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M09/4/PHYSI/SP3/ENG/TZ1/XX+
Option E — Astrophysics
E1. This question is about stars.
(a) Distinguish between apparent magnitude and absolute magnitude.
● apparent magnitude is a measure of how bright a star appears from Earth;
● absolute magnitude is a measure of how bright a star would appear from a distance of 10 pc;
[2]
(b) The table gives information on three stars, Achernar, EG 129 and Mira.
(i) State which one of the three stars appears brightest from Earth.
[1]
● Achernar;
(ii) Estimate the ratio


where LA is the luminosity of Achernar and LE is the luminosity of EG 129.
[3]
● stars differ by ΔM = 16;
● for ΔM = 1 we have a ratio of luminosities by a factor ≈ 2.51
● so


= 2.51
16
≈ 2.5 x 106
(iii) Show that the distance of the star Achernar from Earth is approximately 50 pc.
● m – M = 5 log
≈ 10 × 10
3.5
5
d
10
≈ 50 
⟹  = 10 × 10
−
5
[2]
(c) The surface temperature of Mira is 5 times lower than that of Achernar.

Estimate the ratio  where RM is the radius of Mira and RA is the radius of Achernar.

[3]
●


=
2 4
4

2 (5)4
4
=1 ⟹


= 25
E2. This question is about cosmic microwave background radiation.
The graph shows the spectrum of the cosmic microwave background radiation.
The shape of the graph suggests a black body
spectrum i.e. a spectrum to which the Wien
displacement law applies.
(a) Use the graph to estimate the black body temperature.
[2]
2.9 × 10−3
● =
1.07 × 10−3
● T = 2.7 K
● Accept wavelengths in the range 1.05 to 1.10 for a temperature range 2.64 to 2.76 K.
(b) Explain how your answer to (a) is evidence in support of the Big Bang model.
[2]
● according to the Big Bang model the temperature of the universe (and the radiation it
contained) in the distant past was very high;
● the temperature falls as the universe expands and so does the temperature
of the radiation in the universe;
(c) State and explain another piece of experimental evidence in support of the Big Bang model.
[2]
● (Hubble’s law shows that) the universe is expanding;
● therefore in the distant past the universe must have been a very small/hot/dense point-like object;
or
● Doppler shift of spectral lines;
● indicates galaxies moving away so in the past they were close to each other;
M09/4/PHYSI/SP3/ENG/TZ2/XX+
Option E — Astrophysics
E1. This question is about the star Antares.
The star Antares is a red supergiant star in the constellation Scorpius.
(a) Describe three characteristics of a red supergiant star and state what is
meant by a constellation.
[4]
Red supergiant star:
●
●
●
●
●
appears red in colour;
(has a very) large luminosity;
(relatively) low (surface) temperature;
(very) large mass;
(very) large surface area;
[3 max]
Constellation:
● a group of stars that form a recognizable pattern (as viewed from Earth) / OWTTE;
[1 max]
(b) The apparent magnitude of Antares is + 1.1 and its absolute magnitude is – 5.3.
(i) Distinguish between apparent magnitude and absolute magnitude.
[2]
(ii) Show that the distance of Antares from Earth is 3.9x107 AU.
● m – M = 5 log
d
10
⟹ 1.1 − (−5.3) = 5 log
[3]
d
10
● ⟹  = 190 
190 × 3.26 × 9.46 × 1015
● ⟹ =
= 3.9 × 107 
11
1.5 × 10
(iii) State the name of the method that is used to measure the distance of Antares
from Earth.
[1]
● stellar / spectroscopic parallax
(c) The apparent brightness of Antares is 4.3 x10-11 times the apparent brightness of the Sun.
(i) Define apparent brightness.
[1]
● the power per square meter received at the surface of Earth / observer
(ii) Using the answer to (b)(ii), show that Antares is 6.5 x104 times more luminous than the Sun.
▪
2
2
 4 
 
(4.3 × 10−11  )(3.9 × 107  )2
=
=
=
2
2
2

4 
 
 
▪

= 4.3 × 10−11 × 3.92 × 1014 = 6.5 × 104

[3]
E2. This question is about models of the universe.
Observations of the night sky indicate that there are many regions of the universe that do not contain
any stars.
(a) Explain why this observation contradicts Newton’s model of the universe.
[3]
● Newton’s model states that the universe is infinite (static) and uniform;
● this means that stars are uniformly spaced;
● and that if it is infinite there must be a star at every point in space
/ a star along every line of sight;
● since there are regions without stars, Newton’s model must be inadequate;
[3 max]
(b) Outline how the Big Bang model of the universe is consistent with this observation.
[3]
● both space and time originated with the Big Bang;
● the universe is expanding (and not infinite);
● due to the expansion, light from the Big Bang is red-shifted to the microwave region
so regions between stars will not appear bright;
● light from very distant stars will not have reached us yet;
● the universe has not existed for all time;
[3 max]
Option E — Astrophysics
N09/4/PHYSI/SP3/ENG/TZ0/XX+
E1. This question is about the star Becrux and Cepheid variables.
(a) Describe what is meant by
(i) the apparent magnitude scale.
[2]
● gives the relative (visual) brightness of stars as seen from Earth;
● e.g. a magnitude 1 star is 100 times brighter than a magnitude 6 star;
To award [2] the idea of a relative scale must be clear.
(ii) absolute magnitude.
● the apparent magnitude a star would have if it were 10 pc from Earth;
[1]
(b) Becrux is a main sequence star and is one of the stars that make up the Southern Cross.
The following data are available for Becrux.
Apparent magnitude = 1.25
Absolute magnitude = –3.92
Apparent brightness = 7.00 x 10-12 bSun
bSun is the apparent brightness of the Sun. Use the data to deduce that the
(i) distance of Becrux from Earth is 108 pc.
▪ m – M = 5 log
d
10
⟹ log
d
10
=
[3]
−
5
= 1.03
▪ d = 10 x 101.03 = 108 pc
(ii) luminosity of Becrux is 3.43 x 103 LSun where LSun is the luminosity of the Sun.
(1 pc = 2.05 x105 AU)
▪ L = 4π d2 b

 2
(4.3 × 10−11  )(3.9 × 107  )2
▪
=
=
2
2
  
 
▪  = 3.43 × 103 
[3]
(c) Becrux is a spectral class B star. On the axes of the Hertzsprung–Russell diagram label
with the letter B the approximate position of Becrux.
B
● in the region [30 → 50, – 2.55 → – 5.0]
[1]
(d) On the axes of the Hertzsprung–Russell diagram , draw the approximate region
in which Cepheid variable stars are located.
[1]
● Cepheid as shown;
● Judge by eye for reasonable range
of magnitude and temperature.
(e) State the reason for the periodic variation in luminosity of a Cepheid variable.
● the outer layers undergo a periodic expansion and contraction/periodic fluctuations in
temperature;
(f) State the two quantities that need to be measured in order to use a Cepheid variable
as a “standard candle” to determine the distance to the galaxy in which the Cepheid
is located.
[2]
● period/frequency with which luminosity varies
● apparent brightness / apparent magnitude;
[1]
E2. This question is about cosmology.
(a) The diagram below represents a spherical region of space based on Newton’s model of the
universe. Earth is at the centre of the region. The dark line represents a very thin spherical
shell of space distance R from Earth.
With reference to the diagram and Newton’s model of the universe explain quantitatively
Olbers’ paradox.
[4]
●
●
●
●
Newton’s model assumed a uniform infinite (and static) universe;
therefore number of stars in shell is proportional to R2
intensity of radiation/light from shell reaching Earth is proportional to 1/R2
since according to Newton’s model such shells stretch to infinity
/ the sky can never be dark/willalways be light
/ OWTTE;
(b) Outline how the Big Bang theory provides a resolution to Olbers’ paradox.
● the universe is expanding;
● the universe has a beginning;
● the stars (and galaxies) are not uniformly distributed;
[2]
[2 max]
M10/4/PHYSI/SP3/ENG/TZ1/XX+
Option E — Astrophysics
E1. This question is about determining some properties of the star Wolf 359.
(a) The star Wolf 359 has a parallax angle of 0.419 arcseconds.
(i) Describe how this parallax angle is measured.
[4]
● angular position of star measured;
● relative to the background of fixed stars;
● in two positions six months apart;
● p is ½ of the angle of separation
/ p indicated on diagram;
(ii) Calculate the distance in light-years from Earth to Wolf 359.
● =
1
1
=
= 2.3866  ≈ 2.39 
 0.419
● = 2.3866 x 3.26 ly = 7.78 ly
[2]
(iii) State why the method of parallax can only be used for stars at a distance of less than
a few hundred parsecs from Earth.
[1]
● beyond this distance the parallax angle is too small to be measured (accurately)
/ OWTTE
 ℎ   359
 3.7 × 10−15
 ℎ 
(b) The ratio
Show that the ratio
   359
= 8.9 × 10−4 (1  = 6.3 × 104 )
 
▪ L = 4 π d2b
2
 

▪
= 2

 
▪ dS =1 AU
▪


dW = 7.78x 6.3 x 104 = 4.9 x 105 AU
= [4.9 x 105]2 x 3.7 x 10-15 = 8.9 x 10-4
[4]
(c) The surface temperature of Wolf 359 is 2800 K and its luminosity is 3.5×1023 W.
Calculate the radius of Wolf 359.
[2]

3.5 × 1023
▪ =
=
= 1.0 × 1017 2
4
−8
4

5.7 × 10 × 2800
▪ =

=
4
1.0 × 1017
= 8.9 × 107 
4
(d) By reference to the data in (c), suggest why Wolf 359 is neither a white dwarf nor a
red giant.
● temperature too low to be white dwarf;
● luminosity too low to be red giant;
● radius too small to be a red giant; [2 max]
Answer must be consistent with answer in (c) for third marking point.
[2]
E2. This question is about the density of the universe.
(a) Define critical density.
[1]
● density at which universe will expand forever but rate of expansion will approach zero
/ the density at which the universe will begin to contract after infinite amount of time
/ the density for which the curvature of the universe is zero
/ OWTTE;
Reference to “flat” model without definition does not gain mark.
(b) Explain how the future of the universe may be predicted by comparing the estimated
density of the universe to the critical density.
[3]
● value of density determines whether or not universe will expand forever, or
at some point, begin to contract;
● at density less than critical density, universe will expand forever;
● at density greater than critical density, universe will stop expanding and contract;
If second and third marks gained, first mark is also gained by implication.
(c) Explain why the existence of dark matter makes it difficult to measure the density of
the universe.
● dark matter does not radiate
/cannot be directly measured/seen;
[1]
M10/4/PHYSI/SP3/ENG/TZ2/XX
Option E — Astrophysics
E1. This question is about the relative population density of stars and galaxies.
The number of stars around the Sun, within a distance of 17 ly, is 75. The number of galaxies in the
local group, within a distance of 4.0 x 106 ly from the Sun, is 26.
(a) Calculate the average population density, per ly3, of stars and galaxies.
Stars:
Galaxies:
▪
75
=
[2]
75
= 3.6 × 10−3  −3
3
4.19 (17)
4
()3
3
26
26
▪
=
= 9.7 × 10−20  −3
6
3
4
()3 4.19 (4.0 x 10 )
3
● Award [1 max] if the response does not use the volume of the sphere but uses the cube instead
(b) Use your answer to (a) to determine the ratio
average population density of stars
average population density of galaxies
10−3
▪
= 3.8 × 1016
−20
10
     ℎ  
       16.
[1]
E2. This question is about the luminosity, size and distance of stars.
The Hertzsprung–Russell (HR) diagram shows the variation with spectral class of the
absolute magnitude of stars.
The star Capella and the Sun are in the same spectral class (G). Using the HR diagram,
(a) (i) suggest why Capella has a greater surface area than the Sun.
[2]
● luminosity is a function of surface and temperature (of star);
● (same class) same temperature (therefore greater surface area);
(ii) estimate the luminosity of Capella in terms of that of the Sun.
● LC = 80 LS
(accept answer in the range of 60 to 100)
(iii) calculate the radius of Capella in terms of that of the Sun.
▪


[1]
=
 2

= 80
▪  2 = 80  2
⟹
 = 8.9 
[2]
(b) The spectroscopic parallax method can be used to measure the distance of star Vega.
(i) Using the HR diagram, state the absolute magnitude of Vega.
[1]
● 0.6 (accept answer in the range of 0.4 to 0.8)
(ii) The apparent magnitude of Vega is 0.0. Determine (in parsec) the distance of Vega from Earth.
● m – M = 5 log
●
d
10
⟹ 0.0-0.6 = 5log
[3]
d
10
d
= 10−0.12
10
● d = 7.6 pc
(iii) Light from Vega is absorbed by a dust cloud between Vega and Earth. Suggest the effect,
if any, this will have on determining the distance of Vega from Earth.
● Vega appears dimmer;
● hence distance over-estimated;
accept:
● Vega will look redder (because blue light scatters more in dust);
● so Vega looks cooler/lower apparent temperature (thus wrong position on HR diagram);
(c) Vega is a very massive star. State why Vega does not undergo gravitational collapse.
● the inward gravitational pressure is balanced by the outward radiative pressure;
[1]
[2]
E3. This question is about cosmic microwave background radiation (CMB)
and the density of the universe.
The graph shows the relative intensity of the CMB as a function of wavelength.
(a) Explain how this graph is consistent with the Big Bang model of the universe.
● (Big Bang theory predicts that CMB will) correspond to the black-body at 3K
● the graph is of a black-body curve;
2.9 × 10−3 2.9 × 10−3
● =
=
≈3

10−3
[3]
(b) The density of the universe will determine its ultimate fate. Outline the problems
associated with determining the density of the universe.
[2]
● measurement of mass in a given volume is (very) uncertain/difficult;
● there exists dark matter that is difficult to observe;
● measurement of distances is uncertain/difficult;
● matter not uniformly distributed;
[2 max]
Option E — Astrophysics
N10/4/PHYSI/SP3/ENG/TZ0/XX
E1. This question is about the characteristics of the stars Procyon A and Procyon B.
(a) The stars Procyon A and Procyon B are both located in the same stellar cluster in the
constellation Canis Minor. Distinguish between a constellation and a stellar cluster.
[2]
Constellation:
● a collection/group of stars that form a recognizable pattern (as viewed from Earth)
/ a group/pattern of stars not close together (in space);
Stellar cluster:
● a group of stars (including gas and dust) held together by gravity/forming a globular/open
arrangement / a group of stars close to each other (in space);
The table shows some data for Procyon A and Procyon B.
Apparent
magnitude
Absolute
magnitude
Apparent brightness
/ W m–2
Procyon A (PA )
+ 0.400
+ 2.68
2.06 x 10–8
Procyon B (PB )
+ 10.7
+ 13.0
1.46 x 10–12
(b) Explain, using data from the table, why
(i) as viewed from Earth, PA is much brighter than PB.
(ii) the luminosity of PA is much greater than that of PB.
(i) • the apparent magnitude of PA is (much) smaller than that of PB;
• the smaller apparent magnitude, the brighter the star
or
• apparent brightness of PA is greater than PB;
• apparent brightness is intensity at surface of Earth;
[2]
(ii) • the absolute magnitude of PA is smaller than that of PB;
• the absolute magnitude is the apparent magnitude at a distance of 10 pc from the Earth;
• so at the same distances from Earth PA is much brighter than PB so must be more luminous;
or
[3]
• absolute magnitude of PA is less than absolute magnitude of PB;
• absolute magnitude is a measure of luminosity;
• lower values of absolute magnitude refer to brighter/more luminous star;
or
• Accept answer based on answer to (c).
• distances are the same from (c);
• since L = 4πd 2b PA is brighter than PB;
Apparent
magnitude
Absolute
magnitude
Apparent brightness
/ W m–2
Procyon A (PA )
+ 0.400
+ 2.68
2.06 x 10–8
Procyon B (PB )
+ 10.7
+ 13.0
1.46 x 10–12
(c) Deduce, using data from the table, that PA and PB are approximately
the same distance from Earth.
[2]
• for PA
m – M = – 2.28 for PB
m – M = – 2.30

• since m – M = 5 lg
then d for each star is very nearly same;
10
or
−
5
• d = 10 x 10
•  = 10 × 10−0.456 ≈ 3.5 pc
 = 10 × 10−0.45 ≈ 3.5 pc
(d) State why PA and PB might be binary stars.
[1]
• same distance from Earth and in the same region of space
(e) Calculate, using data from the table, the ratio
●

.

[2]
recognize that the ratio of the luminosities is the same as the ratio of apparent brightness;
 42 
2.06 × 10−8
•
=
=
= 1.41 × 104
2
−12

1.46 × 10
4 
(f ) The surface temperature of both PA and PB is of the order of 104 K. The luminosity of PA
is of the order of 10 LS, where LS is the luminosity of the Sun. The diagram shows the grid
of a Hertzsprung–Russell diagram.
● PA → 10 000 K at 10; (labelled A)
● PB → 10 000 K at 10–3; (labelled B)
Label, on the grid above, the approximate position of
(i) star PA with the letter A.
(ii) star PB with the letter B.
(g) Identify the nature of star PB.
● white dwarf;
Allow ECF from diagram.
[1]
[1]
[1]
E2. This question is about the Big Bang model and red-shift.
(a) Describe what is meant by the Big Bang model.
[1]
● space and time originated from a single point in a large explosion
/ an expanding universe that originated from a single point / OWTTE;
(b) In the 1960s, Penzias and Wilson discovered a uniform cosmic background radiation
(CMB) in the microwave region of the electromagnetic spectrum.
(i) Explain how the CMB is consistent with the Big Bang model.
[3]
● temperature of the universe immediately after the Big Bang was very high;
as it expanded it cooled down;
● the wavelength of the CMB corresponds to a temperature consistent with
this cooling down / OWTTE;
(ii) State why the red-shift of light from galaxies supports the Big Bang model.
● indicates that the universe is expanding;
[1]
Option E — Astrophysics
M11/4/PHYSI/SP3/ENG/TZ2/XX
E1. This question is about the properties of a star.
(a) Describe what is meant by a
(i) constellation.
[2]
● a collection of stars that form a recognizable group (as viewed from Earth);
● that need not be/are not close to each other/gravitationally bound;
(ii) stellar cluster.
● stars that are gravitationally bound/forming an open arrangement
/close to each other (in space);
[1]
(b) Some data for the variable star Betelgeuse are given below.
Average absolute magnitude = – 5.1
Average apparent magnitude = + 0.60
Average apparent brightness = 1.6 × 10–7 W m–2
Radius = 790 solar radii
The luminosity of the Sun is 3.8 × 1026 W and it has a
surface temperature of 5700 K.
(i) Show that the distance from Earth to Betelgeuse is about 4 ×1018 m.
● m – M = 5 log
d
10
0.6 + 5.1 = 5 log
● d =138 pc
d
10
1.14 = log
d
10
1 pc = 3.1x 1016 m
● d = 138 x 3.1x 1016 m = 4.279 x 1018 ≈ 4 x 1018 m
[3]
(ii) Determine, in terms of the luminosity of the Sun, the luminosity of Betelgeuse.
=
[2]

4 2
●  = (4 2 ) = 1.6 × 10−7 × 4 × ( 4.279  1018 )2 = 3.68 × 1029 

3.68 × 1029
●
=
= 9.7 × 104
26

3.8 × 10
⟹
 = 9.7 × 104 
(iii) Calculate the surface temperature of Betelgeuse.

 4(790 ) 2 4 (790) 2 4
●
=
=
= 9.7 × 104
4
4
2

 4( ) 

4
●  =  9.7 × 104 ×
= 3578  ≈ 3600 
1
7902
[2]
● position labelled B within shaded area;
Award [1] if label B is missing but point is clear.
● generally the correct shape; (allow broad line)
(c) On the Hertzsprung–Russell diagram above,
(i) label the position of Betelgeuse with the letter B.
[1]
(ii) sketch the position of main sequence stars.
[1]
(d) Some stars, such as Betelgeuse, are in combination with a companion star forming
a spectroscopic binary system. Describe and explain the characteristics of a
spectroscopic binary system.
[3]
● over time spectral lines regularly split into two lines and then recombine;
● as one star approaches observer the other recedes;
● leading to Doppler shifts in opposite directions;
E2. This question is about the density of the universe.
(a) Explain, with reference to the possible fate of the universe, the significance of the
critical density of matter in the universe.
[3]
● if less than critical density, universe expands without limit;
● if equal to critical density universe stops expanding after an infinite amount of time;
● if greater than critical density, universe expands first then contracts; [3]
Award [1 max] if terms open, flat and closed are used and not defined.
(b) Suggest one reason why it is difficult to estimate the density of matter in the universe.
● there is matter that cannot be detected;
● which is likely to consist of dark matter/neutrinos;
or
● difficulty of measuring volume accurately;
● because of difficulty of measuring distances accurately;
or
● matter is not evenly distributed;
● so density may vary from place to place;
[2]

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