Astrophysics and Cosmology

Q: Do I need to memorize the value of the Hubble constant?

No, the value of will always be supplied in exam questions for this topic.

Edexcel International A-Level Physics· 2018 Spec Issue 3, Statements 154–171, WPH15· 45 min read

1. Gravitational Fields and Orbital Motion★★★☆☆⏱ 10 min

📘 Definition

Gravitational field strength

g = \frac{F}{m}

Force per unit mass acting on a small test mass placed in a gravitational field

Example:

Earth's surface gravitational field strength is ~9.81 N kg⁻¹

Newton's law of universal gravitation describes the attractive force between two point masses: $F = \frac{G m _{1} m _{2}}{r ^{2}}$ . To derive gravitational field strength from this law, equate $F = m g$ to the gravitational force, cancel the test mass $m$ , to get $g = \frac{GM}{r ^{2}}$ , where $M$ is the mass of the body producing the field. Radial gravitational potential is given by $V_{g r a v} = - \frac{GM}{r}$ , always negative because work is required to move a mass away from the field, with zero defined at infinity. Gravitational fields are always attractive, unlike electric fields which can be attractive or repulsive.

📐 Worked Example

Calculate the gravitational field strength at the surface of Mars, given $M_{M a r s} = 6.42 \times 1 0^{23}$ kg, $r_{M a r s} = 3.39 \times 1 0^{6}$ m, $G = 6.67 \times 1 0^{- 11}$ N m² kg⁻².

1
Recall the formula for gravitational field strength:
$g = \frac{GM}{r^2}$
2
Substitute the given values into the formula:
$g = \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^6)^2}$
3
Calculate the result: $g \approx 3.72$ N kg⁻¹

Exam tip:

Always use centre-to-centre distance (not surface separation) for all gravitational force and field calculations.

2. Stellar Radiation and Radiation Laws★★★☆☆⏱ 10 min

📘 Definition

Black body radiator

An ideal object that absorbs all incident radiation and emits a continuous spectrum of radiation whose shape depends only on its absolute temperature.

Stars approximate black body radiators. The Stefan-Boltzmann law relates luminosity (total power output) to surface area and temperature: $L = σ A T^{4}$ . Wien's displacement law relates peak emission wavelength to temperature: $λ_{ma x} T = 2.898 \times 1 0^{- 3}$ m K, so hotter stars emit shorter wavelength (bluer) light. Intensity of radiation received at a distance $d$ from a star follows the inverse square law: $I = \frac{L}{4 π d ^{2}}$ .

📐 Worked Example

The Sun has a surface temperature of 5778 K and peak emission wavelength of 500 nm. A red giant star has a peak emission wavelength of 700 nm. Calculate the surface temperature of the red giant.

1
Use Wien's law, which states $λ_{ma x} T$ is constant:
$\lambda_1 T_1 = \lambda_2 T_2$
2
Rearrange to solve for $T_{2}$ :
$T_2 = \frac{\lambda_1 T_1}{\lambda_2}$
3
Substitute values (units cancel, so no need to convert nm to m):
$T_2 = \frac{500 \times 5778}{700} \approx 4127 \text{ K}$

Exam tip:

Always convert temperature to Kelvin before using Stefan-Boltzmann or Wien's law formulas.

3. Astronomical Distance Measurement★★☆☆☆⏱ 8 min

Two primary methods are used to measure astronomical distances: trigonometric parallax for nearby stars, and standard candles for distant objects. Parallax uses the apparent shift of a star against the background of distant stars as Earth orbits the Sun. The distance $d$ in parsecs is given by $d = \frac{1}{p}$ , where $p$ is the parallax angle in arcseconds. Standard candles are objects with known luminosity (e.g. Type Ia supernovae, Cepheid variables): measure the received intensity, then use the inverse square law to calculate distance.

📐 Worked Example

A nearby star has a measured parallax angle of 0.2 arcseconds. Calculate its distance from Earth in parsecs.

1
Recall the parallax distance formula:
$d = \frac{1}{p}$
2
Substitute $p = 0.2$ arcseconds:
$d = \frac{1}{0.2} = 5 \text{ parsecs}$

Exam tip:

Trigonometric parallax is only reliable for stars within ~100 parsecs of Earth, as further stars have unmeasurably small parallax angles.

4. Hertzsprung-Russell Diagram and Stellar Life Cycles★★☆☆☆⏱ 8 min

📘 Definition

Hertzsprung-Russell (HR) Diagram

A scatter plot of stars with luminosity (logarithmic y-axis) plotted against surface temperature (reversed x-axis, measured in Kelvin or spectral class).

Most stars fall on the main sequence, where they fuse hydrogen into helium in their core. Low mass stars leave the main sequence to become red giants, then eventually cool to form white dwarfs. High mass stars become red supergiants, explode as supernovae, and leave behind either a neutron star or black hole.

📐 Worked Example

A star is located in the top right region of the HR diagram. State its type, approximate temperature, and luminosity relative to the Sun.

1
Top right region = high luminosity, low temperature: red giant or supergiant
2
Approximate temperature: 3000–4000 K, ~100–10,000 times more luminous than the Sun

Exam tip:

The x-axis of the HR diagram is reversed: hotter stars are plotted on the left, cooler stars on the right.

5. Redshift, Hubble's Law and Cosmology★★★☆☆⏱ 9 min

📘 Definition

Redshift

z = \frac{\Delta\lambda}{\lambda} \approx \frac{v}{c}

The fractional increase in wavelength of radiation from a receding object, used to calculate recessional speed for non-relativistic velocities.

Edwin Hubble observed that almost all galaxies are redshifted, meaning they are receding from Earth. Hubble's law states $v = H_{0} d$ , where $v$ is recessional speed and $d$ is distance. This observation supports the Big Bang theory of an expanding universe. The approximate age of the universe is $\frac{1}{H _{0}}$ , but discrepancies in measured values of $H_{0}$ , and evidence for dark matter (to explain galactic rotation curves) and dark energy (to explain accelerating expansion) are ongoing areas of cosmological controversy.

📐 Worked Example

A galaxy has a redshift $z = 0.02$ . Calculate its recessional speed, and distance from Earth if $H_{0} = 70$ km s⁻¹ Mpc⁻¹.

1
Calculate recessional speed using $z \approx \frac{v}{c}$ :
$v = zc = 0.02 \times 3 \times 10^8 = 6 \times 10^6 \text{ m s}^{-1} = 6000 \text{ km s}^{-1}$
2
Rearrange Hubble's law to solve for distance:
$d = \frac{v}{H_0} = \frac{6000}{70} \approx 85.7 \text{ Mpc}$

Exam tip:

Ensure units match when using Hubble's law: if $H_{0}$ is given in km s⁻¹ Mpc⁻¹, convert velocity to km s⁻¹ to get distance directly in Mpc.

6. Common Pitfalls

Wrong move:

Using surface separation instead of centre-to-centre distance for gravitational calculations

Why:

Gravitational formulas assume point masses located at the centre of spherical objects

Correct move:

Add the radii of both objects to surface separation to get the correct value of $r$

Wrong move:

Using Celsius instead of Kelvin for radiation law calculations

Why:

Stefan-Boltzmann and Wien's laws use absolute temperature, so Celsius values produce invalid results

Correct move:

Add 273.15 to Celsius temperature to convert to Kelvin before substituting into formulas

Wrong move:

Forgetting the HR diagram x-axis is reversed

Why:

Hotter stars are plotted on the left, cooler on the right, opposite to standard graph conventions

Correct move:

Check axis labels first, and remember OBAFGKM spectral classes go from hottest to coolest left to right

Wrong move:

Omitting the negative sign for gravitational potential

Why:

Gravitational potential is defined as zero at infinity, so all finite values are negative

Correct move:

Always include the negative sign in gravitational potential calculations and interpretations

Wrong move:

Mixing units in Hubble's law calculations

Why:

$H_{0}$ is typically given in km s⁻¹ Mpc⁻¹, so using m s⁻¹ for velocity produces wrong distance values

Correct move:

Convert all units to match the units of $H_{0}$ before substituting into $v = H_{0} d$

7. Quick Reference Cheatsheet

Formula	Key Variables	Units	Notes
$F = \frac{G m _{1} m _{2}}{r ^{2}}$	$G = 6.67 e - 11$ , $m$ =mass, $r$ =centre distance	N	Attractive force only
$g = \frac{GM}{r ^{2}}$	$M$ =mass of field-producing body	N kg⁻¹	Equals acceleration due to gravity
$V_{g r a v} = - \frac{GM}{r}$	$V$ =gravitational potential	J kg⁻¹	Always negative, zero at infinity
$L = σ A T^{4}$	$σ = 5.67 e - 8$ , $T$ =temp (K)	W	Stefan-Boltzmann law
$λ_{ma x} T = 2.898 e - 3$	$λ_{ma x}$ =peak wavelength	m K	Wien's displacement law
$I = \frac{L}{4 π d ^{2}}$	$I$ =intensity, $d$ =distance	W m⁻²	Inverse square law
$z = \frac{Δ λ}{λ} \approx \frac{v}{c}$	$z$ =redshift, $v$ =recessional speed	Dimensionless	Non-relativistic only
$v = H_{0} d$	$H_{0}$ =Hubble constant	km s⁻¹ / Mpc	$H_{0}$ value supplied in question

8. Frequently Asked

Why is gravitational potential always negative?

Gravitational potential is defined as zero at an infinite distance from a mass. Work must be done to move a mass away from a gravitational field, so all potential values at finite distances are negative.

Do I need to memorize the value of the Hubble constant?

No, the value of $H_{0}$ will always be supplied in exam questions for this topic.

Can I use Celsius temperature for radiation law calculations?

No, Stefan-Boltzmann and Wien's laws use absolute temperature, so you must convert Celsius values to Kelvin by adding 273.15 before substituting into formulas.

Going deeper

official_documentEdexcel IAL Physics Data SheetContains all formulas for this topic

What's Next

Now that you have mastered Astrophysics and Cosmology for Edexcel IAL Physics Unit 5, you are ready to tackle synoptic questions that combine gravitational orbital motion with circular mechanics, and practice past paper questions to refine your exam technique. Make sure you memorize the sign conventions for gravitational potential and the reversed axis of the HR diagram, as these are common marks lost in exams. Next, review the remaining Unit 5 topics to build a complete understanding of the unit content before attempting full WPH15 past papers.

→Thermodynamics (Edexcel IAL Physics U5)