v P
rA  A
2
M A v B 10 1



M B v A 20 2
v P
rB  B
2
P = 3.00×(3.16×107 s) = 9.48×107 s
rA

20 km/s 9.48  107 

 3.02  108 km
rB

10 km/s 9.48  107 

 1.51  108 km
2π
rA 
2π
3.02  108 km
1.50  10 km/AU
rB 
8
 2.01AU
1.51  108 km
1.50  10 km/AU
8
 1.00 AU
a = rA + rB = (2.01 + 1.00) AU = 3.01 AU
3
3.01

The total mass is M  M 
 3.03M
B
A
2
 3.00 
MA  2MA  3.03M
3MA  3.03M
MA 1.01M
MB  2.02M
Flux
Eclipsing Binaries
MPEG Animation of the Algol
System from a Paper by Blondin,
Richards, and Malinowski
Time
t A: eclipse of hotter star begins
tB : eclipse of hotter star complete
tC : The hotter star is about to emerge from behind
the cooler one.
tD = eclipse of hotter star ends.
v = relative orbital velocity
RH = radius of the hotter star
2 RH = v (tB - t A )
RC = radius of the cooler star
2 RC = v (tC - t A )
Example 3
The orbital velocity of an eclipsing binary system is 85 km/s, and the time for the
eclipse of the hotter star to be complete is 4.0 hours, what is the radius of the hotter
star?
1
1
RH =
v (t B - t A ) =
v = 85km/s
´ 85km/s ´ 4 ´ 3600 s = 612000km
2
2
Example 4
In the same system, the hotter star is eclipsed for 8 hours, what is the radius if the
cooler star?
RC =
1
v (tC - t A )
2
=
1
´ 85km/s ´ 8´ 3600s = 1.22 ´ 106 km
2
Properties of Stars
Some Important Results
The Mass-Luminosity Relation
A graph of absolute visual magnitude (a measure of luminosity) is plotted as a function of
the logarithm of mass, the result is almost a straight line as shown below.
Absolute Visual Magnitude
-10
-5
L=L

 M
M


3.5





0
L  the luminosity of theSun
5
M  the mass of the Sun.
10
15
0.1
1
M/M
10
100
Hertzsprung-Russel Diagram