E3 – Stellar distances

Parallax

Parallax

Parallax angle

Parallax angle P (in rads) R (=1 AU) d 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) = 1 p (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 radians 360 Multiply by 1 to convert to arcsecs 3600

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).

Let’s try some questions! ‘Parallax angle and parsec worksheet’

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 between a magnitude 1 star and a magnitude 6 star is ‘5 steps’ on the magnitude scale and the scale is logarithmic. This means that each ‘step’ equated to a brightness decrease of 2.512 since (2.512)5=100

r5 = 100 r = 2.512 Magnitude 1 Magnitude 2 Magnitude 3 Magnitude 4

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

Difference in magnitudes is 0.5 - -26.7 = 27.2 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)5=100 ) Therefore the difference in brightness = 2.51227.2 = 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. Which of these is brightest? By what factor?

Difference in apparent magnitudes = 8.4 – 4.8 = 3.6 Galaxy is brighter Difference in apparent magnitudes = 8.4 – 4.8 = 3.6 Difference 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 is the apparent magnitude of a star when viewed from a distance of 10 pc.

Absolute magnitude

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?

Let’s try some questions! ‘Apparent and absolute magnitude questions’

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

Cepheid variables At distances greater than Mpc, neither parallax nor spectroscopic parallax can be relied upon to measure the distance to a star. When we observe another galaxy, all of the stars in that galaxy are approximately the same distance away from the earth. What we really need is a light source of known luminosity in the galaxy. If we had this then we could make comparisons with the other stars and judge their luminosities. In other words we need a ‘standard candle’ –that is a star of known luminosity. The outer layers of Cepheid variable stars undergo periodic expansion and contraction, producing a periodic variation in its luminosity.

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

The process of estimating the distance to a galaxy (in which the individual stars can be imagined) might be as follows: Locate a Cepheid variable in the galaxy Measure the variation in brightness over a given period of time. Use the luminosity-period relationship for Cepheids to estimate the average luminosity. Use the average luminosity, the average brightness and the 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 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

Question Page 513 Question 16.