Flashcards · H556 · Topic 5

Newtonian world and astrophysics flashcards

50 cards on Topic 5, subtopics 5.1 to 5.5. Pick a subtopic or study them all.

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5.1 Thermal physics

Thermal equilibrium: No net energy transfer between objects in thermal contact; they are at the same temperature.
Absolute zero, and converting °C to K: The lowest limit of temperature, 0 K, where a substance has minimum internal energy. T (K) ≈ θ (°C) + 273; a temperature change is the same in K and °C.
Internal energy: The total of the randomly distributed kinetic and potential energies of the molecules in a system.
Specific heat capacity and specific latent heat: c: energy to raise the temperature of 1 kg by 1 K, E = mcΔθ. l: energy to change the phase of 1 kg at constant temperature, E = ml (fusion: solid to liquid; vaporisation: liquid to gas).
During melting, what happens to temperature, internal energy and molecular energies?: Temperature is constant; internal energy rises; the potential energy of the molecules rises as they separate, while their mean kinetic energy stays the same.
Brownian motion: what is seen and what it shows: Smoke particles move in random zig-zags, from unbalanced impacts of fast-moving air molecules. It shows molecules are in continuous random motion.
Kinetic theory assumptions: Large number of molecules in rapid random motion; negligible volume compared with the gas; perfectly elastic collisions of negligible duration; negligible forces between molecules except during collisions.
Ideal gas equations: pV = nRT = NkT (T in kelvin), with k = R/N_A. pV = ⅓Nm⟨c²⟩.
Mean kinetic energy of a molecule, and internal energy of an ideal gas: ½m⟨c²⟩ = (3/2)kT. U = (3/2)NkT = (3/2)nRT: only kinetic energy, since there are no intermolecular forces.
How does c_rms depend on temperature? What happens if T in kelvin is quadrupled?: c_rms ∝ √(T/m). Quadrupling T doubles c_rms.
Effect of a higher temperature on the Maxwell–Boltzmann distribution: The peak moves to a higher speed and becomes lower and broader; more molecules have high speeds. The area (number of molecules) is unchanged.

5.2 Circular motion

The radian: The angle at the centre of a circle subtended by an arc equal in length to the radius. 2π rad = 360°.
Angular velocity and speed in a circle: ω = Δθ/Δt = 2π/T = 2πf (rad s⁻¹). v = ωr.
Centripetal acceleration and force: a = v²/r = ω²r, towards the centre. F = mv²/r = mω²r, supplied by a real force (tension, friction, gravity, magnetic force…).
Why does an object in uniform circular motion keep a constant speed?: The net force is perpendicular to the velocity, so it has no component along the motion, does no work, and cannot change the kinetic energy.
A 0.90 kg mass moves in a circle of radius 0.35 m at 2.0 Hz. Find the centripetal force.: ω = 2π × 2.0 = 12.57 rad s⁻¹. F = mω²r = 0.90 × 12.57² × 0.35 ≈ 50 N.
In F = mv²/r, what happens to F if v doubles, or r doubles (v fixed)?: v doubles: F × 4. r doubles at fixed v: F halves.

5.3 Oscillations

Simple harmonic motion: Acceleration is proportional to displacement and always directed towards the equilibrium position: a = −ω²x.
SHM equations for displacement and speed: x = A cos ωt (timing from maximum displacement) or x = A sin ωt (from equilibrium). v = ±ω√(A² − x²); vmax = ωA; amax = ω²A.
Isochronous oscillator: One whose period is independent of its amplitude.
Phase of velocity and acceleration relative to displacement in SHM: Velocity leads displacement by π/2. Acceleration is in antiphase (π) with displacement.
Energy in SHM: Total E = ½mω²A². Ek is maximum at x = 0, Ep is maximum at x = ±A. Ek and Ep each vary at twice the oscillation frequency.
k = 36 N m⁻¹, m = 0.25 kg. Find the period.: ω² = k/m = 144 s⁻², so ω = 12 rad s⁻¹ and T = 2π/12 = 0.52 s.
Light, critical and heavy damping: Light: oscillates, amplitude decays steadily. Critical: settles at equilibrium in the least time with no oscillation (car suspension). Heavy: no oscillation, very slow return.
Resonance, and the effect of damping on the resonance curve: Maximum amplitude when the driving frequency equals the natural frequency. More damping gives a lower, broader peak at a slightly lower frequency.

5.4 Gravitational fields

Gravitational field strength: The force per unit mass on a small test mass, g = F/m, in N kg⁻¹. A vector, pointing towards the source.
Newton's law of gravitation and g in a radial field: F = GMm/r² (attractive). g = GM/r², with r measured from the centre of the body, not the surface.
Kepler's three laws: 1. Orbits are ellipses with the Sun at one focus. 2. The line from the Sun to a planet sweeps equal areas in equal times. 3. T² ∝ r³.
Derive T² = (4π²/GM)r³: Gravity provides the centripetal force: GMm/r² = mv²/r, so v² = GM/r. Substitute v = 2πr/T: 4π²r²/T² = GM/r, giving T² = (4π²/GM)r³.
Geostationary orbit: conditions and radius: Above the equator, moving west to east, circular, period 24 h, so it stays above one point. For the Earth r ≈ 4.22 × 10⁷ m.
Gravitational potential: The work done per unit mass in bringing a small mass from infinity to the point. Vg = −GM/r (J kg⁻¹). Zero at infinity, negative everywhere else.
Gravitational potential energy of a mass m at distance r: E = mVg = −GMm/r. Work done moving between two points = mΔVg = area under the force–distance graph.
Escape velocity: definition and equation: The minimum launch speed for an object to just reach infinity with no further propulsion. ½mv² − GMm/r = 0, so v = √(2GM/r) = √(2gR) at the surface.
g vs G: g is the field strength at a place; it varies with r. G is the universal gravitational constant.

5.5 Astrophysics and cosmology

How does a star form?: A cloud of dust and gas collapses under gravity; gravitational potential energy becomes thermal energy, forming a protostar. When the core is hot enough, hydrogen fuses into helium and the star is on the main sequence.
What keeps a main sequence star stable?: Gravity inwards is balanced by gas pressure plus radiation pressure outwards.
Evolution of a low-mass star: Main sequence → red giant → planetary nebula (outer layers ejected) → white dwarf.
Electron degeneracy pressure: Electrons cannot share quantum states (Pauli exclusion principle), so compressing them forces them into higher energy states. This gives an outward pressure that does not depend on temperature.
Chandrasekhar limit: The maximum mass of a stable white dwarf, about 1.4 solar masses.
Evolution of a massive star: Main sequence → red supergiant (fusing heavier nuclei up to iron) → core collapse and supernova → neutron star, or a black hole if the remnant core is massive enough.
HR diagram: axes and regions: Luminosity up the vertical axis; surface temperature along the horizontal, decreasing left to right. Main sequence: diagonal band. Red giants and supergiants: upper right. White dwarfs: bottom left.
Why are atomic energy levels negative?: Zero is the energy of a free electron at rest; a bound electron has less energy than that. Ionisation from a level needs energy equal to its magnitude.
Emission line vs absorption line spectrum: Emission: bright lines on dark, from hot low-pressure gas. Absorption: dark lines on a continuous spectrum, from light passing through cooler gas. An element's lines are at the same wavelengths in both.
Wien's law and Stefan's law: Wien: λmax T = 2.90 × 10⁻³ m K. Stefan: L = 4πr²σT⁴. Use Wien for T, then Stefan for r.
Astronomical unit, light-year and parsec: AU: mean Earth–Sun distance, 1.50 × 10¹¹ m. Light-year: distance light travels in a vacuum in one year, about 9.5 × 10¹⁵ m. Parsec: distance at which a star has a parallax of 1 second of arc, about 3.1 × 10¹⁶ m (Data booklet values).
A star has a parallax angle of 0.20 seconds of arc. How far away is it?: d = 1/p = 1/0.20 = 5.0 pc (about 16 ly).
Cosmological principle: On a large scale the universe is homogeneous, isotropic, and the laws of physics are the same everywhere.
Doppler red shift and Hubble's law: Δλ/λ ≈ Δf/f ≈ v/c (v ≪ c); a receding source is red shifted. Hubble: v ≈ H₀d, so recession speed is proportional to distance.
Two pieces of evidence for the Big Bang: Galactic red shift and Hubble's law (expansion). The cosmic microwave background, observed from all directions at about 2.7 K.
Estimate the age of the universe from H₀ = 70 km s⁻¹ Mpc⁻¹: 1 Mpc ≈ 3.1 × 10²² m, so H₀ = 70 × 10³ ÷ 3.1 × 10²² ≈ 2.3 × 10⁻¹⁸ s⁻¹. t ≈ 1/H₀ ≈ 4.4 × 10¹⁷ s ≈ 1.4 × 10¹⁰ years.