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 + ats = ut + ½at²v² = 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
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
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
- State and rearrange Hubble’s law to obtain the age estimate
d / v = 1 / H₀. - Calculate the Hubble constant from simple distance–velocity data.
- Convert
H₀ into an age (years) and discuss the underlying assumptions. - 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)
- Take a typical value:
H₀ = 70 km s⁻¹ Mpc⁻¹. - 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
- Invert to obtain the age:
1 / H₀ ≈ 1 / 2.27 × 10⁻¹⁸ s⁻¹ = 4.4 × 10¹⁷ s
- 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:
H₀ = v / d = 700 km s⁻¹ / 10 Mpc = 70 km s⁻¹ Mpc⁻¹.- Convert:
H₀ ≈ 2.27 × 10⁻¹⁸ s⁻¹ (as shown above). - Age:
t = 1 / H₀ ≈ 4.4 × 10¹⁷ s ≈ 14 billion yr. - 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
| H₀ (km s⁻¹ Mpc⁻¹) | H₀ (s⁻¹) | 1 / H₀ (s) | Age (billion yr) |
|---|
| 67 | 2.17 × 10⁻¹⁸ | 4.61 × 10¹⁷ | 14.6 |
| 70 | 2.27 × 10⁻¹⁸ | 4.40 × 10¹⁷ | 14.0 |
| 74 | 2.40 × 10⁻¹⁸ | 4.17 × 10¹⁷ | 13.2 |
Interpretation & significance
- Hubble’s law shows the Universe is expanding.
- Extrapolating the linear relationship back to time = 0 gives a single origin point – the core idea of the Big‑Bang model.
- The simple estimate
1 / H₀ yields an age (~14 Gyr) that agrees remarkably with detailed cosmological models (≈ 13.8 Gyr). - 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:
- Plot v (vertical) against d (horizontal) on graph paper or a spreadsheet.
- Draw the best‑fit straight line; determine its slope (≈ H₀) using two well‑spaced points.
- Convert the slope to SI units and calculate the age of the Universe.
- 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.