Quasars and Exoplanets

SPEC 3.9.3.3, 3.9.3.4 • COSMOLOGY

Understanding the most distant objects in the universe and detecting planets beyond our solar system

What are Quasars?

Definition

Quasar stands for Quasi-Stellar Radio Source. Quasars are extremely luminous objects that appear star-like but are actually the active cores of distant galaxies powered by supermassive black holes.

Key Facts

Quasars are the most distant measurable objects in the universe

Distances range from hundreds of millions to billions of light years

Intensities between 10¹² and 10¹⁵ times that of the Sun

A single quasar can emit about 1000 times as much energy as our whole galaxy

Their size is not much bigger than a star despite their enormous power output

Example: Quasar 3C 273 has an absolute magnitude of -27

Discovery of Quasars

Quasars were first discovered in the early 1960s as bright radio sources that appeared point-like, similar to stars.

Historical Discovery

3C 48 (1960) - First quasar identified, showed unusual spectral lines

3C 273 (1963) - Maarten Schmidt identified its spectrum as highly red-shifted hydrogen lines

Initially puzzling - looked like stars but had spectra unlike any known star

Why Radio Sources?

Early radio telescopes detected intense radio emissions from these objects. The name "quasi-stellar radio source" reflects their star-like appearance combined with strong radio emission. We now know many quasars emit across the entire electromagnetic spectrum.

Red Shifts of Quasars

Quasars show large optical red shifts, indicating they are receding from us at high velocities due to the expansion of the universe.

Red Shift Formula

z = \frac{Δ λ}{λ} = \frac{v}{c}

(valid for

v ≪ c

)

SymbolMeaningUnits

z

Red shift (dimensionless)-

Δ λ

Change in wavelengthm or nm

λ

Original (rest) wavelengthm or nm

v

Recession velocitym s⁻¹

Important

Typical quasar red shifts range from z = 0.5 to z > 7. These large values indicate enormous distances - the light from distant quasars has been travelling for billions of years.

Exam Tip

The formula

z = v / c

cannot be used for quasars with very high red shifts because the high value of v means relativistic effects become significant. For exam calculations, you will be given values where v << c.

Calculating Quasar Distance and Power

Once we measure a quasar's red shift, we can calculate its distance using Hubble's Law:

Key Equations

v = zc

v = H_{0} d

Therefore:

d = \frac{zc}{H _{0}}

Once distance is known, we can calculate the luminosity (power output) from the apparent brightness:

Inverse Square Law

L = 4 π d^{2} F

SymbolMeaningUnits

L

Luminosity (power output)W

d

Distance to quasarm

F

Flux (apparent brightness)W m⁻²

H_{0}

Hubble constants⁻¹ (or km s⁻¹ Mpc⁻¹)

01Worked Example

A quasar is observed with a red shift of z = 0.158. Calculate:

(a) The recession velocity of the quasar

(b) The distance to the quasar in metres

Take c = 3.00 × 10⁸ m s⁻¹ and H₀ = 2.4 × 10⁻¹⁸ s⁻¹

1

Part (a): Calculate recession velocity using

v = zc

2

v = 0.158 \times 3.00 \times 1 0^{8}

3

v = 4.74 \times 1 0^{7} m s^{- 1}

4

Part (b): Calculate distance using

d = \frac{v}{H _{0}}

5

d = \frac{4.74 \times 1 0 ^{7}}{2.4 \times 1 0 ^{- 18}}

6

d = 1.98 \times 1 0^{25} m

Answer(a) v = 4.74 × 10⁷ m s⁻¹ (b) d = 1.98 × 10²⁵ m (about 2.1 billion light years)

Worked Example02

A quasar at a distance of 1.5 × 10²⁵ m has an observed flux (apparent brightness) of 2.8 × 10⁻¹² W m⁻². Calculate the luminosity of the quasar and compare it to the Sun's luminosity (L☉ = 3.83 × 10²⁶ W).

1

Use the inverse square law:

L = 4 π d^{2} F

2

L = 4 π \times (1.5 \times 1 0^{25})^{2} \times 2.8 \times 1 0^{- 12}

3

L = 4 π \times 2.25 \times 1 0^{50} \times 2.8 \times 1 0^{- 12}

4

L = 7.92 \times 1 0^{39} W

5

Ratio to Sun:

\frac{L}{L _{⊙}} = \frac{7.92 \times 1 0 ^{39}}{3.83 \times 1 0 ^{26}} = 2.07 \times 1 0^{13}

AnswerL = 7.9 × 10³⁹ W, approximately 20 trillion times the Sun's luminosity

Formation of Quasars

Quasars are formed from active supermassive black holes (SMBHs) at the centres of galaxies.

Formation Mechanism

A quasar forms when a supermassive black hole (10⁶ to 10¹⁰ solar masses) actively accretes matter. The infalling material forms an accretion disk that heats to extreme temperatures, converting gravitational potential energy into electromagnetic radiation.

Why So Luminous?

Matter falling into the black hole is compressed and heated

Accretion disk temperatures reach millions of degrees

Gravitational potential energy → electromagnetic radiation

More efficient than nuclear fusion (~10% mass-energy conversion vs ~0.7% for fusion)

Relativistic jets may be ejected perpendicular to the disk

Exam Tip

Quasars emit all wavelengths of the electromagnetic spectrum strongly, from gamma rays to radio waves, but especially in the radio frequency range. This is why they were first discovered by radio telescopes.

Important

Because of their enormous rate of energy emission, the lifetimes of quasars are relatively short on cosmological timescales. They cannot sustain such intense output indefinitely.

Quasar Structure

Exam Note

The appearance of a quasar depends on viewing angle. When viewed "face-on" (looking down the jet), it appears as a "blazar" with rapid brightness variations. When viewed "edge-on", the dusty torus may obscure the central engine.

What are Exoplanets?

Definition

Exoplanets (or extrasolar planets) are planets that orbit stars other than our Sun. They exist beyond our solar system.

Key Facts

First confirmed discovery: 1992 (pulsar planets), 1995 (51 Pegasi b - around a main sequence star)

Over 5,500 confirmed exoplanets discovered to date

Types include: hot Jupiters, super-Earths, mini-Neptunes, and terrestrial planets

Some orbit in the "habitable zone" where liquid water could exist

Why Study Exoplanets?

Finding exoplanets helps us understand planetary formation, the prevalence of Earth-like worlds, and the potential for life elsewhere in the universe.

Difficulties in Direct Detection

Directly imaging exoplanets is extremely challenging. The AQA specification requires you to understand why.

Key Difficulties

Brightness contrast: Stars are typically 10⁶ to 10¹⁰ times brighter than their planets - the planet is obscured by the light of its host star

Small objects at great distance: Exoplanets are relatively small objects at distances of at least 10 light years

Angular separation: At interstellar distances, the planet appears very close to its star

Glare: The star's light overwhelms the faint reflected/emitted light from the planet

When Direct Imaging Works

Direct detection is only possible for:

Very large planets (Jupiter-sized or larger)

Wide orbits (far from the star's glare)

Young, hot planets (emit their own infrared radiation)

Using coronagraphs to block starlight

Due to these difficulties, astronomers use indirect methods to detect most exoplanets.

Radial Velocity Method

Also called the "Doppler wobble" method. This technique detects the gravitational effect of a planet on its host star.

Principle

A star and its planet both orbit their common centre of mass. As the star moves towards us, its light is blue-shifted. As it moves away, its light is red-shifted. This periodic Doppler shift reveals the presence of the planet.

Doppler Shift Formula

\frac{Δ f}{f} = \frac{v}{c}

What We Measure

Periodic shifts in stellar spectral lines

The period of the shift equals the orbital period

The amplitude relates to the star's velocity (and hence planet mass)

Radial Velocity Method - How It Works

Key Point

The radial velocity method only measures the component of velocity towards or away from Earth. This is why we obtain a minimum mass (m sin i), since the orbital inclination is usually unknown.

Radial Velocity - What We Learn

PropertyHow Determined

Orbital periodTime between successive red/blue shift cycles

Minimum planet mass (m sin i)From amplitude of stellar wobble (larger planet → bigger wobble)

Orbital eccentricityShape of the velocity curve (circular orbit = sinusoidal)

Semi-major axisFrom orbital period using Kepler's third law

Limitation

The orbital inclination (i) is usually unknown. We only measure m sin i, the minimum mass. If the orbit is edge-on (i = 90°), we get the true mass. If face-on (i = 0°), we detect no wobble at all.

Detection Bias

Radial velocity favours detection of massive planets in close orbits - they produce larger, faster stellar wobbles.

03Worked Example

A star shows periodic Doppler shifts in its spectrum with a period of 4.2 days. The maximum change in wavelength observed for a spectral line at 656.3 nm is 0.030 nm. Calculate the radial velocity of the star.

Take c = 3.00 × 10⁸ m s⁻¹

1

Use the Doppler shift formula:

\frac{Δ λ}{λ} = \frac{v}{c}

2

Rearrange for v:

v = \frac{Δ λ}{λ} \times c

3

Substitute values:

v = \frac{0.030}{656.3} \times 3.00 \times 1 0^{8}

4

v = 4.57 \times 1 0^{- 5} \times 3.00 \times 1 0^{8}

5

v = 1.37 \times 1 0^{4} m s^{- 1}

AnswerThe radial velocity of the star is approximately 13.7 km s⁻¹ (or 1.4 × 10⁴ m s⁻¹)

Transit Method

The transit method detects planets when they pass in front of their host star as viewed from Earth.

Principle

When a planet transits (crosses in front of) its star, it blocks a small fraction of the star's light. This causes a periodic dip in brightness that we can measure.

Transit Depth Formula

\frac{Δ L}{L} = (\frac{R _{p}}{R _{⋆}})^{2}

SymbolMeaning

Δ L / L

Fractional decrease in brightness (transit depth)

R_{p}

Radius of the planet

R_{⋆}

Radius of the star

Transit Method - Typical Light Curve

Reading a Light Curve

Ingress: Planet begins to cross stellar disk (brightness decreases)

Flat bottom: Planet fully in front of star (minimum brightness)

Egress: Planet exits stellar disk (brightness recovers)

Period: Time between consecutive transits = orbital period

Transit Method - What We Learn

PropertyHow Determined

Planet radius (size)From transit depth (amount intensity falls by): larger planet → deeper dip

Orbital periodTime between consecutive transits (duration of the cycle)

Orbital inclinationMust be close to 90° (edge-on) for transit to occur

Planetary atmosphereChanges in observed spectrum due to absorption of starlight by the exoplanet's atmosphere during transit

Limitation

The transit method only works when the orbital plane is nearly edge-on to our line of sight (the line of sight to the star must be in the plane of the planet's orbit). This is more likely for planets with small orbits - smaller orbits mean parts of the planet are more likely to cross in front of the star and block some of its light.

Detection Bias

Transit method favours detection of large planets in close orbits - they produce larger dips and have higher transit probability.

Worked Example04

During a transit, a star's brightness decreases by 1.0%. If the star has a radius of 7.0 × 10⁸ m (similar to the Sun), calculate the radius of the transiting planet.

1

Transit depth:

\frac{Δ L}{L} = 1.0% = 0.010

2

Use:

\frac{Δ L}{L} = (\frac{R _{p}}{R _{⋆}})^{2}

3

Rearrange:

\frac{R _{p}}{R _{⋆}} = \frac{Δ L}{L} = 0.010 = 0.10

4

Calculate planet radius:

R_{p} = 0.10 \times R_{⋆} = 0.10 \times 7.0 \times 1 0^{8}

5

R_{p} = 7.0 \times 1 0^{7} m

AnswerThe planet radius is 7.0 × 10⁷ m (approximately the size of Jupiter, which has radius 7.0 × 10⁷ m)

Comparison of Detection Methods

AspectRadial VelocityTransit

What is measuredDoppler shift of stellar spectral linesPeriodic brightness dip

What we determineMinimum mass (m sin i)Planet radius

Geometry requiredAny inclination (except face-on)Edge-on alignment (~90°)

Detection biasMassive, close-in planetsLarge, close-in planets

Combining Methods

When both methods detect the same planet:

Transit gives radius (Rₚ)

Radial velocity gives mass (m) - now true mass since i ≈ 90°

We can calculate density:

ρ = \frac{m}{\frac{4}{3} π R _{p}^{3}}

Density tells us about composition (rocky vs gaseous)

Practice

Question 1: A quasar has a red shift of z = 0.25. Calculate the distance to this quasar in metres.

Take c = 3.00 × 10⁸ m s⁻¹ and H₀ = 2.4 × 10⁻¹⁸ s⁻¹

1

Calculate velocity:

v = zc = 0.25 \times 3.00 \times 1 0^{8} = 7.5 \times 1 0^{7} m s^{- 1}

2

Calculate distance:

d = \frac{v}{H _{0}} = \frac{7.5 \times 1 0 ^{7}}{2.4 \times 1 0 ^{- 18}} = 3.1 \times 1 0^{25} m

Answerd = 3.1 × 10²⁵ m

Practice

Question 2: A star's spectrum shows a hydrogen line at 486.5 nm when the star is stationary. When observed, the line appears at 486.8 nm. Calculate the radial velocity of the star and state whether it is moving towards or away from Earth.

Take c = 3.00 × 10⁸ m s⁻¹

1

Δ λ = 486.8 - 486.5 = 0.3 nm

(positive = red shift)

2

v = \frac{Δ λ}{λ} \times c = \frac{0.3}{486.5} \times 3.00 \times 1 0^{8}

3

v = 1.85 \times 1 0^{5} m s^{- 1} = 185 km s^{- 1}

Answerv = 1.85 × 10⁵ m s⁻¹, moving away from Earth (red shift indicates recession)

Practice

Question 3: An exoplanet transiting a star causes a 0.25% decrease in the star's brightness. The star has a radius of 8.0 × 10⁸ m. Calculate the radius of the exoplanet.

1

Transit depth:

\frac{Δ L}{L} = 0.25% = 0.0025

2

\frac{R _{p}}{R _{⋆}} = 0.0025 = 0.050

3

R_{p} = 0.050 \times 8.0 \times 1 0^{8} = 4.0 \times 1 0^{7} m

AnswerRₚ = 4.0 × 10⁷ m (about 0.6 Jupiter radii, or 6 Earth radii - a "mini-Neptune" type planet)

Summary

Quasars

ConceptKey Points

DefinitionMost distant measurable objects; active galactic nuclei powered by supermassive black holes

DiscoveryFirst detected as bright radio sources in the 1960s

Key equations

v = zc

,

d = v / H_{0}

,

L = 4 π d^{2} F

Exoplanet Detection

MethodMeasuresKey Equation

Radial VelocityMinimum mass (m sin i)

Δ f / f = v / c

TransitPlanet radius

Δ L / L = (R_{p} / R_{⋆})^{2}

Key Exam Skills

Calculate quasar distance from red shift using Hubble's law

Calculate quasar luminosity from distance and flux

Explain difficulties in direct exoplanet detection

Describe and apply both radial velocity and transit methods

Interpret light curves from transit observations