Astrophysics

Astrophysics

:::info Board Coverage AQA Paper 2 (Option 9) | Edexcel CP5 (Option 9) | OCR (A) Paper 2 (Option D) :::

1. Astronomical Distances and Measurements

Astronomical Units of Distance

Definition. The astronomical unit (AU) is the mean Earth—Sun distance, defined as exactly 149,597,870,700 m (approximately $1.50 \times 10^{11}$ m).

Definition. The light-year (ly) is the distance travelled by light in a vacuum in one Julian Year:

$1\;\mathrm{ly} = c \times 1\;\mathrm{year} = 3.00 \times 10^8 \times 3.156 \times 10^7 = 9.46 \times 10^{15}\;\mathrm{m}$

Definition. The parsec (pc) is the distance at which one astronomical unit subtends an angle Of one arcsecond:

$1\;\mathrm{pc} = \frac◆LB◆1\;\mathrm{AU}◆RB◆◆LB◆\tan(1'')◆RB◆ \approx \frac◆LB◆1.50 \times 10^{11}◆RB◆◆LB◆4.848 \times 10^{-6}◆RB◆ = 3.09 \times 10^{16}\;\mathrm{m}$

Key conversions:

Unit	Metres	Light-years	Parsecs
1 AU	$1.50 \times 10^{11}$	$1.58 \times 10^{-5}$	$4.85 \times 10^{-6}$
1 ly	$9.46 \times 10^{15}$	1	$0.307$
1 pc	$3.09 \times 10^{16}$	$3.26$	1

Stellar Parallax

As Earth orbits the Sun, nearby stars appear to shift position against the background of more Distant stars. This apparent angular displacement is called stellar parallax.

Proof of the Parallax Formula

Consider a nearby star at distance $d$ from the Sun. As Earth moves from one side of its orbit to The other (separation $2\;\mathrm{AU}$), the star appears to shift by an angle $2p$Where $p$ is The parallax angle measured in arcseconds.

For small angles, $\tan p \approx p$ (in radians):

$\tan p = \frac◆LB◆1\;\mathrm{AU}◆RB◆◆LB◆d◆RB◆ \approx p$

Converting $p$ from arcseconds to radians:

$p\;(\mathrm{rad}) = p\;('') \times \frac◆LB◆\pi◆RB◆◆LB◆648,000◆RB◆ = \frac{p\;('')}{206,265}$

Therefore:

$d = \frac◆LB◆1\;\mathrm{AU}◆RB◆◆LB◆p\;(\mathrm{rad})◆RB◆ = \frac◆LB◆206,265\;\mathrm{AU}◆RB◆◆LB◆p\;('')◆RB◆$

By definition, when $p = 1''$, $d = 1$ pc $= 206,265$ AU. Hence:

$\boxed{d\;(\mathrm{pc}) = \frac{1}{p\;('')}}$

$\square$

Intuition. The parsec is defined so that the formula becomes simple: a star with a Parallax angle of 1 arcsecond is at a distance of 1 parsec. The inverse relationship means that Closer stars have larger parallax angles, making them easier to measure.

Limitations of the Parallax Method

• Parallax angles are extremely small ($p \lt 1''$ for all stars), making precise measurement difficult.
• Ground-based observations are limited to $d \lt 100$ pc (atmospheric turbulence limits angular resolution).
• The Hipparcos satellite extended this to $\sim 1000$ pc; Gaia extends to $\sim 10,000$ pc.
• Beyond these distances, parallax is too small to measure reliably.

Standard Candles and the Cosmic Distance Ladder

Definition. A standard candle is an astronomical object of known intrinsic luminosity $L$ Whose distance can be determined by comparing its apparent brightness $b$ with its luminosity.

The cosmic distance ladder uses overlapping methods to measure distances across the universe:

1. Parallax --- direct geometric measurement ($0$—$10,000$ pc)
2. Cepheid variables --- period—luminosity relation (up to $\sim 30$ Mpc)
3. Type Ia supernovae --- nearly constant peak luminosity (up to $\sim 1000$ Mpc)

Cepheid Variables

Cepheid variables are pulsating stars whose luminosity is directly related to their pulsation Period. The period—luminosity relation (discovered by Henrietta Leavitt, 1912) states that Brighter Cepheids have longer periods. By measuring the period of variability, astronomers can Determine the absolute luminosity and hence the distance.

Type Ia Supernovae

Type Ia supernovae occur when a white dwarf accreting matter from a companion star exceeds the Chandrasekhar limit ($1.4\;M_\odot$). The resulting thermonuclear explosion releases a remarkably Consistent peak luminosity, making them excellent standard candles for cosmological distances.

Luminosity and Apparent Brightness

Definition. The luminosity $L$ of a star is the total power it radiates in all directions (units: watts, W).

Definition. The apparent brightness $b$ is the power received per unit area at a distance $d$ from the star (units: W m$^{-2}$).

By conservation of energy, the luminosity is spread uniformly over a sphere of surface area $4\pi d^2$:

$\boxed{b = \frac◆LB◆L◆RB◆◆LB◆4\pi d^2◆RB◆}$

This is the inverse square law for intensity: doubling the distance reduces the apparent Brightness by a factor of four.

Stefan-Boltzmann Law

A star approximates a black body --- a perfect absorber and emitter of radiation. The power Radiated per unit surface area of a black body at temperature $T$ is:

$j = \sigma T^4$

Where $\sigma = 5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$ is the Stefan-Boltzmann constant.

For a star of radius $r$ and surface temperature $T$The total luminosity is:

$\boxed{L = 4\pi r^2 \sigma T^4}$

Proof of the Radius Formula

Starting from $b = L/(4\pi d^2)$ and $L = 4\pi r^2 \sigma T^4$:

$b = \frac◆LB◆4\pi r^2 \sigma T^4◆RB◆◆LB◆4\pi d^2◆RB◆ = \frac◆LB◆r^2 \sigma T^4◆RB◆◆LB◆d^2◆RB◆$

Rearranging for the radius:

$\boxed{r = d\sqrt◆LB◆\frac{b}{\sigma T^4}◆RB◆}$

This allows us to determine a star’s radius from its distance, apparent brightness, and surface Temperature. $\square$

Intuition. A star’s luminosity depends on two factors: how hot it is ($T^4$) and how large it is ($r^2$). A cool red giant can be more luminous than a hot blue star if it is sufficiently large --- This is why giants and supergiants occupy the upper-right region of the H-R diagram.

Wien’s Displacement Law

The wavelength at which a black body emits maximum radiation is inversely proportional to its Temperature:

$\boxed{\lambda_{\max} T = 2.898 \times 10^{-3}\;\mathrm{m K}}$

Where $\lambda_{\max}$ is in metres and $T$ is in kelvin.

This result follows from differentiating Planck’s radiation law $B(\lambda, T)$ with respect to $\lambda$ and setting the derivative to zero, which yields the transcendental equation $x = 5(1 - e^{-x})$ where $x = hc/(\lambda_{\max} kT)$. The numerical solution gives the constant $2.898 \times 10^{-3}$ m K.

Intuition. Hotter objects emit radiation peaked at shorter wavelengths. The Sun ($T \approx 5800$ K) peaks in the visible range ($\lambda_{\max} \approx 500$ nm). A cool red star ($T \approx 3000$ K) peaks in the infrared. This is why hotter stars appear bluer and cooler stars Appear redder.

:::info Board-Specific

• AQA requires Wien’s law, Stefan-Boltzmann law, and the inverse square law explicitly.
• Edexcel emphasises the period—luminosity relation for Cepheids and the use of Type Ia supernovae as standard candles.
• OCR (A) includes parallax, standard candles, and black body radiation in the Turning Points option. :::

2. Stellar Evolution

Star Formation

Stars form from the gravitational collapse of regions within nebulae --- vast clouds of gas (mostly hydrogen) and dust. For a cloud to collapse under its own gravity, its gravitational Potential energy must exceed the thermal kinetic energy of the gas:

$|E_{\mathrm{grav}}| \gt E_{\mathrm{thermal}}$

When this condition is met, the cloud fragment collapses and heats up. Conservation of angular Momentum causes it to spin faster and flatten into a protoplanetary disk. The core temperature rises Until hydrogen fusion ignites --- a star is born.

The Hertzsprung—Russell (H-R) Diagram

The H-R diagram plots stars according to their luminosity (or absolute magnitude) on the Vertical axis against their surface temperature (or spectral class) on the horizontal axis. Temperature decreases from left to right.

Key regions:

• Main sequence --- a diagonal band from upper-left (hot, luminous O-type stars) to lower-right (cool, dim M-type stars). Stars spend $\sim 90\%$ of their lifetime here.
• Red giants --- luminous but cool, located in the upper-right.
• White dwarfs --- hot but very dim, located in the lower-left.
• Red supergiants --- extremely luminous and cool, at the top-right.

Main Sequence Stars

Main sequence stars are in hydrostatic equilibrium: gravitational collapse is balanced by the Radiation pressure from nuclear fusion in the core.

Two fusion processes convert hydrogen to helium:

1. Proton—proton (pp) chain --- dominant in stars with $T \lt 1.5 \times 10^7$ K (like the Sun):

   $4\,^{1}\mathrm{H} \rightarrow ^{4}\mathrm{He} + 2e^+ + 2\nu_e + 2\gamma$

Energy released per reaction: $\sim 26.7$ MeV.

2. CNO cycle --- dominant in stars with $T \gt 1.5 \times 10^7$ K (more massive stars). Uses carbon, nitrogen, and oxygen as catalysts. This process is more temperature-sensitive than the pp chain, leading to convective cores in massive stars.

The mass—luminosity relation for main sequence stars:

$\boxed{L \propto M^{3.5}}$

A star of twice the solar mass has a luminosity roughly $2^{3.5} \approx 11$ times that of the Sun, And therefore exhausts its fuel much faster. More massive stars have shorter lifetimes.

Evolution of Low-Mass Stars ($M \lt 8\;M_\odot$)

1. Main sequence --- hydrogen fuses to helium in the core ($\sim 10^{10}$ years for solar-mass stars).
2. Red giant --- hydrogen shell burning causes the envelope to expand and cool. A helium flash ignites helium fusion in the core ($^{4}\mathrm{He}$ to $^{12}\mathrm{C}$ and $^{16}\mathrm{O}$).
3. Planetary nebula --- the outer layers are gently ejected, exposing the hot core.
4. White dwarf --- the remaining core ( $\sim 0.6\;M_\odot$), supported by electron degeneracy pressure. No fusion occurs; it slowly cools over billions of years.

Evolution of High-Mass Stars ($M \gt 8\;M_\odot$)

1. Main sequence --- rapid hydrogen fusion ($\sim 10^7$ years for a $25\;M_\odot$ star).
2. Red supergiant --- successive nuclear burning stages create an onion-like shell structure: He $\rightarrow$ C $\rightarrow$ Ne $\rightarrow$ O $\rightarrow$ Si $\rightarrow$ Fe. Each stage releases less energy and proceeds faster than the last.
3. Supernova --- iron core collapse triggers a catastrophic explosion. The core collapses in milliseconds, rebounds, and drives a shock wave that ejects the outer layers.
4. Neutron star (if remnant mass $\lt 3\;M_\odot$) --- supported by neutron degeneracy pressure. Typical radius $\sim 10$ km, density $\sim 10^{17}$ kg m$^{-3}$.
5. Black hole (if remnant mass $\gt 3\;M_\odot$) --- no known force can halt gravitational collapse.

The Chandrasekhar Limit

Definition. The Chandrasekhar limit ($\approx 1.4\;M_\odot$) is the maximum mass of a white Dwarf that can be supported by electron degeneracy pressure.

If a white dwarf exceeds this mass (e.g., by accreting matter from a binary companion), it undergoes A Type Ia supernova --- the entire star is destroyed in a thermonuclear explosion.

Supernovae

Type Ia supernova:

• White dwarf exceeds the Chandrasekhar limit
• No hydrogen lines in the spectrum (hydrogen has been consumed or stripped)
• Consistent peak luminosity (standard candle)
• Complete destruction of the star

Type II supernova:

• Core collapse of a massive star ($M \gt 8\;M_\odot$)
• Hydrogen lines present in the spectrum
• Variable luminosity (not a reliable standard candle)
• Leaves behind a neutron star or black hole

Both types are crucial for nucleosynthesis --- creating elements heavier than iron and Dispersing them into the interstellar medium, enriching future generations of stars and planets.

Neutron Stars and Black Holes

Neutron stars have radii of $\sim 10$ km and densities comparable to an atomic nucleus. Pulsars are rapidly rotating neutron stars that emit beams of electromagnetic radiation from Their magnetic poles.

Black holes are regions of spacetime from which nothing, not even light, can escape. The Boundary is the event horizon, at the Schwarzschild radius.

Proof of the Schwarzschild Radius

The Schwarzschild radius is found by setting the escape velocity equal to the speed of light $c$:

$v_e = \sqrt◆LB◆\frac{2GM}{r}◆RB◆ = c$

Squaring both sides and solving for $r$:

$\frac{2GM}{r_s} = c^2$

$\boxed{r_s = \frac{2GM}{c^2}}$

$\square$

Intuition. The Schwarzschild radius defines the event horizon --- the boundary within which the Escape velocity exceeds the speed of light. For the Sun, $r_s \approx 3$ km; for Earth, $r_s \approx 9$ mm. This shows how extraordinarily compact a black hole must be: the entire mass of The Sun compressed into a sphere smaller than a small city.

:::info Board-Specific

• AQA requires detailed knowledge of stellar evolution pathways, the H-R diagram, and the Chandrasekhar limit.
• Edexcel emphasises supernovae including light curves, and the use of standard candles.
• OCR (A) covers neutron stars and black holes, including the Schwarzschild radius derivation, in the Turning Points option. :::

3. Cosmology

Olbers’ Paradox

Olbers’ paradox asks: if the universe is infinite, static, and uniformly filled with stars, why Is the night sky dark?

In an infinite, static universe with uniformly distributed stars, every line of sight should Eventually intersect the surface of a star, making the entire night sky as bright as the surface of A typical star. The resolution of the paradox relies on three key facts:

1. The universe has a finite age ($\sim 13.8$ billion years), so we can only observe light from within our observable universe.
2. The universe is expanding, which redshifts the light from distant objects, reducing their energy density.
3. Stars have finite lifetimes and the universe does not contain enough energy to keep every point in the sky illuminated at stellar surface brightness.

Doppler Effect for Light

When a light source moves relative to an observer, the observed wavelength is shifted. For recession Speeds much less than the speed of light ($v \ll c$):

$\boxed{\frac◆LB◆\Delta \lambda◆RB◆◆LB◆\lambda◆RB◆ = \frac{v}{c}}$

Where $\Delta \lambda = \lambda_{\mathrm{obs}} - \lambda_{\mathrm{emit}}$ is the change in Wavelength.

• Redshift ($\Delta \lambda \gt 0$): source receding from the observer
• Blueshift ($\Delta \lambda \lt 0$): source approaching the observer

For cosmological redshifts, the redshift parameter $z$ is defined as:

$z = \frac◆LB◆\lambda_{\mathrm{obs}} - \lambda_{\mathrm{emit}}◆RB◆◆LB◆\lambda_{\mathrm{emit}}◆RB◆$

So that $\Delta \lambda / \lambda = z = v/c$ for $v \ll c$.

Hubble’s Law

Edwin Hubble (1929) discovered a linear relationship between the recession velocity of galaxies and Their distance:

$\boxed{v = H_0 d}$

Where $H_0 \approx 70$ km s$^{-1}$ Mpc$^{-1}$ is the Hubble constant. This law implies that the Universe is expanding uniformly --- more distant galaxies recede faster because there is more space Between them to expand.

Proof of the Hubble Time

If the expansion has been at a constant rate, the time since all galaxies were at a single point (the Big Bang) is:

$t_H = \frac{1}{H_0}$

With $H_0 = 70$ km s$^{-1}$ Mpc$^{-1}$:

$H_0 = \frac◆LB◆70 \times 10^3\;\mathrm{m s}^{-1}◆RB◆◆LB◆3.09 \times 10^{22}\;\mathrm{m}◆RB◆ = 2.27 \times 10^{-18}\;\mathrm{s}^{-1}$

$t_H = \frac◆LB◆1◆RB◆◆LB◆2.27 \times 10^{-18}◆RB◆ = 4.41 \times 10^{17}\;\mathrm{s} \approx 14.0\;\mathrm{billion years}$

$\boxed{t_H \approx \frac{1}{H_0} \approx 14\;\mathrm{Gyr}}$

$\square$

Intuition. Hubble’s law tells us that more distant galaxies recede faster. If we “rewind” the Expansion, all matter converges to a single point at a finite time in the past --- the Big Bang. The Hubble time gives a rough upper estimate of the age of the universe. The actual age is slightly less Because the expansion rate has not been constant (deceleration due to gravity, then acceleration due To dark energy).

Evidence for the Big Bang

Three major lines of evidence support the Big Bang model:

1. Hubble’s law --- the expansion of the universe implies a hot, dense beginning.
2. Cosmic Microwave Background (CMB) --- the afterglow of the initial hot, dense phase, discovered by Penzias and Wilson (1965).
3. Abundance of light elements --- the observed ratios of hydrogen ($\sim 75\%$), helium ($\sim 25\%$), deuterium, and lithium match predictions from Big Bang nucleosynthesis. No other model successfully predicts these abundances.

The Cosmic Microwave Background (CMB)

Definition. The Cosmic Microwave Background is the remnant electromagnetic radiation from The early universe, emitted when the universe became transparent at the recombination epoch ($\sim 380,000$ years after the Big Bang, $T \approx 3000$ K).

Key properties:

• Temperature: $T \approx 2.725$ K (the radiation has been redshifted by a factor of $\sim 1100$ since emission)
• Spectrum: a near-perfect black body curve peaking in the microwave region
• Isotropy: uniform to one part in $10^5$With tiny anisotropies that are the seeds of large- scale structure formation

Using Wien’s law to find the peak wavelength:

$\lambda_{\max} = \frac◆LB◆2.898 \times 10^{-3}◆RB◆◆LB◆2.725◆RB◆ = 1.06 \times 10^{-3}\;\mathrm{m} = 1.06\;\mathrm{mm}$

This confirms the microwave nature of the CMB, which was initially detected as excess noise at a Wavelength of 7.35 cm by Penzias and Wilson.

Expansion of the Universe

The expansion of the universe is not galaxies moving through space, but rather the expansion of Space itself. This is described by the scale factor $a(t)$Where $a = 1$ today. As the universe Expands, photon wavelengths are stretched, producing cosmological redshift:

$1 + z = \frac◆LB◆a_{\mathrm{now}}◆RB◆◆LB◆a_{\mathrm{then}}◆RB◆$

:::info Info

• AQA requires understanding of Hubble’s law, the CMB, and evidence for the Big Bang.
• Edexcel includes the Doppler effect for electromagnetic radiation and redshift calculations.
• OCR (A) covers Olbers’ paradox and its resolution in the Turning Points option. :::

4. Telescopes and Observational Astronomy

Refracting Telescopes

A refracting telescope uses a converging (convex) objective lens to form a real image, which is Then magnified by a converging eyepiece lens.

The angular magnification is the ratio of the angle subtended by the image to the angle Subtended by the object at the unaided eye:

$\boxed{M = \frac◆LB◆\theta'◆RB◆◆LB◆\theta◆RB◆ = \frac{f_o}{f_e}}$

Where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.

Advantages of refracting telescopes:

• Sharp images with good contrast
• Rugged, sealed tube design (no air currents inside)
• Durable alignment

Disadvantages of refracting telescopes:

• Chromatic aberration --- different wavelengths refract by different amounts, producing colour fringes (corrected with achromatic doublets, but not perfectly)
• Spherical aberration --- marginal rays focus at different points from paraxial rays
• Expensive to manufacture large, defect-free lenses
• Heavy lenses can distort under their own weight, limiting practical sizes to $\sim 1$ m diameter

Reflecting Telescopes

A reflecting telescope uses a concave primary mirror to collect and focus light. The most common Designs are:

• Newtonian --- flat secondary mirror at 45 degrees redirects light to the eyepiece at the side
• Cassegrain --- convex secondary mirror reflects light through a hole in the primary mirror

Advantages of reflecting telescopes:

• No chromatic aberration (mirrors reflect all wavelengths equally)
• Cheaper to manufacture large mirrors than large lenses
• Easier to support structurally (mirrors can be supported from behind)
• Practically no limit on aperture size (modern telescopes exceed 10 m)

Disadvantages of reflecting telescopes:

• Spherical aberration (corrected by using parabolic mirrors)
• Central obstruction from the secondary mirror reduces light-gathering and introduces diffraction spikes
• Requires regular alignment of optics (collimation)
• Mirror surfaces degrade over time and require re-coating

Ray Diagrams

For a converging lens, the three principal rays are:

1. A ray parallel to the principal axis refracts through the far focal point $F'$.
2. A ray through the optical centre passes straight through undeviated.
3. A ray through the near focal point $F$ refracts parallel to the principal axis.

For a concave (converging) mirror:

1. A ray parallel to the principal axis reflects through the focal point $F$.
2. A ray through the centre of curvature $C$ reflects back on itself.
3. A ray through the focal point $F$ reflects parallel to the principal axis.

Resolving Power and the Rayleigh Criterion

Definition. The resolving power of a telescope is its ability to distinguish between two Closely spaced objects.

Two point sources are just resolved when the central maximum of one diffraction pattern coincides With the first minimum of the other. For a circular aperture, this gives the Rayleigh criterion:

$\boxed{\theta = \frac◆LB◆1.22\lambda◆RB◆◆LB◆D◆RB◆}$

Where $\theta$ is the minimum angular resolution (in radians), $\lambda$ is the wavelength of the Observed radiation, and $D$ is the diameter of the aperture.

Proof of the Rayleigh Criterion

For a circular aperture of diameter $D$The diffraction pattern is an Airy disk. The angular Position of the first minimum is given by:

$\sin\theta = 1.22\frac◆LB◆\lambda◆RB◆◆LB◆D◆RB◆$

For small angles ($\theta \ll 1$ rad), $\sin\theta \approx \theta$:

$\boxed{\theta \approx \frac◆LB◆1.22\lambda◆RB◆◆LB◆D◆RB◆}$

$\square$

Intuition. A larger aperture produces a narrower diffraction pattern, allowing finer detail to Be resolved. Shorter wavelengths also improve resolution --- this is why electron microscopes (de Broglie wavelengths of $\sim 0.01$ nm) resolve far finer detail than optical microscopes ($\lambda \approx 500$ nm).

Why Large Telescopes Are Needed

Large telescopes serve two fundamental purposes:

1. Collecting power --- proportional to $D^2$. A larger aperture collects more light, enabling the observation of fainter objects. The collecting power relative to the human eye (diameter $\sim 5$ mm) is:

   $\mathrm{Collecting power ratio} = \left(\frac◆LB◆D◆RB◆◆LB◆d_{\mathrm{eye}}◆RB◆\right)^2$

2. Resolving power --- proportional to $1/D$. A larger aperture gives smaller minimum angular resolution $\theta$Allowing finer detail to be distinguished.

These are the two fundamental reasons why astronomers continually push for larger telescopes.

Radio Telescopes

Radio telescopes detect radio waves (wavelengths from $\sim 1$ mm to $\sim 10$ m) using large Parabolic dishes.

Advantages:

• Operate 24 hours a day (unaffected by daylight or clouds)
• Detect non-thermal emission (e.g., synchrotron radiation from cosmic-ray electrons)
• Radio waves penetrate interstellar dust clouds that block visible light
• Interferometry --- linking multiple dishes (e.g., the Very Large Array) achieves angular resolution equivalent to a single dish of diameter equal to the maximum baseline separation

Disadvantages:

• Poor inherent resolution due to long wavelengths ($\theta \propto \lambda/D$)
• Require very large single dishes or interferometer arrays to achieve useful resolution
• Susceptible to radio frequency interference (RFI) from terrestrial sources

Ground-Based vs Space-Based Telescopes

Ground-based telescopes:

• Cheaper to build, launch, and maintain
• Larger apertures are possible (currently up to $\sim 39$ m for the Extremely Large Telescope)
• Limited by atmospheric absorption (UV, X-rays, gamma rays are absorbed)
• Atmospheric turbulence causes seeing (image blurring) --- partially mitigated by adaptive optics, which deform the mirror surface in real time to compensate

Space-based telescopes:

• No atmospheric absorption --- full access to the electromagnetic spectrum
• No atmospheric turbulence --- diffraction-limited resolution
• Much more expensive to build, launch, and maintain (no servicing missions for most)
• Limited aperture size (constrained by launch vehicle fairings; JWST’s primary mirror is 6.5 m)

:::info Board-Specific

• AQA requires comparison of reflecting and refracting telescopes, and the Rayleigh criterion.
• Edexcel emphasises angular magnification and the advantages of large-diameter telescopes.
• OCR (A) covers radio telescopes and the comparison of ground-based and space-based observatories. :::

5. Problems

Problem 1. A star has a parallax angle of $0.05''$. Calculate its distance in parsecs and Light-years.

Hint

Using $d = 1/p$: $d = 1/0.05 = 20$ pc.

Converting to light-years: $20 \times 3.26 = 65.2$ ly.

Problem 2. A star has an apparent brightness of $2.0 \times 10^{-8}$ W m$^{-2}$ and a luminosity Of $5.0 \times 10^{29}$ W. Calculate its distance in metres and in parsecs.

Hint

From $b = L/(4\pi d^2)$:

$d^2 = \frac◆LB◆L◆RB◆◆LB◆4\pi b◆RB◆ = \frac◆LB◆5.0 \times 10^{29}◆RB◆◆LB◆4\pi \times 2.0 \times 10^{-8}◆RB◆ = \frac◆LB◆5.0 \times 10^{29}◆RB◆◆LB◆2.513 \times 10^{-7}◆RB◆ = 1.99 \times 10^{36}$

$d = \sqrt◆LB◆1.99 \times 10^{36}◆RB◆ = 1.41 \times 10^{18}\;\mathrm{m}$

Converting to parsecs: $d = 1.41 \times 10^{18}/(3.09 \times 10^{16}) = 45.6$ pc.

Problem 3. The Sun has a surface temperature of $5800$ K and a luminosity of $3.85 \times 10^{26}$ W. Calculate its radius.

Hint

From $L = 4\pi r^2 \sigma T^4$:

$r^2 = \frac◆LB◆L◆RB◆◆LB◆4\pi \sigma T^4◆RB◆ = \frac◆LB◆3.85 \times 10^{26}◆RB◆◆LB◆4\pi \times 5.67 \times 10^{-8} \times (5800)^4◆RB◆$

$(5800)^4 = 1.133 \times 10^{15}$ K$^4$

$r^2 = \frac◆LB◆3.85 \times 10^{26}◆RB◆◆LB◆4\pi \times 5.67 \times 10^{-8} \times 1.133 \times 10^{15}◆RB◆ = \frac◆LB◆3.85 \times 10^{26}◆RB◆◆LB◆8.07 \times 10^{8}◆RB◆ = 4.77 \times 10^{17}$

$r = 6.91 \times 10^{8}\;\mathrm{m} \approx 6.91 \times 10^{5}\;\mathrm{km}$

This matches the accepted solar radius of $6.96 \times 10^{8}$ m.

Problem 4. A star has a surface temperature of $12,000$ K. Calculate the wavelength at which it Emits maximum radiation. In which part of the electromagnetic spectrum does this peak lie?

Hint

Using Wien’s displacement law:

$\lambda_{\max} = \frac◆LB◆2.898 \times 10^{-3}◆RB◆◆LB◆T◆RB◆ = \frac◆LB◆2.898 \times 10^{-3}◆RB◆◆LB◆12,000◆RB◆ = 2.42 \times 10^{-7}\;\mathrm{m} = 242\;\mathrm{nm}$

This is in the ultraviolet region. The star appears blue-white to the human eye, with most of Its visible output at shorter (blue) wavelengths.

Problem 5. A galaxy is observed to have a redshift $z = 0.05$. Calculate its recession velocity (assuming $v \ll c$) and its distance, using $H_0 = 70$ km s$^{-1}$ Mpc$^{-1}$.

Hint

From $z = v/c$: $v = 0.05 \times 3.0 \times 10^8 = 1.5 \times 10^{7}$ m s$^{-1}$ $= 15,000$ km S$^{-1}$.

From Hubble’s law: $d = v/H_0 = 15,000/70 = 214$ Mpc.

Problem 6. A reflecting telescope has a primary mirror of diameter $200$ mm. Calculate its Minimum angular resolution for light of wavelength $550$ nm. Express your answer in arcseconds.

Hint

Using the Rayleigh criterion:

$\theta = \frac◆LB◆1.22\lambda◆RB◆◆LB◆D◆RB◆ = \frac◆LB◆1.22 \times 550 \times 10^{-9}◆RB◆◆LB◆0.200◆RB◆ = 3.36 \times 10^{-6}\;\mathrm{rad}$

Converting to arcseconds: $\theta = 3.36 \times 10^{-6} \times 206,265 = 0.69''$.

Problem 7. A Cepheid variable in a nearby galaxy has a pulsation period of $10$ days. Its Absolute magnitude is $M = -4.0$ and its apparent magnitude is $m = 20.0$. Calculate the distance to The galaxy using the distance modulus.

Hint

Using the distance modulus $m - M = 5\log_{10}(d/10)$ where $d$ is in parsecs:

$20.0 - (-4.0) = 5\log_{10}\!\left(\frac{d}{10}\right)$

$24.0 = 5\log_{10}\!\left(\frac{d}{10}\right)$

$\log_{10}\!\left(\frac{d}{10}\right) = 4.8$

$\frac{d}{10} = 10^{4.8} = 6.31 \times 10^{4}$

$d = 6.31 \times 10^{5}\;\mathrm{pc} = 631\;\mathrm{kpc}$

Problem 8. Calculate the Schwarzschild radius of a black hole with mass $10\;M_\odot$. ($M_\odot = 1.99 \times 10^{30}$ kg)

Hint

$r_s = \frac{2GM}{c^2} = \frac◆LB◆2 \times 6.67 \times 10^{-11} \times 10 \times 1.99 \times 10^{30}◆RB◆◆LB◆(3.0 \times 10^8)^2◆RB◆$

$r_s = \frac◆LB◆2.653 \times 10^{21}◆RB◆◆LB◆9.0 \times 10^{16}◆RB◆ = 2.95 \times 10^{4}\;\mathrm{m} \approx 29.5\;\mathrm{km}$

This is comparable to the size of a large city, containing ten times the mass of the Sun.

Problem 9. Two stars A and B have the same luminosity. Star A has a surface temperature of $6000$ K and Star B has a surface temperature of $3000$ K. Calculate the ratio of their radii $r_A / r_B$.

Hint

Since $L_A = L_B$: $4\pi r_A^2 \sigma T_A^4 = 4\pi r_B^2 \sigma T_B^4$

$r_A^2 T_A^4 = r_B^2 T_B^4$

$\left(\frac{r_A}{r_B}\right)^2 = \left(\frac{T_B}{T_A}\right)^4 = \left(\frac{3000}{6000}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

$\frac{r_A}{r_B} = \frac{1}{4} = 0.25$

Star A is four times smaller in radius than Star B, despite being twice as hot. Both emit the same Total power --- Star B compensates for its lower temperature with a much larger surface area.

Problem 10. Estimate the age of the universe using $H_0 = 67.4$ km s$^{-1}$ Mpc$^{-1}$ (Planck Satellite value).

Hint

Convert $H_0$ to SI units:

$H_0 = \frac◆LB◆67.4 \times 10^{3}◆RB◆◆LB◆3.09 \times 10^{22}◆RB◆ = 2.181 \times 10^{-18}\;\mathrm{s}^{-1}$

$t_H = \frac{1}{H_0} = \frac◆LB◆1◆RB◆◆LB◆2.181 \times 10^{-18}◆RB◆ = 4.585 \times 10^{17}\;\mathrm{s}$

Converting to years:

$t_H = \frac◆LB◆4.585 \times 10^{17}◆RB◆◆LB◆3.156 \times 10^7◆RB◆ = 1.453 \times 10^{10}\;\mathrm{years} \approx 14.5\;\mathrm{Gyr}$

This is a reasonable estimate of the age of the universe. The accepted value from the Planck data is $\sim 13.8$ Gyr; the difference arises because the expansion rate has not been constant.

Problem 11. A radio telescope has a dish diameter of $30$ m and operates at a wavelength of $21$ Cm (the hydrogen line). Compare its resolving power with an optical telescope of diameter $1.0$ m Operating at $\lambda = 550$ nm. Give the ratio of their minimum resolvable angles.

Hint

Radio telescope:

$\theta_{\mathrm{radio}} = \frac◆LB◆1.22 \times 0.21◆RB◆◆LB◆30◆RB◆ = 8.54 \times 10^{-3}\;\mathrm{rad} \approx 29.3'$

Optical telescope:

$\theta_{\mathrm{optical}} = \frac◆LB◆1.22 \times 550 \times 10^{-9}◆RB◆◆LB◆1.0◆RB◆ = 6.71 \times 10^{-7}\;\mathrm{rad} \approx 0.14''$

Ratio:

$\frac◆LB◆\theta_{\mathrm{radio}}◆RB◆◆LB◆\theta_{\mathrm{optical}}◆RB◆ = \frac◆LB◆8.54 \times 10^{-3}◆RB◆◆LB◆6.71 \times 10^{-7}◆RB◆ \approx 12,700$

The optical telescope resolves about 12,700 times finer detail despite its much smaller aperture, Because the resolving power depends on $\lambda/D$ and the radio wavelength is $\sim 400,000$ times Longer. This illustrates why radio astronomers use interferometry with baselines of many kilometres.

Problem 12. A Type Ia supernova in a distant galaxy has an apparent brightness of $3.2 \times 10^{-15}$ W m$^{-2}$. Given that Type Ia supernovae have a peak luminosity of Approximately $1.0 \times 10^{36}$ W, calculate the distance to the galaxy in megaparsecs.

Hint

From $b = L/(4\pi d^2)$:

$d^2 = \frac◆LB◆L◆RB◆◆LB◆4\pi b◆RB◆ = \frac◆LB◆1.0 \times 10^{36}◆RB◆◆LB◆4\pi \times 3.2 \times 10^{-15}◆RB◆ = \frac◆LB◆1.0 \times 10^{36}◆RB◆◆LB◆4.02 \times 10^{-14}◆RB◆ = 2.49 \times 10^{49}$

$d = 4.99 \times 10^{24}\;\mathrm{m}$

Converting to megaparsecs: $d = 4.99 \times 10^{24}/(3.09 \times 10^{22}) = 161$ Mpc.

Common Pitfalls

1. Incorrectly applying $\vec{F} = m\vec{a}$ when forces are not collinear — resolve into components first.

2. Confusing displacement with distance, or velocity with speed, particularly in graphs and calculations.

3. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

4. Misidentifying the system boundary when applying conservation laws — define what is included before writing equations.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.