1. E3 – Stellar distances

2. Parallax

4. Parallax

5. Parallax angle

7. Parallax angle P (in rads) R (=1 AU) d tan p = R/d
for small p, tan p ≈ p
so d = R/p

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

9. Parsec
If the parallax angle is one arcsecond (1 “) the distance to the star is called a parsec
d (parsecs) = 1
p (in arcsecs)

10. Parsec
1 pc = 3.26 ly

11. 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 = = 2.9pc
p (in arcsecs)
Distance in light years = 2.9 x 3.26 = 9.5 ly

12. Converting degrees to arcsecs in radians
Multiply by 2π to convert to radians
360
Multiply by 1 to convert to arcsecs
3600

13. 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).

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

15. Apparent and absolute magnitudes

16. Hipparchus
Greek astronomer
Lived 2000 years ago

17. Hipparchus compared the relative brightness of stars (as seen from earth)
Brightest star – magnitude 1
Faintest star – magnitude 6

18. Apparent magnitude and brightness
Magnitude 1 star is
100 times brighter than
a magnitude 6 star

19. 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 since
(2.512)5=100

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

21. * Under what conditions?
Clear sky
When viewed from earth
As visible to the naked eye

22. Can a star have a magnitude greater than 6?

23. Can a star have a magnitude greater than 6?
Yes, but these stars are only visible through a telescope

24. A star of apparent magnitude less than 1

25. Negative apparent magnitude?
They are very bright!!

26. Guess the apparent magnitude of Sun
It is -26.7

27. 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 = m

28. Question
Apparent magnitude of Sun is and that of Betelgeuse is 0.5. How much brighter is Sun than Betelgeuse?

29. Difference in magnitudes is 0.5 - -26.7 = 27.2
Apparent magnitude of Sun is and that of Betelgeuse is 0.5. How much brighter is Sun than Betelgeuse?
Difference in magnitudes is = 27.2
Each difference in magnitude is a difference of in brightness ((2.512)5=100 )
Therefore the difference in brightness =
= 7.6 x 1010

30. Sun is 76 billion times brighter than Betelgeuse

31. 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?

32. 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 = = 27.5 times

33. 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.

34. Apparent magnitude Is it a fair way of measuring brightness of a star?
Brightness depends on distance and obeys inverse square law

35. Absolute magnitude
is the apparent
magnitude of a star
when viewed from
a distance of 10 pc.

36. Absolute magnitude

37. Absolute magnitude M and apparent magnitude m
m – M = 5 log (d/10)
d is in parsecs!

38. Question Calculate the absolute magnitude of Sun.
Apparent magnitude = -26.7
Distance from earth = 4.9 x 10-6 pc

39. m – M = 5 log(d/10)
M = 5 log (4.9 x 10-6/10)
M =-26.7 – 5log(4.9 x 10-7)
M = 4.85

40. M = 4.85
This means at a standard distance of 10 parsecs the sun would appear to be a dim star.

41. Can absolute magnitude be
Positive ?
Negative ?
Any value?

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

43. Spectroscopic parallax

44. 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

45. 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 = 12000K

46. 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)

47. 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

48. Using cepheids to measure distance

49. 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.

50. 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.

51. 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.

52. Cepheid calculation - Example

53. 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).

54. 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 = m
b = 2.52 x 10-8 x = 9.15 x W.m-2

55. 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

56. Question
Page 513 Question 16.