Download 1b91: answers to problem sheet no 1

1B91: ANSWERS TO PROBLEM SHEET NO 1
QUESTION ONE
Briefly (not more than half a page) describe four of the discoveries that Galileo made by
observing the heavens with a telescope. For each discovery, state why they were
“revolutionary”compared to the beliefs at the time.
From the lectures notes:
•The Moon is not smooth but has mountains and craters
•The Sun is not perfect but has sun spots
•The Milky Way is composed of thousands of stars
•Venus exhibits phases like those of the Moon which means it has to be between the
Earth and the Sun
•The apparent size of Venus is related to the planet’s phase which means that Venus
and the Earth both orbit the Sun
•Jupiter has four moons (known as the Galilean satellites) which are in orbit around it
- a Copernican system in miniature
First two were revolutionary because they showed that heavenly bodies were not pure or
perfect – third must also show this to some extent. Last three supported revolutionary view
that the Earth and the other planets must orbit the Sun.
Marks – total 8 (1 for each discovery and 1 for showing why it was revolutionary).
QUESTION TWO (Total marks = 8)
(a) What is the wavelength of a radio signal of frequency 200 MHz?
(b) The blackbody spectrum of star A peaks at a wavelength of 200 nm. That of star B peaks
at 650 nm. (i) What observational wave-bands do these peaks occur in? (ii) Which star is
hotter and by what factor? (iii) How many more times more energy per unit area does the
hotter star radiate per second?
(a) Using equation ν= c/λ where c=velocity of light = 3 x 108 m/s and the frequency
ν= 200 MHz = 200 x 106 Hz, then the wavelength λ = (3 x 108)/( 200 x 106) = 1.5 m.
Answer – the wavelength of this radio signal is 1.5 metres. [2 marks]
(b) (i) Star A - 200 nm is in the ultraviolet waveband; star B peaks in the visible waveband.
[2 marks]
(b) (ii) Using Wien’s law λmax = (3 x 106)/T, then star A has to be hotter since its energy
distribution peaks at a shorter wavelength. Factor = λmax(star B)/ λmax(star A) = 650/200 =
3.25. Star A is hotter than star B by a factor of 3.25 [2 marks]
(b) (iii) From lecture notes, energy emitted per second ∝ R2 T4. Thus the energy emitted per
unit area per second ∝ T4. Thus the hotter star will emit (3.25)4 times more energy per unit
area per unit second = 111.6. [2 marks]
QUESTION THREE (Total marks = 8)
Outline the main phenomena observed in the Sun during its most active phases.
Determine the fraction of energy emitted per unit area by a sunspot of temperature 4500 K
compared to the surrounding photosphere of temperature 5800 K.
When the Sun is in an active phase, the phenomena that are seen are:
•
•
•
Increased number of sun spots – this is due to the increasing complexity of the solar
magnetic field
Solar flares – brief eruption of hot, ionized gases from a sunspot group
Coronal mass ejections - much larger eruption that involves immense amounts of
gas erupting from the corona. [6 marks]
Using the same equation as given in Q2, energy emitted per unit area per second ∝ T4
and thus fraction of energy emitted per unit area by the sunspot = (4500/5800)4 = 0.36.
[2 marks]
QUESTION FOUR (Total marks = 4)
Briefly explain what is meant by the term hydrostatic equilibrium. Why do nuclear reactions
in the Sun occur only in the core?
A star is in hydrostatic equilibrium if, at every point within the star, the inward force of the
material above due to gravity exactly balances the outward pressure force from the material
below. [2 marks]
Nuclear reactions can only occur in the core of the Sun because very high temperatures (in
excess of 10 million K) and therefore high pressures/densities are required for four protons
to fuse to form a helium nucleus. [2 marks]
QUESTION FIVE (Total marks = 6)
ζ Puppis is an O4 I star and τ Sco is a B1 V star. State what is meant by these classification
terms and give an indication of the surface temperatures of these stars.
Stars are classified according to their surface temperatures with the spectral sequence
defined as OBAFGKM in order of decreasing temperature. Each spectral class is divided
into subclasses from 0-9. [2 marks]
The first star is an O4 star which is one of the hottest spectral types with a temperature of ~
35,000 K. The second star is a B1 star which is slightly cooler than an O star with a
temperature of ~ 30,000 K. [2 marks]
Stars are also classified according to their luminosities with three main classes I, III and V
standing for supergiants, giants and dwarfs. The first star is a supergiant and the second is a
dwarf. [2 marks]
QUESTION SIX (Total marks = 5)
The maximum in the flux distribution of a white dwarf star and a red giant star occurs at 30
nm and 1500 nm, respectively. If their corresponding radii are 106 m and 1011 m, determine
their stellar temperatures and the ratio of their luminosities.
Using Wien’s law, λmax = (3 x 106)/T, then the temperature of the white dwarf = (3 x
106)/30 = 100,000 K and the temperature of the red giant is (3 x 106)/1500 = 2000 K.
[2 marks]
The luminosity is ∝ R2 T4, and thus the ratio of the luminosities is given by:
LWD/LRG = RWD2 TWD4/ RRG2 TRG4 = (1012 x 1020)/(1022 x (2x103)4 = 6.25 x 10-4. [3 marks]
QUESTION SEVEN (Total marks = 6)
(a)What does the main-sequence represent in the H-R diagram?
(b) Briefly explain why the more massive a star is, the shorter its lifetime.
The band called the main sequence in the H-R diagram runs diagonally from hot, luminous
blue stars in the upper left hand corner to cool, red, faint stars in the lower right hand
corner. It represents the part of a star’s evolution when it is fusing hydrogen to helium in its
core. [2 marks]
A higher mass star has stronger gravity. This means the star must have a higher pressure to
be stable. This in turns means a higher interior temperature. A higher interior temperature
means hydrogen fusion occurs at a faster rate. This means that the star is more luminous and
has a shorter lifetime. Main sequence lifetime ∝ available fuel supply (=mass)/rate of
consumption (= luminosity) ∝ M/L. [4 marks]
TOTAL MARKS = 45