Astrophysics

Question 1

Apparent Magnitude, m

Answer 1

Apparent magnitude is a measure of the brightness of a star in the night sky when it is seen from earth which depends on the intensity of the light received from the star.

Question 2

Hipparcos Scale

Answer 2

Ranks stars on their brightness from 1 to 6: the lower the number, the brighter the star is

1 is the brightest

6 is the dimmest star visible to the naked eye

Every difference of 5 magnitudes corresponds to 100 times more light

I_x/I_y = 100^ςm/5

Question 3

Absolute magnitude, M

Answer 3

The value of the apparent magnitude if a star is at a distance of 10 parsecs from the observer

Parsec: distance from a star that subtends 1 arc second to the line between the sun and the earth

Light Year: distance travelled by light through space in one year

m - M = 5logd/10

Question 4

Astronomical Distances

Answer 4

Parsec: distance to a star which subtends to an angle of 1 arc second to the line from the earth

Astronomical Unit: the mean earth to sun distance at 149,597,871km

Metre: distance travelled by light in a vacuum in 1/29979347s

Light Year: distance travelled by light through space in one earth year

(see data booklet for conversions)

Question 5

Stellar Parallax

Answer 5

The shifting of nearby stars against the background of more distant stars due to the orbital movement of the earth about the sun

As we are travelling in an orbit, stars appear to shift position. The closer an object is, the more apparent the shift is

d=R/θ

Question 6

Stellar Spectral Classes

Answer 6

O: > 30,000 (He⁺, He, H)

B: 10,000 - 30,000 (He, H)

A: 7,500 - 10,000 (H, ionized metals)

F: 6,000 - 7,500 (ionized metals)

G: 5,000 - 6,000 (ionized and neutral metals)

K: 3,500 - 5,000 (neutral metals)

M: 2,000 - 3,500 (neutral atoms, TiO)

The temperature is related to absorbtion spectra limited to Hydrogen Balmer absorbtion lines: which is a requirement for atoms in an n = 2 state.

• Heavier stars go further up the main sequence
• Super giants are very bright and large
• Giants are cooler giants and will be red
• White dwarfs are very dim and small
• Stars are much larger and heavier towards the top, and are smaller at the bottom

A star will evolve from the main sequence to become one of the other types on the diagram

Question 7

Stellar Luminosity

Answer 7

Luminosity is the power output of a star (releasing energy in all directions)

L (watts) = E (joules) /T (seconds)

Question 8

Inverse Square Law

Answer 8

The energy radiated by a star is spread spherically, hence the intensity of radiation decays as 1/r²

The brightness of a star depends on its luminosity and distance to earth:

B = L/4πr²

Question 9

Classification of Stars

Answer 9

The stars behave like a black body (perfect absorbers and emitters)

Stars emit electromagnetic radiation in a spectrum of wavelengths, the power emitted changes with the temperature

Stefan’s Law states that the power emitted by a black body depends on its temperature and surface area

P = σAT⁴

Wien’s Law states that the black-body radiation curve for different temperatures will peak at different wavelengths that are inversely proportional to the temperature

λ_max T = 2.9 x 10^-3 mK

Question 10

Supergiants

Answer 10

Radius = 1000 x radius of sun

Very bright - M ranging from: -10 to -5

Huge range of temperatures

Question 11

Giants

Answer 11

Radius = 10-100 x radius of sun

Brighter than the sun - M randing from: -2 to +2

Cooler than the sun

Question 12

White Dwarf

Answer 12

Radius = 0.01 x radius of sun

Very dim - M ranging from: +5 to +10

Very hot and small

Question 13

Star Formation

Answer 13

Star Formation

1. A protostar is formed - a gas cloud that is contracting due to its own gravity
2. When the gas cloud is contracting, the potential energy is converted to kinetic energy and the temperature increases
3. Once the temperature reaches 10⁶ Kelvin, hydrogen fusion is able to occur within the core of the star
4. The star must also have enough mass for the temperature to increase to the required magnitude
5. Fusion in the core converts hydrogen to helium
6. The photosphere surrounding the core, has a lower temperature than the core - 2500 K to 50,000 K

Question 14

Moving into the Main Sequence

Answer 14

Main Sequence

1. Once fusion has occurred, the star becomes stable and joins the main sequence
2. The star remains in the sequence for 10⁹ years - large stars remain in the main sequence for a shorter amount of time as they have a higher rate of fusion (they are more luminous and hotter)
3. In a stable star, the forces are in equilibrium

• Gravitational field forces act towards the core in all directions
• Fusion created radiational forces that act outwards

4. As time goes on, the radiation pressure decreases as the star runs out of fuel, meaning the graviational forces are greater than the radiational forces and it is no longer in equilibrium

Question 15

Formation of Red Giant

Answer 15

Red Giant

1. The formation of a red giant takes about 1/3 the time it spends in the main sequence
2. The core will shrink and collapse due to the graviational forces being larger than the radiational forces
3. As the core collapse, the temperature increases, meaning there is enough energy for helium nuclei to begin fusing and producing heavier elements (C, O, Be)
4. In the outer area of the core, there is a small area that continues to burn hydrogen known as the hydrogen-burning shell
5. This causes the photosphere to expand and increase in temperature, and it starts emitting red light, becoming a red giant and exiting the main sequence
6. When the helium begins to fuse, the forces will return to equilibrium

Question 16

White Dwarf Formation

Answer 16

White Dwarf

1. A white dwarf will be formed when the star runs out of helium and the core collapses again.
2. The temperature increases and starts fusing heavier elements to produce elements up to iron
3. The photosphere will expand until the star loses control of it forming a planetary nebula
4. If the mass of the core is less than 1.4 x the mass of the sun then the core will stop contracting and turn into a white dwarf
5. The temperature increases causing it to be white and the luminosity decreases

Question 17

Stellar Evolution

Answer 17

1. Protostar
2. Star
3. Star joins main sequence
4. Red Giant
5. Planetary Nebula
6. White Dwarf

Question 18

Gamma Ray Bursts

Answer 18

Gamma ray bursts occur due to the collapse of supergiant stars to form neutron stars or black holes

• Intense jets of gamma rays have been observed in distant galaxies
• Short-livedbursts (lasting 0.01-1s) are associated with the merger ofneutron starsto form ablack holeor aneutron star falling into a black hole
• Long-lived bursts (10-1000s) are associated with the collapse of a massive star in a supernova

The energy output from a typical gamma ray burst is the same as the total energy output of the sun over its entire lifetime

Question 19

Black Holes

Answer 19

When the mass of the core is three times the mass of the sun, the star will develop into a black hole

It is so dense that nothing can escape from it, not even light

If the velocity > c, they they will be able to escape the black hole

The event horizon of a black hole is the time where the speed that would be required to escape the black holes graviational pull is greater than or equal to c.

Schwarzchild Radius:

Gives the area around the star where all matter will be engulfed

R_s = 2GM/c²

Question 20

Supermassive Black Holes

Answer 20

Supermassive black holes are black holes of unimaginable mass that are believed to exist in the centre of galaxies

They can gain enourmous amounts of mass as they draw in matter from orbiting stars

Orbital speeds and distance from galaxies such as Andromeda and the Milky Way suggest they are orbiting a supermassive centre

Question 21

Occurance of Supernova and the Fate of Massive Stars

Answer 21

Fate of Massive Stars

1. Once the mass of the core is greater than 1.4 x mass of the sun, iron will be produced in the core
2. Iron is produced via fusion
3. Electrons are pushed very close together by the graviational force and they begin reacting with protons
4. As electrons are pushed together, density increases to around 10¹⁷ kgm^-3 which is the density of nuclei leaving a very rigid core
5. Any infalling matter will bounce off in an explosion called a supernova

Supernova

• ​They are very luminous - M from -50 to -20
• The increase in luminosity happens within 24 hours
• This is when some heavier elements can be formed
• Neutron stars are the cores of supernovae, and have a diameter of around 30km

Question 22

Standard Candle

Answer 22

Standard candles are objects of known absolute magnitude/luminosity that can be used to find its distance from earth to tell us distances to the galaxies it belongs to

Cepheids and type 1a supernova are standard candles

Question 23

Classification of Supernovae

Answer 23

Classified based on their line absorption spectra

Type I Supernova:

• No H lines
• Luminosity decreases steadily

Type 1a:

• Silicon absorption lines
• Act as standard candles
• Luminosity of 10⁹ x luminosity of sun
• M ranges from -19.3 to + 0.03
• Result from a white dwarf attracting a companion giant star, carbon fuses to form silicon and the star becomes unstable resulting in an explosion

Type 1b:

• Strong He Lines
• Originates from the collapse of a supergiant

Type 1c:

• No H or He lines
• Originates from the collapse of supergiants that have completely exhaisted their supply of helium

Type 2 Supernova:

• Strong H lines
• Luminosity decreases unsteadily

Question 24

Doppler Effect

Answer 24

The apparent change in frequency and wavelength of a wave caused by the relative motion between the source (star) and the observer (astronomers)

z = v/c

When the wave (source) is moving towards the observer, the frequency increases and the wavelength decreases, and a Blueshift will be seen

• shift in frequency = +v/c
• shift in wavelength = -v/c

When the wave is moving away from the observer, the frequency decreases and the wavelength increases and a Redshift will be seen

• shift in frequency = -v/c
• shift in wavelength = +v/c

∆λ = (v/c) x λ

Question 25

Binary Systems

Answer 25

Binary systems occur when there are two stars in a system orbiting around the centre of mass

Spectroscopy binary stars cannot be resolved by telescopes, spectral lines will show a blue and red shift

V = 2πr/T

• If m₁ > m₂, the centre of mass will be closer to m₁, causing it to have an orbit with r₁ < r₂
• At any time, one of the stars will be redshifted and the other will be blueshifted

Question 26

Hubble’s Law

Answer 26

The further away galaxies are, the larger their recessional speed is

V = H₀d

Redshift from stars and Hubble’s Law suggest that the universe is expanding and that is must have started from a common point

Question 27

Estimating the Age of the Universe

Answer 27

Method 1:

For the galaxy to be receding at the speed of light, its distance would be:

d = s/t

d = (300,000 kms^-1)/(65 kms^-1 Mpc^-1) = 4615 Mpc = 15 x 10⁹ light years

The Universe has an upper age limit of 15 billion years

• Galaxies cannot travel this fast so light cannot have been travelling for this length of time
• Therefore, the universe must be less than 15 billion years old

Method 2:

Assume a constant expansion rate (H₀ constant)

The time taken for a galaxy to travel from the origin of the universe:

t = s/v = d/H₀d = 1/H₀

t = 1 s Mpc/65 km

Question 28

Big Bang Theory

Answer 28

Space time started as a single point and as time passed space expanded

Question 29

Evidence for the Big Bang

Answer 29

Hubble’s Law and Redshift

• V = H₀d
• The further away, the quicker they are moving

Cosmic Microwave Background Radiation

• There is an isotopic glow of radiation, with a peak in the microwaves, as the radiation was detected from all directions, with little variation in intensity, it was realised it must be universal or cosmic in origin
• The expansion of the universe has increased the wavelengths of radiation, so that it now occurs in the microwave region
• The spectrum fitted a black body of temperature around 2.7 K at the peak - supporting that it is heat remaining for the radiation emitted by the universe at the big bang suggesting that it was a hot point at the origin and cooled down while expanding

Relative Abundance of Hydrogen and Helium

• Stars and galaxies contain a 3:1 ratio of hydrogen mass to helium mass
• The proton to neutron ratio is 7:1
• This is because the rest energy for the neutron is larger than the proton, so protons were more readily created during the big bang
• The big bang had a brief period of Hydrogen fusion, when there was a hot temperature, producing helium but then quickly expanding and cooling

Question 30

Quasars

Answer 30

Quasi-stellar radio source

• They were first discovered as a source of radio waves
• The spectrum is very redshifted so they have a large recessional speed and are the furthest observable objects: 2 to 10 billion light years
• 1000 x the power of the host galaxy (10⁴² W)
• Small
• Fast moving clouds of gases and jets of matter being ejected

They are found around super massive black holes at the centre of galaxies and form from the merging of 2 galaxies

Question 31

Exoplanets

Answer 31

Planets that are outside the solar system and are always found orbiting a star

Question 32

Direct Observation of Exoplanets

Answer 32

This method is difficult because:

• the planets are not brigh compared to the brightness of the star that they orbiting, and will be drowned out
• the star is relatively small and too cool to be detected against the star

Question 33

Radial Velocity Method of Detecting Exoplanets

Answer 33

• There is a periodic shift in the spectra of a star due to the star’s movement around the common centre of mass with the planet
• When the spectral lines are shifted towards the short wavelength, the star moves towards earth, blueshift
• When the spectral lines are shifted towards the long wavelength, the star moves awa from earth, redshift

Difficulties

• The earth like planet has a much smaller mass than the star, so the effect on it is much smaller and it has a smaller radial velocity
• To see the shift, a very sensitive spectrometer is needed since the change in wavelenght is so small.

Question 34

Transit Method of Detecting Exoplanets

Answer 34

When an exoplanet crosses the star it blocks some of the light leading to a change in apparent magntiude and a dip in the light curve seen from earth.

Using these measurements, the radius can be found

There is a very low chance that the planet’s path will cross between earth and the star, due to the angle of tilt of the planet’s orbit, so it can only be used as a method of confirmation

Question 35

Converging Lens

Answer 35

• Brings light rays together by refraction
• Rays parallel to the principle axis converge onto a point called the principle focus
• parallel rays that aren’t parallel to the principle focus converge somewhere else on the focal plane
• Focal length is the distance between the lens axis and focal plane

The image produced is real and inverted

A real image is formed when light rays appear to come from another point in space - the light rays aren’t where the image appears to be so can’t be captured on a screen

If the object is futher than the focal length away from the lens, the image is real. If it is closer, then it is virtual

Question 36

Ray Diagrams

Answer 36

Ray diagrams are used to find the position and nature of the image being formed

Images can be:

• real or virtual
• upright or inverted
• magnified or diminshed

1/f = 1/u + 1/v

• u is the distance between the objects and lens axis
• v is the distance between the image and the lens axis
• f is the focal length

Question 37

Refracting Telescope

Answer 37

In normal adjustment, 2 converging lenses are placed with their principle focuses meeting

• distance between the 2 lenses is the sum of their focal lengths
• f_o + f_e

1. The objective lens converges the rays to form a real image
2. The eye lens acts as a magnifying glass to form a magnified virtual image
3. The object is assumed to be at infinity so the rays from it are parallel and the real image is formed on the focal plane
4. A telescope is set up so that the principle focus of the objective lens is in the same position as that of the eye lens, to that the final magnified image appears to be at infinity

Magnification = fo/f_e

Magnification = angle subtended by image/angle subtended by object

Question 38

Cassegrain Reflecting Telescope

Answer 38

A converging lens collects light rays which are then reflected back onto a secondary convex mirror which then focuses the light to a point in the primary converging mirror, where it forms a real image at the eye piece

The focal length is increased by having a convex mirror instead of a plane

Question 39

Collecting/Resolving Power

Answer 39

The collecting power of a telescope is related to the amount of light that goes through it.

It is proportional to the area of the lens

CP = d²

When comparing lenses:

CP₁/CP₂ = d₁²/d₂²

Stars appear brighter through telescopes as point objects, but they are not magnified

Planets remain with the same brightness as they are magnified so additional light is spread over a larger image

Resolving power is a measure of how much detail you can see and is dependent on the minimum angular resolution - the smallest angular separation at which the telescope can distinguish and resolve two points

• the smaller the angular resolution, the better the resolving power
• the two light sources can be distinguished is the centre of the airy disc from one source, is at least as far away as the first minimum of the other source - Raleigh’s Criterion

θ = λ​/D

Question 40

Analysis of Refracting Telescopes

Answer 40

Problems

Glass refracts different colours (wavelengths) of light by different amounts so the image for each colour is in a slightly different position called chromatic aberration

Any bubbles or impurities in the glass mean that faint objects can’t be seen, so building large lenses that are of a high enough quality is difficult and expensive

Large lenses are heavy and can only be supported from their edges so the shape becomes distorted

For a larger magnification, a very large focal length is required meaning the telescopes will be even more expensive

Advantages

They have a wider field of view

Question 41

Analysis of Reflecting Telescopes

Answer 41

Advantages

Large mirrors of good quality are cheaper to build than large lenses, and can also be supported from underneath

Mirrors don’t experience chromatic aberration

Disadvantages

Mirrors suffer from spherical aberration that occurs if the shape of the mirror isn’t quite parabolic, so the parallel rays that refract off different parts of the mirror do not all converge on the same point