E3 – Stellar distances

Parallax

Parallax

Parallax angle

Parallax angle

» Tan p

R (=1 AU)

d

P (in rads)

tan p = R/d

for small p, tan p ≈ p

so d = R/p

Parsec

• If the parallax angle is one arcsecond (1 “) the distance to the star is called a parsec

Parsec

• If the parallax angle is one arcsecond (1 “) the distance to the star is called a parsec

• d (parsecs) = 1p (in arcsecs)

Parsec

• 1 pc = 3.26 ly

Example

• A star has a parallax angle of 0.34 arcsecs. How far is the star away from earth in light years?d (parsecs) = 1 = 1 = 2.9pc

p (in arcsecs) 0.34

Distance in light years = 2.9 x 3.26 = 9.5 ly

Converting degrees to arcsecs in radians

• Multiply by 2π to convert to radians360

• Multiply by 1 to convert to arcsecs3600

Parallax method

• Only useful for close stars (up to 300 ly (100 pc) as further than that the parallax angle is too small (space based telescopes can use this method to measure stars up to distances of 500 pc).

Apparent and absolute magnitudes

Hipparchus

• Greek astronomer

• Lived 2000 years ago

Hipparchus compared the relative

brightness of stars (as seen from earth)

Brightest star – magnitude 1 Faintest star – magnitude 6

Apparent magnitude and brightness

Magnitude 1 star is 100 times brighter than

a magnitude 6 star

• The difference The difference between a magnitude between a magnitude 1 star and a magnitude 1 star and a magnitude 6 star is ‘5 steps’ on 6 star is ‘5 steps’ on the magnitude scale the magnitude scale and the scale is and the scale is logarithmic. This logarithmic. This means that each ‘step’ means that each ‘step’ equated to a equated to a brightness decrease of brightness decrease of 2.5122.512 since since

• (2.512(2.512)5=)5=100 100

Magnitude 1

Magnitude 2

Magnitude 3

Magnitude 4

Magnitude 5

Magnitude 6

r5 = 100

r = 2.512

* Under what conditions?

• Clear sky

• When viewed from earth

• As visible to the naked eye

Can a star have a magnitude greater than 6?

Can a star have a magnitude greater than 6?

Yes, but these stars are only visible through a telescope

A star of apparent magnitude less than 1

Negative apparent magnitude?

They are very bright!!

Guess the apparent magnitude of Sun

It is -26.7

Apparent magnitude

The apparent magnitude m, of a star of apparent brightness b is defined by

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

where b0 is taken as a reference value of 2.52 x 10-8 W.m-2

This can also be written as b/b0 = 2.512-m

Question Apparent magnitude of Sun is -26.7 and that of

Betelgeuse is 0.5. How much brighter is Sun than Betelgeuse?

Apparent magnitude of Sun is -26.7 and that of Betelgeuse is 0.5. How much brighter is Sun than

Betelgeuse?

• Difference in magnitudes is 0.5 - -26.7 = 27.2• Each difference in magnitude is a difference of

2.512 in brightness ((2.512(2.512)5=)5=100 )100 )• Therefore the difference in brightness = Therefore the difference in brightness =

2.5122.51227.227.2

• = 7.6 x 1010

Sun is 76 billion times brighter than Betelgeuse

Question 2

Apparent magnitudes of Andromeda galaxy and Crab nebula are 4.8 and 8.4 respectively.

a) Which of these is brightest?

b) By what factor?

Galaxy is brighter

Difference in apparent magnitudes = 8.4 – 4.8 = 3.6Difference in brightness therefore = 2.5123.6 = 27.5 times

The Andomeda Galaxy is a vast collection

of stars

• The Crab Nebulae is a debris of supernova and is the birth place of the new star.

Apparent magnitude

- Is it a fair way of measuring brightness of a star?

- Brightness depends on distance and obeys inverse square law

ABSOLUTE MAGNITUDE

Let the standard distance be 10 pc

1 pc = 3.086 x 1016 m= 3.26 ly= 206265 AU

Absolute magnitude

is the apparent

magnitude of a star

when viewed from

a distance of 10 pc.

Absolute magnitude M and apparent magnitude m

m – M = 5 log (d/10)

d is in parsecs!

Question

Calculate the absolute magnitude of Sun.

Apparent magnitude = -26.7

Distance from earth = 4.9 x 10-6 pc

m – M = 5 log(d/10)

-26.7- M = 5 log (4.9 x 10-6/10)

M =-26.7 – 5log(4.9 x 10-7)M = 4.85

M = 4.85

• This means at a standard distance of 10 parsecs the sun would appear to be a dim star.

Can absolute magnitude be

Positive ?

Negative ?

Any value?

Spectroscopic parallax

Spectroscopic parallax

• This refers to the method of finding the distance to a star given the star’s luminosity and apparent brightness. It doesn’t use parallax! Limited to distances less than 10 Mpc

• We know that b = L/(4πd2) so d = (L/(4 πb))½

Spectroscopic parallax - Example

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

• λ 0T = 2.9 x 10-3 Km

• T = 2.9 x 10-3 / 4.3 x 10-9 = 12000K

• T = 12000K. From an HR diagram we can see this corresponds to a brightness of about 100x that of the sun (= 100 x 3.9 x 1026 = 3.9 x 1028 W)

Spectroscopic parallax - Example

Thus d = (L/(4 πb))½

d = (3.9 x 1028/(4 x π x 4.3 x 10-9))½

d = 8.5 x 1017 m = 90 ly = 28 pc

Using cepheids to measure distance

• At distances greater than Mpc, neither parallax nor At distances greater than Mpc, neither parallax nor spectroscopic parallax can be relied upon to measure spectroscopic parallax can be relied upon to measure the distance to a star. the distance to a star.

• When we observe another galaxy, all of the stars in When we observe another galaxy, all of the stars in that galaxy are approximately the same distance that galaxy are approximately the same distance away from the earth. What we really need is a light away from the earth. What we really need is a light source of known luminosity in the galaxy. If we had source of known luminosity in the galaxy. If we had this then we could make comparisons with the other this then we could make comparisons with the other stars and judge their luminosities. In other words we stars and judge their luminosities. In other words we need a ‘standard candle’ –that is a star of known need a ‘standard candle’ –that is a star of known luminosity.luminosity.

• The outer layers of Cepheid variable stars undergo The outer layers of Cepheid variable stars undergo periodic expansion and contraction, producing a periodic expansion and contraction, producing a periodic variation in its luminosity.periodic variation in its luminosity.

Cepheid variables

Cepheid variable Cepheid variable stars are useful to stars are useful to astronomers because of the astronomers because of the periodperiod of their of their variation in luminosity turns out to be related variation in luminosity turns out to be related to the average to the average absolute magnitude absolute magnitude of the of the Cepheid. Thus the luminosity of the Cepheid Cepheid. Thus the luminosity of the Cepheid can be calculated by observing the variation in can be calculated by observing the variation in brightness. brightness.

The process of estimating the distance to a galaxy (in which the The process of estimating the distance to a galaxy (in which the individual stars can be imagined) might be as follows:individual stars can be imagined) might be as follows:

• Locate a Cepheid variable in the galaxyLocate a Cepheid variable in the galaxy• Measure the variation in brightness over a given period of Measure the variation in brightness over a given period of

time.time.• Use the luminosity-period relationship for Cepheids to Use the luminosity-period relationship for Cepheids to

estimate the average luminosity.estimate the average luminosity.• Use the average luminosity, the average brightness and the Use the average luminosity, the average brightness and the

inverse square law to estimate the distance to the star.inverse square law to estimate the distance to the star.

Cepheid calculation - Example

• From the left-hand graph we can see that the period of the cepheid is 5.4 days. From the second graph we can see that this corresponds to a luminosity of about 103 suns (3.9 x 1029 W).

• From the left hand graph we can see the peak apparent magnitude is 3.6 which means we can find the apparent brightness from

• b/b0 = 2.512-m

• b = 2.52 x 10-8 x 2.512-3.6 = 9.15 x 10-10 W.m-2

• Now using the relationship between apparent brightness, luminosity and distance

• d = (L/(4πb))½ • d = (3.9 x 1029/(4 x π x 9.15 x 10-10))½

• d = 5.8 x 1018 m = 615 ly = 189 pc

Questions

Page 512 questions 3, 4, 6, 7, 8, 10, 11, 13, 14, 15, 16.