E1 Astronomy and Space Science Chapter 4 Stars and the Universe

4 Stars and the Universe

Practice 4.1 (p. 98) 1 A 2 D 3 (a) Star X is closer to the Earth. (b) Distance to star X 1 = = 10 pc 0.1 Distance to star Y 1 = = 20 pc 0.05 Star X is 10 pc closer than star Y. 4 Star Parallax Distance Distance Distance (in ) (in pc) (in AU) (in ly) Sirius A 0.379 2.64 5.44  8.60 105 Altair 0.194 5.15 1.06  16.8 106 Vega 0.125 8 1.65  26.1 106  half the width of paper 5 tan = 2 length of arm (Put in their own data into the formula.) Sample result: 21.5  cm tan = 2 2 80 cm  = 15.3 This angle is much larger than the parallax of Proxima Centauri. 6 The angular shift  of the star is about half the width of the photo, i.e.  = 4.  p = = 2 2 1 1 Distance to the star = = = 0.5 pc p 2

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

7 (a) Distance to the star Apparent magnitude of star X 1 1 100  = = = 83.3 pc = 1 + 5 log   = 4 p 0.012  10  (b) In this case, we have Apparent magnitude of star Y 30.1 AU 10  = tan p = 3 + 5 log   = 3 d 10  30.1  p (in arcsecs) = Since the apparent magnitude of star Y is d (in parsecs) smaller, it appears brighter. Therefore, 6 D parallax measured on Neptune 7 (a) Star X 30.1 = (in arcsecs) = 0.361 (b) Star A 83.3 8 (a) Venus orbits the Sun. Its distance to us 8 (a) The parallaxes of stars farther away are changes and therefore its apparent too small to be measurable by the magnitude varies. Also, Venus has parallax method. different phases. With a phase showing (b) The distance between the two a larger part, it looks brighter. measurements is larger (about 3 AU). (b) Sirius looks brighter. The parallax measured would be larger IS and thus more accurate. Ratio of intensity I A 9 Distance to the star m m  = 2.512 A S 1 = = 6.67 pc = 11.4 0.15 2.512 Time between sending and receiving signal = 9.12 The ratio of the intensity of Sirius to = 6.67  3.26  2 yr = 43.5 yr that of Antares is 9.12 : 1. IV 10 Parallax of the farthest star measured (c) Ratio of intensity I 1 S = = 0.005 m m  200 = 2.512 S V = 2.5121.44.4 Practice 4.2 (p. 116) = 15.9 1 B The ratio of the maximum intensity of 2 C Venus to that of Sirius is 15.9 : 1. 3 A 9 (a) Intensity received energy received 4 C = area  time 5 A 8109 2 =  5.5103  π   2     2 

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

= 1.68  104 W m2 (b) Energy received = intensity received  area  time 2  610 3  4   = 1.68  10       1  2  = 4.75  109 J 10 (a) Object A mainly emits ultra-violet radiation. Object B mainly emits red light. (b) Object A 11 The x-axis represents surface temperature or spectral class, while the y-axis represents luminosity, absolute magnitude, or luminosity relative to that of the Sun. 12 From the H-R diagram, the surface temperature of Betelgeuse is about 3000 K and its luminosity is about 70 000 times the Sun’s. L Radius of Betelgeuse = T 2 70 000 2 =  3000     5800  = 989 solar radii

2 4 10   1  =      1   2  = 6.25 Its luminosity is 6.25 times that of the Sun. I B m m  31 14 (a) = 2.512 A B = 2.512 = 39.8 I A A star of apparent magnitude 1 is 39.8 times brighter than another one of apparent magnitude 3.

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

(b) A difference of 4 magnitudes means difference in intensities by a factor of 39.8, as calculated in (a). As a result, a star of apparent magnitude 3 is 39.8 times brighter than another one of apparent magnitude 7. Therefore, a star of apparent magnitude 1 is (39.8  39.8 =) 1580 times brighter than another one of apparent magnitude 7. 15 (a) Since I  T4, intensity of the star = 24 = 16 intensity of the Sun Each cm2 of the star’s surface radiates 16 times more than the Sun does. intensity surface area (b) star intensity surface areaSun 2 I star rstar =  I 2 Sun rSun 2  5  = 16     1  = 400 The whole star radiates 400 times more than the Sun does. 1 16 (a) Distance of Canopus = = 96.2 pc 0.0104  d  (b) Absolute magnitude = m  5 log  10  = 0.62   96.2  5log   10  = 5.54

(c)

= 2.5124.85.54 = 13 700

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

The luminosity of Canopus is 13 700 7 (a) Orbital speed of the Earth 2πr times that of the Sun. = 13 700 T 11 L 2 2π (1.510 ) (d) R = =  7400  = 71.9 solar radii = T 2   365 24 60 60  5800  = 29 900 m s1 Practice 4.3 (p. 128) Minimum speed of the Earth as seen by the space traveller 1 D = 100 000  29 900 2 B 1 λ v = 70 100 m s 3 By  r , λ c Maximum speed of the Earth as seen by radial velocity of the galaxy the space traveller  λ  = 100 000 + 29 900 =  c  λ  = 129 900 m s1 λ v = 0.2  3  108 (b) By  r , λ c = 6  107 m s1 (away from the Earth) for minimum wavelength, λ vr 4 By  , v λ c  = λ  r c v  25103  = λ  r = 500  = 0.0417 nm 129 900 c 3108 = 700  3108 We see the spectral line at (500  0.0417 =) = 0.301 nm 499.9583 nm. Minimum wavelength = 700  0.301 λ v 5 By  r , λ c = 699.699 nm 800 For maximum wavelength, vr  = λ  = 650  3.6 = 0.000 481 nm vr c  = λ  3108 c 70 100 We would see it at (650 + 0.000 481 =) = 700  3 108 650.000 481 nm. λ v = 0.164 nm 6 By  r , λ c Maximum wavelength = 700 + 0.164 v 1000103 = 700.164 nm  = λ  r = 656.3  = 2.19 nm c 3108 8 (a) Red shift of a galaxy means that visible We would see it at (656.3 + 2.19 =) light from it is slightly shifted to the 658.49 nm. long-wavelength side when received by us. (b) When a galaxy recedes from our galaxy, the visible light from it is stretched and

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

the wavelength increases when received by us, causing the red shift. The red shift suggests that the universe is expanding. 9 (a) Time / h  / 1015 m Radial velocity / m s1 76 114 52.1 85 77.9 35.6 94 15.4 7.04 100 30.6 14.0 (b)

(c) The graph repeats in 96 hours. Therefore the time T for the planet to make one complete revolution is 96 hours. 4π 2 r 3 (d) By T2 = , GM 1  2  3 r =  GMT   2   4π  Radius of the planet’s orbit r 1  11 30 2  3 = (6.6710 )(410 )(963600)  2   4π 

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

= 9.31  109 m Revision exercise 4 Multiple-choice (p. 132) 1 D 2 C By Stefan’s law, I  T4, therefore the energy emitted would become (34 =) 81 times its present value. 3 A 4 D 5 C 6 D

Conventional (p. 132) 1 1 (a) Distance to the star = (1M) 0.08 = 12.5 pc (1A) (b) 12.5 pc = (12.5  3.26) ly (1M) = 40.75 ly (1A) 2 A difference of 5 magnitudes are different in intensities by a factor of 100 times. I B = 1 000 000 = 1003 I A Apparent magnitude of star B = +14  (5  3) (1M) = 1 (1A) 3  = 525.17  525 = 0.17 nm λ v By  r , λ c radial velocity  λ  =  c (1M)  λ   0.17  =    3  108  525  = 9.71  104 m s1 (away from the Earth) (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

4 The whole spectrum would be shifted to = M  31.6 (1M) longer wavelengths compared with that if it is at rest. (1A) In this case, the star spears to emit radiation with longer wavelengths, (1A) so it would appear to be cooler. (1A) 5 (a) Absolute magnitude  d  = m  5 log   (1M) 10   30  = +4  5 log    10  = 1.61 (1A)  d  (b) By M = m  5 log   , (1M) 10   d  5 = +5  5 log   10  d = 1000 pc (1A) The star is 1000 pc from the Earth. 6 (a) Star B (1A) L B M M  (b) = 2.512 A B LA

4.5 M M  = 2.512 A B 0.5

MA  MB = 2.39 (1M)

  d A    d B  m A  5 log   mB  5log  = 2.39   10    10   d   B  (mA  mB) + 5 log   = 2.39  d A  d B = 3.01 (1A) d A The distance to star B is 3.01 times that to star A. 7 Apparent magnitude as seen on the Earth  d  = M + 5 log   10   1  = M + 5 log   10 206 265 

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

Apparent magnitude as seen on Saturn  = 5.72 (1A)  d  = M + 5 log   10   9.54  = M + 5 log   10 206 265  = M  26.7 (1M) Difference in apparent magnitude = (M  31.6)  (M  26.7) = 4.9 (1A) The apparent magnitude as seen on the Earth is smaller than that as seen on Saturn by 4.9. 8 (a) Distance of the star from the Earth 1 = (1M) 0.05 = 20 pc (1A) (b) Absolute magnitude  d  = m  5 log   (1M) 10   20  = 5  5 log    10  = 3.49 (1A) I m m (c) By = 2.512 0 , (1M) I 0  I    log  m = m0   I 0  log 2.512 log 60 = 5  (1M) log 2.512 = 0.555 (1A) Its apparent magnitude is 0.555. 9 (a)

 100 tan = (1M) 2 2000

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

The angular diameter of the nebula is 5.72. The size of the nebula is larger than that of the Andromeda as seen from the Earth. (1A) (b)

3 100 tan = (1M) 2 d d = 3820 pc (1A) The nebula would have to be at 3820 pc from the Earth. I m m By = 2.512 0 , (1M) I 0 10 = 2.51226.8m m = 29.3 (1A) The apparent magnitude of the nebula is 29.3.  d  10 (a) By M = m  5 log   , (1M) 10  minimum apparent magnitude m 100 000  = +5 + 5 log    10  = +25 (1A) (b) (1M)

log(3.8107 ) =  5  log 2.512 = 13.9 (1A) The absolute magnitude of LBV 1806- 20 is 13.9.

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

(c) From (a), we know that the minimum apparent magnitude of stars that the HST can see is +25.  d  By M = m  5 log   , (1M) 10   d  13.9 = +25  5 log   10  d = 6.03  108 pc (1A) 11 (a) Star L T R Sun 1 1 1 Barnard’s 0.0045 0.56 0.21 Star Sirius A 23 1.6 1.9 Vega 51 1.6 2.8 Arcturs 160 0.78 21 (3 correct answers.) (1A) (4 correct answers.) (2A) (b) Star Spectral class Sun G Barnard’s Star M Sirius A A Vega A Arcturs K (3 correct answers.) (1A) (5 correct answers.) (2A) (c) Star Type Sun main sequence star Barnard’s Star main sequence star Sirius A main sequence star Vega main sequence star Arcturs giant (3 correct answers.) (1A) (5 correct answers.) (2A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

12 (a) (i) Mass contributed to the gravitational force 2  R  π  =  2   N  m πR 2 1 = Nm 4 Orbital speed GM = r  1  G Nm 4 =   R 2 GNm = (1A) 2R (ii) Mass contributed to the gravitational force = Nm GM GNm Orbital speed = = r R (1A) (iii) Mass contributed to the gravitational force = Nm GM GNm Orbital speed = = r 2R (1A) (b)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

(Correct labelled axes.) (1A) (Correct curve.) (1A) (Passing through the three points in (a).) (1A)

(c) The curve for RX  R will shift upwards, depending on the density and distribution of the invisible stars. For same density as that within the disc of radius R, the rotational curve will be as follow:

(Correct curve.) (1A) 4.000 42  3.999 58 13  = = 0.000 42 nm 2 4.000 42  3.999 58  = = 4.000 00 nm 2 λ v By  r , (1M) λ c radial velocity of the spacecraft  λ  =  c  λ   0.000 42  =  3108  4.000 00  = 3.15  104 m s1 GM By v = , (1M) r

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

mass of the planet (All points plotted correctly.) (2A) v 2 r = (Points plotted correctly except one.) G (1A) (3.15104 ) 2 (85 000103 ) = (Smooth curve through points.) (1A) 11 6.6710 (b) This is because the relative motion 27 = 1.26  10 kg (1A) between source and observer (1A) 30 14 Mass of the star = 1.99  10  0.32 causes the Doppler effect. (1A) = 6.368  1029 kg λ v (c) By  r , (1M) For the inner planet, λ c 6 Tinner = 30  24  3600 = 2.592  10 s star’s orbital speed 2 3  λ  2 4π ainner Tinner = (1M) =  c GM  λ   393.69  393.40  ainner =  3108 (1M) 1  393.40   GMT 2  3 =  inner  = 2.21  105 m s1 (1A)  4π 2    (d) (i) Period measured from the graph 3 6.671011  6.3681029  (2.59106 ) 2  = 70 hours (1M) =  2   4π  = (70  3600) s 5 = 1.93  1010 m = 2.5  10 s (1A) 2 2 2πr Tinner Touter (ii) By v = , (1M) By = , (1M) T a 3 a 3 inner outer radius of the orbit 1  2  3 vT  Touter  3 = aouter =    a  2π  T  inner   inner   2.21105  2.5105 = (1M) 1 2 2π  2   3 =    (1.931010 )3  = 8.79  109 m (1A)  1     16 (a) (i) Doppler effect (1A) 10 = 3.06  10 m (1A) (ii) The universe is expanding. (1A) The semimajor axis of the orbit of the outer (b)  = 650  590 = 60 nm planet is 3.06  1010 m. λ v By  r , (1M) 15 (a) λ c velocity of the galaxy relative to the Earth  λ  =  c  λ   60  =  3108  590 

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

= 3.05  107 m s1 (1A) (c) (i) For diffraction grating, d sin  = n (1M) nλ sin  = d = 2  (590  109)  (4.5  105) (1M)  = 32.1 (1A) (ii)

(Appropriate diagram.) (1A) Light from slits is coherent. (1A) At positions with path difference equal to a multiple of one wavelength, (1A) the light waves arrive in phase, (1A) and constructive interference occurs to produce bright spectral lines. (1A) (For effective communication.) (2C) 17 (a) The recessional speed increases with distance. (1A) (b) Red shift / examined line emission spectra / Doppler effect (1A) (c) Galaxies moving away from each other (1A) suggests that they started out from a single point / explosion. (1A) (d) 1000 (1A) million light years (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

18 (a) (i) Galaxies A and B. (1A) The wavelength is longer or the light is red-shifted. (1A) (ii) Red shift /Doppler effect (1A) (iii) Milky Way Galaxy (1A) (b) (i)

(3 points plotted correctly.) (1A) (All points plotted correctly.) (2A) (Best fit line.) (1A) (ii) The graph is a straight line through speed the origin, which means distance is a constant. (1A) 19 (a) The planets orbit the Sun in ellipses, with the Sun at one focus. (1A) The line joining the Sun and a planet sweeps out equal areas in equal intervals of time. (1A) The square of a planet’s orbital period is proportional to the cube of its semimajor axis of its orbit (or planet’s average distance from the Sun). (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

λ v 3 2 (b) (i) By  r , (1M) GMT λ c (vi) By r = , 4π 2 change in wavelength radius of the planet’s orbit  vr  3 =  λ (6.671011)(41030 )(72 3600)2  c  = 4π2  6.1  9 =   656.310 (1M) (1M)  3108  = 7.69  109 m (1A) = 13.3  1015 m (1A) (ii) Physics in articles (p. 137) (a) From Figure b, the percentage drop in the I intensity during the eclipse is about I

0.25%, where I = Ip+ Is is the total intensity of

the planet (Ip) and star (Is) together. (1M) I s   0.9975 (1M) I p  I s

I p  I s I p 1   1   1.0025 I s I s 0.9975

 Ip : Is = 0.0025 : 1 (1A) (b) (i) The intensity drops at the beginning from t = 1.5 h to t = 1 h. This represents the period for the partial eclipse as shown in the figure below. (4 points plotted correctly.) (1A) (1A) (All points plotted correctly.) (2A) (iii) (Reasonable curve.) (1A) (iv)

(Correct position.) (1A) From t = 1 h to t = 1 h, the planet (v) T = 72 h (1A) travels a distance that is about the diameter of the star as shown in the figure below. During this period, the intensity stays low. (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

The intensity increases back to the normal value from t = 1 h to t = 1.5 h. This represents the period for the partial eclipse as shown in the figure below. (1A)

(Correct positions drawn.) (1A) (ii) From (b)(i), we have:

D p = 0.5 hr (1M) v p

Ds  D p = 2 hr (1M) v p Therefore,

Ds  D p 2 = D p 0.5

Ds  Dp = 4Dp

D p 1 = (1A) Ds 5 (c) The star is hot and emits both visible and infra-red radiation. (1A) However, the planet is not hot enough to emit visible light. (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011 E1 Astronomy and Space Science Chapter 4 Stars and the Universe

Hence, the contrast in visible light between the star and the planet is enormous. On the other hand, the planet is warm and emits significant amount of infra-red radiation. (1A) Therefore, the contrast in infrared is much lower. (d) The hotter the object, the more infra-red radiation it emits. (1A) Therefore, by measuring the infra-red intensity of an extrasolar planet, one can get a rough estimate of its surface temperature. (1A) To get a more accurate result, one needs to measure the distribution of the emission intensity in wavelength  the spectrum, and one can compare the emission spectrum of an extrasolar planet to those of the objects heated up to known temperatures in the laboratory. (1A)

New Senior Secondary Physics at Work 1  Oxford University Press 2011