Know that the equation d / v = 1 / H_0 represents an estimate for the age of the Universe and that this is evidence for the idea that all the matter in the Universe was present at a single point

6 Physics – IGCSE (Cambridge 0625) Core Topics

6.1 Space Physics (Core)

• Earth’s rotation

  • Period ≈ 24 h → day/night cycle.
  • Angular speed ω = 2π / T ≈ 7.27 × 10⁻⁵ rad s⁻¹.

• Earth’s tilt (obliquity)

  • Axis tilted ≈ 23.5° to orbital plane.
  • Causes seasons – varying solar altitude and daylight length.

• Earth’s orbit

  • Nearly circular, semi‑major axis a ≈ 1 AU = 1.50 × 10¹¹ m.
  • Period ≈ 365 d (≈ 3.16 × 10⁷ s).
  • Orbital speed (circular approximation):

    
    v = 2πa / T ≈ 30 km s⁻¹

• Lunar phases & eclipses

  • Moon orbits Earth in ≈ 27.3 d (sidereal) → synodic month ≈ 29.5 d.
  • Phases result from Sun–Earth–Moon geometry.
  • Solar eclipse: Moon blocks Sun (requires near‑perfect alignment).
  • Lunar eclipse: Earth’s shadow falls on Moon (only full Moon).

• Solar‑system structure

  • Eight planets (Mercury → Neptune), dwarf planets (Pluto, Eris, …), asteroid belt, Kuiper belt, comets.
  • Typical size & distance scales (e.g., Earth radius ≈ 6.4 × 10⁶ m, Jupiter radius ≈ 7.1 × 10⁷ m, nearest star ≈ 4.2 ly).

Worked example – Orbital speed of Earth

Given a = 1 AU = 1.50 × 10¹¹ m and T = 365 d = 3.16 × 10⁷ s:

v = 2πa / T = 2π × 1.50 × 10¹¹ / 3.16 × 10⁷ ≈ 2.98 × 10⁴ m s⁻¹ ≈ 30 km s⁻¹

6.2 Motion, Forces & Energy

• Quantities & units – distance (m), time (s), speed (m s⁻¹), velocity (vector), acceleration (m s⁻²).
• Equations of motion (constant a)

  • v = u + at
  • s = ut + ½at²
  • v² = u² + 2as

• Forces

  • Resultant force = mass × acceleration (Newton’s 2nd law): F = ma.
  • Weight = mg, normal reaction, tension, friction (static µₛ, kinetic µₖ).

• Energy & power

  • Kinetic energy KE = ½mv².
  • Gravitational potential energy PE = mgh (near Earth).
  • Work W = F s cosθ, power P = W/t.

Worked example – Stopping distance of a car

A 1500 kg car travelling at 20 m s⁻¹ brakes with a constant deceleration of 5 m s⁻². Find the stopping distance.

Using s = v² / (2a) (with a = 5 m s⁻²):

s = (20)² / (2 × 5) = 400 / 10 = 40 m

6.3 Thermal Physics

• Kinetic particle model – temperature proportional to average kinetic energy.
• Temperature scales – Celsius, Kelvin (K = °C + 273.15).
• Specific heat capacity

  
  Q = mcΔT

• Phase changes

  
  Q = mL (L = latent heat of fusion/vapourisation).

• Heat transfer

  • Conduction: P = kAΔT / d.
  • Convection and radiation (Stefan‑Boltzmann law: P = εσAT⁴).

Worked example – Energy to melt ice

How much energy is needed to melt 2 kg of ice at 0 °C? (L_fusion = 3.34 × 10⁵ J kg⁻¹)

Q = mL = 2 × 3.34 × 10⁵ = 6.68 × 10⁵ J

6.4 Waves

• General wave properties

  • Wave speed v = fλ (f = frequency, λ = wavelength).
  • Transverse vs longitudinal.

• Reflection, refraction & diffraction – basic ray diagrams.
• Electromagnetic spectrum

  • Radio → microwave → infrared → visible → UV → X‑ray → γ‑ray.
  • All travel at c ≈ 3.00 × 10⁸ m s⁻¹ in vacuum.

• Sound – speed in air ≈ 340 m s⁻¹, dependence on temperature.

Worked example – Frequency of a radio wave

Given λ = 2 m, find f.

f = v / λ = 3.00 × 10⁸ / 2 = 1.5 × 10⁸ Hz

6.5 Electricity & Magnetism

• Charge & current – elementary charge e = 1.60 × 10⁻¹⁹ C, current I = ΔQ/Δt.
• Potential difference & resistance

  
  V = IR, R = ρℓ/A.

• Power in circuits

  
  P = VI = I²R = V²/R.

• Series & parallel combinations – add resistances appropriately.
• Magnetic fields

  • Force on a moving charge: F = qvB sinθ.
  • Force on a current‑carrying conductor: F = BIL sinθ.

• Electromagnetic induction

  
  ε = -dΦ/dt (Faraday’s law).

• Safety – earthing, fuses, RCDs.

Worked example – Current through a resistor

A 12 V battery is connected across a 4 Ω resistor. Find I and P.

I = V/R = 12/4 = 3 A

P = VI = 12 × 3 = 36 W

6.6 Nuclear Physics

• Atomic structure – nucleus (protons + neutrons) + electrons.
• Radioactive decay

  • α‑decay (He‑2 nucleus), β‑decay (electron or positron), γ‑decay (photon).
  • Half‑life t½ and exponential decay: N = N₀ (½)^{t/t½}.

• Applications & hazards – medical imaging, power generation, shielding.

Worked example – Remaining nuclei after 3 half‑lives

If a sample contains 1.0 × 10⁶ atoms initially, how many remain after 3 t½?

N = N₀ (½)³ = 1.0 × 10⁶ × 1/8 = 1.25 × 10⁵ atoms

6.7 Enrichment – Modern Cosmology

Why include this topic?

It demonstrates how quantitative observations (galaxy red‑shifts) lead to a profound conclusion: the Universe has a finite age and began from a single hot, dense state (the Big‑Bang). The material is optional for IGCSE but provides an excellent bridge to A‑Level physics and contemporary research.

Learning objectives

1. State and rearrange Hubble’s law to obtain the age estimate d / v = 1 / H₀.
2. Calculate the Hubble constant from simple distance–velocity data.
3. Convert H₀ into an age (years) and discuss the underlying assumptions.
4. Explain why a finite age supports the Big‑Bang model (all matter converging to a point when the expansion is extrapolated backwards).

Key concepts

• Hubble’s law: v = H₀ d (v = recession speed, d = distance).
• Hubble constant (H₀): current best estimates 67–74 km s⁻¹ Mpc⁻¹.
• Age estimate: Rearranging gives d / v = 1 / H₀. Since distance ÷ speed has units of time, 1 / H₀ is a first‑order estimate of the time since expansion began.
• Big‑Bang evidence: Extrapolating the linear expansion backwards makes all galaxies converge to a single point, implying a hot‑dense origin.

Derivation of the age estimate

Starting from Hubble’s law:

v = H₀ d

Rearrange:

d / v = 1 / H₀

The left‑hand side is a time (distance divided by speed). If the present expansion rate had been constant since the start, this time equals the age of the Universe.

Numerical conversion (SI units)

1. Take a typical value: H₀ = 70 km s⁻¹ Mpc⁻¹.
2. Convert to metres per second per metre:

   • 1 Mpc = 3.09 × 10²² m.
   • 70 km s⁻¹ Mpc⁻¹ = 70 000 m s⁻¹ / 3.09 × 10²² m ≈ 2.27 × 10⁻¹⁸ s⁻¹

3. Invert to obtain the age:

   
   1 / H₀ ≈ 1 / 2.27 × 10⁻¹⁸ s⁻¹ = 4.4 × 10¹⁷ s

4. Convert seconds to years (1 yr ≈ 3.16 × 10⁷ s):

   
   4.4 × 10¹⁷ s / 3.16 × 10⁷ s yr⁻¹ ≈ 1.4 × 10¹⁰ yr ≈ 14 billion yr

Worked‑example (IGCSE‑style)

Question: A galaxy 10 Mpc away shows a red‑shift corresponding to a recession speed of 700 km s⁻¹. Using these data, estimate the Hubble constant and then the age of the Universe. State one key assumption.

Solution:

1. H₀ = v / d = 700 km s⁻¹ / 10 Mpc = 70 km s⁻¹ Mpc⁻¹.
2. Convert: H₀ ≈ 2.27 × 10⁻¹⁸ s⁻¹ (as shown above).
3. Age: t = 1 / H₀ ≈ 4.4 × 10¹⁷ s ≈ 14 billion yr.
4. Assumption: The expansion rate has remained constant over cosmic time. In reality it has varied (early deceleration, recent acceleration), so the true age differs slightly.

Assumptions & sources of error

• Constant expansion rate – ignores gravitational deceleration and dark‑energy acceleration.
• Peculiar velocities – local motions of galaxies add/subtract from the pure Hubble flow.
• Measurement uncertainty in H₀ (the current “Hubble tension” between CMB and supernova methods).

Table: Hubble constant values and corresponding ages

Interpretation & significance

1. Hubble’s law shows the Universe is expanding.
2. Extrapolating the linear relationship back to time = 0 gives a single origin point – the core idea of the Big‑Bang model.
3. The simple estimate 1 / H₀ yields an age (~14 Gyr) that agrees remarkably with detailed cosmological models (≈ 13.8 Gyr).
4. Current research focuses on refining H₀; the discrepancy between early‑Universe (CMB) and late‑Universe (supernovae) measurements is a major open question.

Suggested classroom activity

Provide students with a small data set of galaxy distances (Mpc) and recession speeds (km s⁻¹). Have them:

1. Plot v (vertical) against d (horizontal) on graph paper or a spreadsheet.
2. Draw the best‑fit straight line; determine its slope (≈ H₀) using two well‑spaced points.
3. Convert the slope to SI units and calculate the age of the Universe.
4. Discuss the impact of outliers (peculiar velocities) and the assumption of constant expansion.

Summary

The rearranged Hubble law d / v = 1 / H₀ provides a quick, model‑independent estimate of the Universe’s age. Using the current range of H₀ values gives ages between 13 and 15 billion years, supporting the Big‑Bang picture that all matter originated from a single hot, dense point. While this enrichment material is not examined in the IGCSE, it reinforces the scientific method, quantitative reasoning, and the connection between observation and theory – skills that are essential across the entire physics syllabus.