**Andrea CLaux**

Estrella Delgado-Parker

Physics SL per. 1

Estrella Delgado-Parker

Physics SL per. 1

**Stellar Distances/Objects**

Stellar Parallax

**Parsec**

A major problem in astrophysics is the accurate determination of the distance to a star.

Nearby stars appear to move with respect to more distant background stars due to the motion of the Earth around the Sun. This apparent motion is called Stellar Parallax.

The parallax movement is very small and is measured in seconds (3600 seconds = 60 minutes = 1 degree = 1/360 of a full circle).

Parallax Second = Parsec (pc)

Fundamental unit of distance in Astronomy

"A star with a parallax of 1 arcsecond has a distance of 1 Parsec."

1 parsec (pc) is equivalent to:

206,265 AU

3.26 Light Years

3.086x10^13 km

The parallax angle is observed and measured as the star position changes over the period of a year.

This formula:

mathematically, this leads to..

The main methods of measure are

the Parallax method

the Spectroscopic Parallax method

the Cepheid Star method

*the parallax method relies on such movement

Therefore, the apparent movement of the star relative to the background of further stars being measured in seconds of an arc leads to the unit Parsec (or per second).

Using basic trigonometry while knowing the distance of the earth to the sun leads us to.....

*theta is the parallax angle in radians.

so, when using parallax to measure distances...

= the parallax angle is proportional to 1/d

& d is the distance to the star

however, the constant in this equation can be changed to 1 by a careful adoption of a distance unit.

Thus:

Calculations show that this will be the case if d is measured in units equivalent to 3.08 x 10^16 m and this defines the distance unit as one parsec (parallax angle of one second)

Therefore to measure the distance to the star, observation will be made of the star “movement” against fixed back of further stars over the period of a year.

Due to the difficulty of measuring small parallax angles, this method is not reliable to stellar distance above 100 parsecs.

Absolute and Apparent Magnitudes in relation to Brightness

The apparent magnitude (m) of a star is a measure of its apparent brightness as seen by an observer on Earth.

Absolute magnitude is a measure of luminosity, how much light a star radiates into space.

The brighter the object appears, the lower the numerical value of its magnitude

*Absolute magnitude can be defined as apparent magnitude a star would have 10 pc away from Earth.*

Thus, the apparent magnitude represents the apparent brightness and the absolute magnitude represents the absolute luminosity.

The magnitude scale

classify stars into 6 categories of brightness

magnitude 1 being the brightest and magnitude 6 being the faintest

The magnitude scale is now defined as magnitude 1 being 100 times brighter than magnitude 6, with the scale being logarithmic.

*this means the each unit decrease of the magnitude scales corresponds to 2.512 times brighter (as 2.5125=100).

Since the apparent brightness of a star depends on the absolute brightness and distance of the star, the relationship is given by the following formula known as the distance modulus.

M = absolute magnitude

m = apparent magnitude

d = distance measured in parsec

If the luminosity is known, using the equation below you can determine an estimate of the distance to a star, in relation to apparent brightness.

b = apparent brightness (m)

L = luminosity

d = distance (m)

r = radius of star (m)

Spectroscopic Parallax

Finding the distance to a star given the star's luminosity and apparent brightness. Misleading because no use of parallax is being made.

1. If we take a spectrum of a star we can determine its spectral class.

2. Knowing the temperature is then allowed so we can use the Hertzsprung-Russell Diagram. Knowing the luminosity can allow us to distinguish between a red supergiant, giant or main sequence star for example.

3. Leading us to find out its absolute magnitude (M) by using the HR diagram or using a reference table.

4. Now knowing m from measurement and inferring M we can use the distance modulus equation:

m - M = 5 log(d/10) to find the distance to the star, d, in parsecs

Another way to use this method (less complicated)....

Following steps 1 and 2, you have enough variable to use:

Example Problem using Parallax

The distance of the Epsolon Eridani is 10.8 ly. What is its parallax?

First, we take our equation into consideration

We are given the distance (d) in light years; however, we know that the appropriate unit is parsec, so it is time for conversion.

1 parsec = 3.26 ly

Thus, we must divide 10.8 by 3.26...which equals to 3.31.

Now, we have the distance in parsecs; 3.31 pc.

We can plug it into the equation in order to find the parallax.

3.31 = 1/p

p = 1/3.31

p = 0.30 arc-second

Example question with absolute and apparent magnitude and distance

The parallax of a star is .025 and its absolute magnitude is M=0.8. Is its apparent magnitude less than or greater than 0.8?

First, we take the equation into consideration...

We are given the absolute magnitude(M) and we are given the parallax; we need d to find m.

We know from before that we can use the parallax to find distance by using d = 1/p, so in this case...

d = 1/.025

d = 40 pc

Now, we are able to plug it back into the equation.

0.8 = m - 5log(40/10)

0.8 = m - 3.01

m = 0.8 + 3.01

m = 3.81

Therefore, m is greater than 0.8.

Example problem using apparent brightness and luminosity

A main sequence star emits most of its energy at a wavelength of 2.4 x 10^-7 m. Its apparent brightness is measured to be 4.3 x 10^-9 W/m^2. How far is the star?

The equation we are going to use for this problem is

However, we only have apparent brightness and wavelength, so we have to find luminosity, somehow.

We know that in order to find L using the HR diagram, the temperature is needed, so we should use Wein's Law.

T = (2.9 x 10^-3)/(2.4 x 10^-7)

T = 12000 K

Now, we can look at the HR diagram in order to find Luminosity.

We can see that at a temperature of 12000 K, the the luminosity is 100 times that of the sun.

L = 3.9 x 10^28 W

now, we can substitute back into the original equation.

can be written as...

so, let's plug it back in...

d = √([3.9x10^28]/[4π][4.3x10^-9])

d = √(L/4πb)

d = √(7.22 x 10^35)

d = 8.5 x 10^17 m

d ≈ 90 ly

d ≈ 28 pc

The magnitude scale is shown by..

b

b0

=

100

-m/5

where b0 = 2.52 x 10

-8

W/m

2

(reference value for apparent brightness)

since 100

-m/5

= 2.512

, it can be written as

b

b0

= 2.512

-m

Taking logarithms (to base 10) gives the equivalent form:

m = -5/2log(b/b0)

Cepheid Variables

Stars whose luminosity is not constant in time but varies from a minimum to a maximum periodically.

The periods usually are from a couple days to a couple months.

The brightness of the star increases sharply then fades off more gradually.

Reason for periodic behavior

interaction of radiation with matter in the atmosphere of the star

This causes the outer layers of the start to undergo periodic expansions and contractions.

brightest-surface expands outward at high velocity

dimmest-when surface moves inward

Cepheid Luminosity-Period Relationship

In 1912, Henrietta Leavitt discovered that the periods of these stars are very closely related to their luminosities.

*The longer the period, the more luminous the star.

This Period-Luminosity Relation is seen to be true because larger (and hence more luminous) Cepheids take longer to pulsate back and forth in size.

Finding its period allows the determination of its luminosity. We can then calculate its distance with the luminosity-distance formula that we have seen before.

30

Example problem:

Here, we have a cepheid with a period of 22 days.

According to the previous diagram, the apparent magnitude is m = 3.7 which corresponds to a luminosity of about L = 2.7 x 10

30

W

We should know determine the apparent brightness by using the magnitude scale.

b

2.52 x 10

-8

= 2.512

-m

b = (2.52 x 10 )(2.512 )

-8

-3.7

b = 8.34 x 10

-10

Now, we are able to plug everything into the equation

in order to find the distance.

8.34 x 10

-10

=

2.73 x 10

30

4πd

2

d = √(2.73 x 10 )/(4π x 8.34 x 10 )

-10

30

d = 1.6 x 10

19

m

= 1700 ly

= 520 pc

Standard Candles

In astronomy, a standard candle is a source that has a known luminosity.

The strong direct relationship between a Cepheid variable's luminosity and period secures for Cepheids their status as important standard candles.

Observing a cepheid star and its period can determine its luminosity.

Thank You!

"Stars shine bright like a diamond!"

-Rihanna