Download presentation

1
Lecture 5 Binary stars

2
Binary stars 85% of all stars in the Milky Way are part of multiple systems (binaries, triplets or more) Some are close enough that they are able to transfer matter through tidal forces. These are close or contact binaries.

3
Examples Two stars are separated by 3 A.U. One star is three times more massive than the other. Plot their orbits for e=0.

4
**Example: binary star system**

Two stars orbit each other with a measurable period of 2 years. Suppose the semimajor axes are measured to be a1=0.75 A.U. and a2=1.5 A.U. What are their masses?

5
**Visual binaries: mass determination**

A perfect mass estimate of both stars is possible if: Both stars are visible Their angular velocity is sufficiently high to allow a reasonable fraction of the orbit to be mapped The distance to the system is known (e.g. via parallax) The orbital plane is perpendicular to the line of sight

6
**Example: Sirius a R a R is the distance to the star.**

Sirius A and B is a visual binary: a period of yr a parallax of p=0.377” The angular extent of its semimajor axis is a=aA+aB=5.52”. aA/aB=0.466 Assume the plane of the orbit is in the plane of the sky: a R a R is the distance to the star.

7
**Visual binaries: inclination effects**

In general the plane of the orbit is not in the plane of the sky. Here is the true orbit Focii

8
**Visual binaries: inclination effects**

In general the plane of the orbit is not in the plane of the sky. Here is the true orbit, which defines the orbital plane

9
**Visual binaries: inclination effects**

In general the plane of the orbit is not in the plane of the sky. Now imagine this plane inclined against the plane of the sky with angle i: i

10
**Visual binaries: inclination effects**

In general the plane of the orbit is not in the plane of the sky. Now imagine this plane inclined against the plane of the sky with angle i: i True major axis=2a 2acosi Instead of measuring a semimajor axis length a, you measure acosi where i is the inclination angle

11
**Visual binaries: inclination effects**

This projection distorts the ellipse: the centre of mass is not at the observed focus and the observed eccentricity is different from the true one. This makes it possible to determine i if the orbit is known precisely enough

12
**Visual binaries: inclination effects**

In practice we don’t measure a physical distance a, but rather an angular distance that we’ll call a. If a is the true angular distance, and a is the measured (projected distance) then: So the ratio of the masses is independent of the inclination effect However, the sum of the masses is not: a cosi R a

13
Example How does our answer for the mass of Sirius A and B depend on inclination? a cosi Sirius A and B is a visual binary: a period of yr a parallax of p=0.377” The angular extent of its observed semimajor axis is a=5.52”. aA/aB=0.466 R i 2acosi a

14
Example How does our answer for the mass of Sirius A and B depend on inclination? Thus our answers are a lower limit on the mass of these stars. The measured inclination is actually i=43.5 degrees. So cos3i=0.38 and mA=2.2 Msun, mB=1.0 Msun

15
Break

16
**Spectroscopic binaries**

Single-line spectroscopic binary: the absorption lines are redshifted or blueshifted as the star moves in its orbit Double-line spectroscopic binary: two sets of lines are visible Java applet:

17
**Spectroscopic binaries: circular orbits**

If the orbit is in the plane of the sky (i=0) we observe no radial velocity. Otherwise the radial velocities are a sinusoidal function of time. The minimum and maximum velocities (about the centre of mass velocity) are given by

18
**Spectroscopic binaries: circular orbits**

We can therefore solve for both masses, depending only on the inclination angle i In general it is not possible to uncover the inclination angle. However, for large samples of a given type of star it may be appropriate to take the average inclination to determine the average mass.

19
**Spectroscopic binaries: non-circular**

If the orbits are non-circular, the shape of the velocity curves becomes skewed in a way that depends on the orientation e.g. e=0.4, i=30°, axis rotation=45° A sinusoidal light curve means orbits are close to circular From analysis of light curve it is possible to determine the eccentricity and orbit orientation, but not the inclination. In practice most orbits are circular because tidal interactions between the stars tend to circularize the orbits Java applet:

20
Example: z Phoenicis Here are raw observations of the velocities of two stars in a double-lined spectroscopic binary system, z Phoenicis: Julian date is a numerical date where each unit is 1 day

21
Example : z Phoenicis This doesn’t look much like a sinewave!! The problem is that observations are taken at discrete points in time, and thus randomly sample the phase of variation. Find the period that gives the “best” curve when plotted against orbital phase: P=1.65 days P= d P=1.68 days Phase Phase Phase

22
**Single-lined spectroscopic binaries**

In general, one star is much brighter than the other (remember faint stars are much more common than bright stars). This means only one set of absorption lines is visible in the spectrum. The Doppler motion of this single set of lines still indicates the presence of a binary system. We can still solve for a function of the two masses: This is the mass function

23
Example: z Phoenicis Imagine only one of the velocity curves in this system was visible. v1=180 km/s Then If we assume m1~m2 then m~4sin3i(1.38)=4.0 MSunsin3i (Recall from both velocity curves we recovered m1=1.9 MSunsin3i and m2=2.8 MSunsin3i )

24
Eclipsing binaries A good estimate of the inclination i can be obtained in the case of eclipsing binaries, separated by distance d: To observer R1+R2 i d If d » R1+R2 (which is usually the case) then i~90 degrees

25
Eclipsing binaries A good estimate of the inclination i can be obtained in the case of eclipsing binaries, separated by distance d: To observer R1+R2 i d Assume i=90 degrees when in reality i=75 degrees. What is the error in sin3i? sin3(75)=0.9 So the error on the masses is only 10% if d > 3.9(R1+R2) If d » R1+R2 (which is usually the case) then i~90 degrees

26
Eclipsing binaries In the system just described, the eclipse just barely happens: Face on To observer So the amount of light blocked is not constant, and the light curve (total brightness as a function of time) looks something like this:

27
Eclipsing binaries However, in the case of total eclipse the smaller star is completely obscured. In this case it is even more likely that the inclination is close to 90 degrees Face on To observer And the light curve shows constant minima:

28
Eclipsing binaries In the case of a total eclipse we can also measure the radii of the stars, and the ratio of their effective temperatures If we assume i~90 degrees and circular orbits that are large relative to the stellar radius, then the radius of the smaller star is: Where v is the relative velocity between the two stars And for the larger star:

29
**Eclipsing binaries Ratio of effective temperatures (0) (1) (2)**

Note that (1) will be the deepest minimum if Ts>Tl. (often the case since the brightest, largest stars are the cool supergiants) Alternatively (2) will be the deepest minimum if Tl>Ts

30
Stellar masses For select star systems, we can therefore measure the mass directly. Luminosity is closely correlated with stellar mass Energy production rate is related to stellar mass If the available energy is proportional to mass, how do stellar lifetimes depend on their main sequence location?

31
**The main sequence revisited**

The main sequence is a mass sequence More massive stars are closer to the top-left (hot and bright) M=30MSun M=MSun M=0.2MSun

32
**The main sequence revisited**

The main sequence is a mass sequence More massive stars are closer to the top-left (hot and bright) Stars on the main sequence have radii 1-3 times that of the Sun Supergiants have R>100 RSun White dwarfs have R~0.01 RSun M=30MSun M=MSun M=0.2MSun

33
Densities Since we know the stellar masses and radii, we can compute their average densities Sun: Recall water has a density of 1000 kg/m3 Dry air at sea level: 1.3 kg/m3

34
Densities Since we know the stellar masses and radii, we can compute their average densities Supergiants (Betelgeuse): or 1/100,000 times less dense than air.

35
Densities Since we know the stellar masses and radii, we can compute their average densities White Dwarfs (Sirius B): or 850,000 times denser than water.

Similar presentations

OK

(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.

(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.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on p&g products brands list Ppt on conceptual art examples Ppt on solid dielectrics capacitors Ppt on supply chain management of dell Appt online Download ppt on mind controlled robotic arms for education Ppt on do's and don'ts of group discussion images Ppt on chapter 3 atoms and molecules ppt Ppt on tunnel diode detector Economics ppt on demand and supply