The Properties of Stars Masses. Using Newton’s Law of Gravity to Determine the Mass of a Celestial Body Newton’s law of gravity, combined with his laws.

The Properties of Stars

Masses

Using Newton’s Law of Gravity to Determine the Mass of a Celestial Body

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.

2

32

JupiterGP

a4M

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.

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.2r

GMmF

s10442.1P

m10883.1a

6Callisto

9Callisto

32 9

Jupiter 211 6

4 1.883 10M

6.673 10 1.442 10

kg10900.1M 27Jupiter Masses Earth

10977.5

10900.1M

24

27

Jupiter

Masses Earth318MJupiter

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

2

32

JupiterGP

a4M

In using to determine the mass of Jupiter, we assumed that the mass

of Jupiter is much greater than the mass of any of its moons.

2

32

21GP

a4MM

where M1 and M2 are the masses of the two stars.

However, in binary star systems, the two bodies have comparable masses and the relevant form of Kepler’s third law is

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

2

3

21P

aMM M1 and M2 are multiples of the Sun’s mass if a is in AU’s and P is in years.

Center of Mass

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.

1 2 2

2 1 1

M r v

M r v

3 Types of Binary Star System

• Visual BinariesBoth stars are visible, so their orbits can be plotted.

• Spectroscopic BinariesThe 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 BinariesThe stars appear as a single star, but we see the orbits edge-on, so the stars periodically eclipse each other.

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.

In order to use M1 M2 a3

P2to find M1 and M2, we have to (1) measure a

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.

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 starsseparated by several arcseconds.

Few observations of stars separated by a fraction of a arcsecond.

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Example 1Consider 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.

1

2

2

1r

r

M

M

2

3

21P

aMM The equations to be used are and

25MM 21

AU 100a years200P We are given and

2

3

21200

100MM So,

We are also told that 5M

M

2

1 , so 25MM5 22

2

25M 4.17 M

6 1 2M 5M 20.8 M

26M 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.

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.

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.

P Q R S

vB

vA

Toward

Earth

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.

t = ¼ of theperiod later.Which star ismoving towardus?

Star

Lab

t = 0 (spectraof A and Bare superimposed)

Star

Lab

The Radial Velocity Graphs

Radial Velocity vs. Time

-150

-100

-50

0

50

100

150

0 1 2 3 4 5 6

Time (days)

Ra

dia

l V

elo

city

(km

/s)

Note: the radial velocity of the center of mass has been subtracted before the graph was drawn.

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.

Star A moving away from Earth.

Stars moving across our line of sight

Star B moving toward Earth.

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

2

Pvr AA

2

Pvr BBand

The semi-major axis is a = rA + rB

The total mass is 2

3

BAP

aMM

The ratio of the masses isA

B

B

Av

v

M

M

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 2A 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 2

1

20

10

v

v

M

M

A

B

B

A

2

Pvr BB

2

Pvr AA(b) P = 3.00×(3.16×107 s) = 9.48×107 s

km103.02

π2

109.48km/s20r 8

7

A

AU2.01km/AU101.50

km103.02r

8

8

A

km1051.1

π2

109.48km/s01r 8

7

B AU00.1

km/AU101.50

km101.51r

8

8

B

a = rA + rB = (2.01 + 1.00) AU = 3.01 AU

The total mass is

33.01M M 3.03MBA 23.00

M 2M 3.03MA A 3M 3.03MA M 1.01MA M 2.02MB

MPEG Animation of the Algol System from a Paper by Blondin, Richards, and Malinowski

Eclipsing BinariesFl

ux

Time

: eclipse of hotter star beginsAt

: eclipse of hotter star completeBt

: The hotter star is about to emerge from behind

the cooler one.Ct

eclipse of hotter star ends.Dt =

relative orbital velocityv=

radius of the hotter starHR =

radius of the cooler starCR =

( )2 vH B AR t t= -

( )2 C C AR v t t= -

Example 3The 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?

85km/sv= ( )1 2H B AR v t t= -

Example 4In the same system, the hotter star is eclipsed for 8 hours, what is the radius if the cooler star?

1 85km/s 4 3600 6120002

s km= ´ ´ ´ =

( )1

2C C AR v t t= - 61 85km/s 8 3600 1.22 102

s km= ´ ´ ´ = ´

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.

-10

-5

0

5

10

15

0.1 1 10 100

Ab

solu

te V

isu

al M

agn

itu

de

3.5ML=L

M

L the luminosity of theSun

M the mass of the Sun.

M / M

Hertzsprung-Russel Diagram