Formula list

v = Δs / Δt

v

average velocitym s⁻¹

Δs

displacementm

Δt

time intervals

A definition, not a booklet equation. Average speed uses distance travelled and average velocity uses displacement; the two differ whenever the path is not straight.

a = Δv / Δt

a

average accelerationm s⁻²

Δv

change in velocitym s⁻¹

Δt

time intervals

A definition, not a booklet equation.

v = u + at

v

final velocitym s⁻¹

u

initial velocitym s⁻¹

a

accelerationm s⁻²

t

times

s = ut + ½at²

s

displacementm

u

initial velocitym s⁻¹

a

accelerationm s⁻²

t

times

v² = u² + 2as

v

final velocitym s⁻¹

u

initial velocitym s⁻¹

a

accelerationm s⁻²

s

displacementm

s = ½(u + v)t

s

displacementm

u, v

initial and final velocitym s⁻¹

t

times

All four of these hold only while the acceleration is constant.

velocity = gradient of s–t · acceleration = gradient of v–t · displacement = area under v–t

gradient

of a tangent where the motion is not uniform

Not printed in the data booklet. Distance travelled is the area under a speed–time graph, which is not the same as displacement once the direction reverses.

horizontal: sx = uxt · vertical: sy = uyt − ½gt²

ux

horizontal component of the launch velocitym s⁻¹

uy

vertical component of the launch velocitym s⁻¹

g

acceleration of free fallm s⁻²

Not a separate booklet equation: it is the four equations above applied to each axis, with zero acceleration horizontally. Fluid resistance is left out here, and its effect on a projectile — shorter range, a steeper descent — is wanted only qualitatively; the drag force itself is in A.2, and terminal speed is where it balances the weight.

F = ma

F

resultant forceN

m

masskg

a

accelerationm s⁻²

The special case of F = Δp/Δt for constant mass. Force and acceleration are in the same direction.

F = Δp / Δt

F

resultant forceN

Δp

change in momentumkg m s⁻¹

Δt

time takens

p = mv

p

momentumkg m s⁻¹, N s

m

masskg

v

velocitym s⁻¹

J = FΔt = Δp

J

impulseN s

F

constant forceN

Δt

time for which it actss

For a varying force, impulse is the area under a force–time graph.

ΣF = 0

ΣF

vector sum of every force on the bodyN

Not printed in the data booklet. It is the condition for constant velocity, drawn from a free-body diagram.

total momentum before = total momentum after

p

summed over the system, with direction

Not printed in the data booklet. It holds for an isolated system, in collisions and in explosions; kinetic energy is conserved only in an elastic collision.

Ff ≤ μs N

Ff

friction forceN

μs

coefficient of static friction

N

normal reaction forceN

An inequality: static friction takes whatever value balances the applied force, up to this maximum.

Ff = μd N

Ff

friction force while slidingN

μd

coefficient of dynamic friction

N

normal reaction forceN

Fb = ρVg

Fb

buoyancy force, the upthrust on the bodyN

ρ

density of the fluidkg m⁻³

V

volume of fluid displacedm³

g

gravitational field strengthN kg⁻¹

Archimedes' principle: the upthrust equals the weight of the fluid displaced. V is the volume of fluid pushed aside, which is the volume of the whole body only when it is fully submerged.

Fd = 6πηrv

Fd

viscous drag forceN

η

viscosity of the fluidPa s

r

radius of the spherem

v

speed of the sphere through the fluidm s⁻¹

For a small sphere in laminar flow only. The drag grows with speed, which is what fixes terminal speed: the body stops accelerating once the drag plus the buoyancy balance the weight.

ω = 2π / T · v = ωr

ω

angular speedrad s⁻¹

T

period of the circular motions

v

linear speedm s⁻¹

r

radius of the circlem

a = v²/r = ω²r = 4π²r / T²

a

centripetal accelerationm s⁻²

v

linear speedm s⁻¹

r

radiusm

Directed towards the centre, so uniform circular motion is accelerated motion even though the speed does not change.

F = mv²/r = mω²r = 4π²mr / T²

F

resultant force towards the centreN

m

masskg

r

radiusm

Not a new force: it is the name for whatever resultant — tension, gravity, friction, the normal force — points at the centre.

W = Fs cos θ

W

work doneJ

F

forceN

s

displacementm

θ

angle between force and displacement°

A force at right angles to the motion does no work, which is why a centripetal force never changes the speed.

work done = area under a force–displacement graph

area

counted negative where the force opposes the motion

Not printed in the data booklet. This is how work is found when the force varies, as it does for a spring.

Ek = ½mv² = p² / (2m)

Ek

kinetic energyJ

m

masskg

v

speedm s⁻¹

p

momentumkg m s⁻¹

The momentum form is worth having: it turns a momentum answer into an energy answer without going back through the speed.

ΔEp = mgΔh

ΔEp

change in gravitational potential energyJ

m

masskg

g

gravitational field strengthN kg⁻¹

Δh

change in heightm

For a uniform field, which is the near-Earth case. D.1 replaces it with Ep = −GMm/r once the field is no longer uniform.

F = kΔx

Rearrangedk = F / Δx · Δx = F / k

F

force applied to the springN

k

spring constantN m⁻¹

Δx

extension or compressionm

Within the limit of proportionality.

Ee = ½kΔx²

Ee

energy stored in the springJ

k

spring constantN m⁻¹

Δx

extension or compressionm

The area under the force–extension graph.

P = W / Δt = Fv

P

powerW

W

work doneJ

Δt

time takens

F

forceN

v

speed in the direction of the forcem s⁻¹

η = useful work out / total work in = useful power out / total power in

η

efficiency (a ratio, no unit)

Multiply by 100% for a percentage. Energy is conserved either way — the shortfall is degraded, not destroyed.

τ = Fr sin θ

τ

torque about the axisN m

F

forceN

r

distance from the axis to the point where the force is appliedm

θ

angle between the force and the line joining the axis to that point°

Higher level only. A body is in rotational equilibrium when the resultant torque about every axis is zero.

I = Σmr²

I

moment of inertiakg m²

m

mass of each particlekg

r

its distance from the axism

Higher level only. It depends on the axis chosen, not only on the body — the rotational counterpart of mass.

τ = Iα

τ

resultant torqueN m

I

moment of inertiakg m²

α

angular accelerationrad s⁻²

Higher level only.

ω = ω₀ + αt

ω

final angular velocityrad s⁻¹

ω₀

initial angular velocityrad s⁻¹

α

angular accelerationrad s⁻²

t

times

Higher level only.

θ = ω₀t + ½αt² · θ = ½(ω₀ + ω)t

θ

angular displacementrad

ω₀, ω

initial and final angular velocityrad s⁻¹

α

angular accelerationrad s⁻²

Higher level only. The four rotational equations mirror the four in A.1 term for term.

ω² = ω₀² + 2αθ

ω, ω₀

final and initial angular velocityrad s⁻¹

α

angular accelerationrad s⁻²

θ

angular displacementrad

Higher level only.

L = Iω · ΔL = τΔt

L

angular momentumkg m² s⁻¹

I

moment of inertiakg m²

ω

angular velocityrad s⁻¹

τ

resultant torqueN m

Higher level only. Angular momentum is conserved when the resultant torque is zero, which is why a skater spins faster with their arms pulled in.

Ek = ½Iω² = L² / (2I)

Ek

rotational kinetic energyJ

I

moment of inertiakg m²

ω

angular velocityrad s⁻¹

L

angular momentumkg m² s⁻¹

Higher level only. A rolling body has this on top of ½mv² for its centre of mass.

x′ = x − vt · u′ = u − v

x′, x

position in the moving and the original framem

u′, u

velocity of an object in each framem s⁻¹

v

speed of one frame relative to the otherm s⁻¹

Higher level only. The low-speed limit, and the thing special relativity replaces.

γ = 1 / √(1 − v²/c²)

γ

Lorentz factor (a ratio, no unit)

v

relative speed of the two framesm s⁻¹

c

speed of light in a vacuumm s⁻¹

Higher level only. γ ≥ 1 always, and every relativistic effect below is γ doing the work.

x′ = γ(x − vt) · t′ = γ(t − vx/c²)

x′, t′

coordinates of an event in the moving framem, s

x, t

its coordinates in the original framem, s

γ

Lorentz factor

Higher level only. The same form works for intervals: Δx′ = γ(Δx − vΔt) and Δt′ = γ(Δt − vΔx/c²).

u′ = (u − v) / (1 − uv/c²)

u′

velocity measured in the moving framem s⁻¹

u

velocity measured in the original framem s⁻¹

v

relative speed of the framesm s⁻¹

Higher level only. Feed in u = c and you get u′ = c, whatever v is — which is the postulate the whole theory is built on.

Δt = γΔt₀

Δt

time interval measured in the frame the clock moves ins

Δt₀

proper time, measured where the two events happen at the same places

γ

Lorentz factor

Higher level only. Getting the answer right is mostly a matter of deciding which observer measures the proper time.

L = L₀ / γ

L

length measured in the frame the object moves inm

L₀

proper length, measured at rest with respect to the objectm

γ

Lorentz factor

Higher level only. Contraction happens along the direction of motion only.

(Δs)² = (cΔt)² − (Δx)²

Δs

spacetime interval between two eventsm

Δt

time between thems

Δx

distance between themm

Higher level only. Every inertial observer disagrees about Δt and Δx and agrees about Δs — the invariant the spacetime diagram is drawn around.

E₀ = mc²

E₀

rest energyJ, MeV

m

rest masskg, MeV c⁻²

c

speed of light in a vacuumm s⁻¹

Higher level only. Working in MeV and MeV c⁻² saves most of the arithmetic.

E = γmc²

E

total energyJ, MeV

γ

Lorentz factor

m

rest masskg

Higher level only.

Ek = (γ − 1)mc²

Ek

kinetic energyJ, MeV

γ

Lorentz factor

m

rest masskg

Higher level only. Total energy minus rest energy; ½mv² is what this collapses to at low speed.

p = γmv

p

momentumkg m s⁻¹, MeV c⁻¹

γ

Lorentz factor

m

rest masskg

v

speedm s⁻¹

E² = p²c² + m²c⁴

E

total energyJ, MeV

p

momentumkg m s⁻¹, MeV c⁻¹

m

rest masskg, MeV c⁻²

Higher level only. It links E and p without needing the speed, and for a massless particle it reduces to E = pc.

T / K = θ / °C + 273

Rearrangedθ / °C = T / K − 273

T

absolute temperatureK

θ

temperature°C

Absolute zero is 0 K. Every gas and radiation equation below needs the temperature in kelvin.

Q = mcΔT

Rearrangedc = Q / (mΔT) · ΔT = Q / (mc)

Q

energy transferredJ

m

masskg

c

specific heat capacityJ kg⁻¹ K⁻¹

ΔT

temperature changeK

A temperature change in kelvin and in degrees Celsius is the same number, so either unit works here.

Q = mL

Q

energy transferredJ

m

mass changing phasekg

L

specific latent heat of fusion or vaporisationJ kg⁻¹

The temperature does not change during a phase change: the energy goes into potential energy between the molecules, not kinetic.

ΔQ / Δt = kA ΔT / Δx

ΔQ/Δt

rate of thermal energy transferW

k

thermal conductivityW m⁻¹ K⁻¹

A

cross-sectional aream²

ΔT/Δx

temperature gradientK m⁻¹

Convection is treated qualitatively, through density differences in a fluid.

L = σAT⁴

L

power radiated by a black bodyW

σ

Stefan–Boltzmann constantW m⁻² K⁻⁴

A

surface aream²

T

surface temperatureK

The fourth power is the point: doubling the absolute temperature multiplies the power by sixteen.

e = power radiated by the body / power radiated by a black body of the same area and temperature

e

emissivity, between 0 and 1 (a ratio, no unit)

A black body has e = 1, so a real surface radiates P = eσAT⁴.

λmax = 2.90 × 10⁻³ / T

λmax

wavelength at which the emitted power peaksm

T

surface temperatureK

The constant is in m K. Hotter bodies peak at shorter wavelengths, which is how a star's colour gives its temperature.

b = L / (4πd²)

b

apparent brightness, the intensity receivedW m⁻²

L

luminosity, the power emittedW

d

distance to the sourcem

An inverse square law, because the power spreads over a sphere of area 4πd². E.5 uses the same equation for stars.

α = total scattered power / total incident power

α

albedo, between 0 and 1 (a ratio, no unit)

The fraction of incoming radiation reflected away. Ice and cloud raise it; open ocean and forest lower it.

S = L / (4πd²)

S

intensity arriving on a surface facing the SunW m⁻²

L

luminosity of the SunW

d

distance from the Sunm

At the Earth's orbit this is the solar constant, about 1.36 × 10³ W m⁻².

average absorbed intensity = S(1 − α) / 4

S

solar intensity at that orbitW m⁻²

α

albedo

Not printed in the data booklet. The factor of four is the disc the planet presents, πr², spread over the whole surface it radiates from, 4πr².

P = eσAT⁴

P

power radiatedW

e

emissivity

σ

Stefan–Boltzmann constantW m⁻² K⁻⁴

A

surface aream²

T

average surface temperatureK

power absorbed = power radiated

T

the equilibrium temperature this fixesK

Not printed in the data booklet. Greenhouse gases — CO₂, H₂O, CH₄, N₂O — absorb outgoing infrared and re-emit it, so equilibrium is reached at a higher surface temperature.

n = N / NA

n

amount of substancemol

N

number of molecules

NA

Avogadro constantmol⁻¹

pV = nRT

p

pressurePa

V

volumem³

n

amount of substancemol

R

molar gas constantJ K⁻¹ mol⁻¹

T

absolute temperatureK

pV = NkBT

N

number of molecules

kB

Boltzmann constantJ K⁻¹

T

absolute temperatureK

The same statement counted in molecules rather than moles; kB = R / NA.

p₁V₁ / T₁ = p₂V₂ / T₂

p, V, T

pressure, volume and absolute temperature before and afterPa, m³, K

Not printed in the data booklet: it is pV = nRT with n fixed. Hold one quantity constant and it becomes Boyle's, Charles's or Gay-Lussac's law.

Ēk = (3/2)kBT

Ēk

average translational kinetic energy of one moleculeJ

kB

Boltzmann constantJ K⁻¹

T

absolute temperatureK

This is what absolute temperature measures — and why 0 K is the floor.

U = (3/2)nRT = (3/2)NkBT = (3/2)pV

U

internal energyJ

n

amount of substancemol

T

absolute temperatureK

N times the average molecular kinetic energy. An ideal gas has no intermolecular potential energy, so this is all of it. B.4 puts it into the first law.

vrms = √(3kBT / m) = √(3RT / M)

vrms

root-mean-square molecular speedm s⁻¹

m

mass of one moleculekg

M

molar masskg mol⁻¹

Not printed in the data booklet: it is Ēk = ½m vrms² set equal to (3/2)kBT. Lighter molecules move faster at the same temperature.

Q = ΔU + W

Q

energy transferred to the gas by heatingJ

ΔU

increase in internal energyJ

W

work done BY the gasJ

Higher level only. Note the sign convention: W here is work done by the gas, so an expansion makes it positive.

W = pΔV

W

work done by the gasJ

p

pressurePa

ΔV

change in volumem³

Higher level only. At constant pressure. In general the work is the area under the p–V curve.

ΔU = (3/2)nRΔT

ΔU

change in internal energyJ

n

amount of substancemol

ΔT

change in absolute temperatureK

Higher level only. No temperature change means no change in internal energy, which is what makes an isothermal process tractable.

pV = constant · W = nRT ln(V₂ / V₁)

W

work done by the gasJ

V₁, V₂

initial and final volumem³

T

the constant absolute temperatureK

Higher level only. ΔU = 0, so Q = W: everything supplied by heating leaves as work.

pV^(5/3) = constant

p

pressurePa

V

volumem³

Higher level only. Q = 0, so ΔU = −W: the gas cools as it expands because the work comes out of its internal energy.

ΔS = ΔQ / T

ΔS

change in entropyJ K⁻¹

ΔQ

energy transferred by heatingJ

T

absolute temperature at which it is transferredK

Higher level only. The second law: the entropy of an isolated system never decreases.

η = W / Qin = 1 − Qout / Qin

η

efficiency (a ratio, no unit)

W

useful work per cycleJ

Qin, Qout

energy taken from the hot reservoir and dumped to the cold oneJ

Higher level only.

ηCarnot = 1 − Tcold / Thot

ηCarnot

the highest efficiency any engine between the two reservoirs can reach

Tcold, Thot

absolute temperatures of the two reservoirsK

Higher level only. A ceiling set by the temperatures alone — no engineering gets past it.

I = Δq / Δt

RearrangedΔq = IΔt

I

currentA

Δq

charge passing a pointC

Δt

time takens

Conventional current runs from positive to negative; electrons drift the other way.

V = W / q

V

potential differenceV

W

work done on or by the chargeJ

q

charge movedC

R = V / I

RearrangedV = IR · I = V / R

R

resistanceΩ

V

potential differenceV

I

currentA

This defines resistance for any component. It is Ohm's law only where R stays constant, which a filament lamp does not.

R = ρL / A

ρ

resistivityΩ m

L

lengthm

A

cross-sectional aream²

I = nAvq

n

number density of charge carriersm⁻³

A

cross-sectional aream²

v

drift speedm s⁻¹

q

charge on each carrierC

P = IV = I²R = V² / R

P

power dissipatedW

I

currentA

V

potential differenceV

R

resistanceΩ

Pick the form that matches what is held constant — I²R for a fixed current, V²/R for a fixed supply.

Rs = R₁ + R₂ + …

Rs

combined resistanceΩ

1 / Rp = 1/R₁ + 1/R₂ + …

Rp

combined resistanceΩ

The combined resistance is always smaller than the smallest branch.

Σ I = 0 at a junction · Σ V = 0 around a loop

I

currents in and out of the junction, with signA

V

e.m.f.s and potential drops around the loop, with signV

Not printed in the data booklet as equations. The first is conservation of charge, the second conservation of energy.

ε = I(R + r)

RearrangedV = ε − Ir

ε

electromotive force of the cellV

I

current drawnA

R

external resistanceΩ

r

internal resistance of the cellΩ

The terminal p.d. V is what a voltmeter across the cell reads, and it falls below ε as soon as current flows.

a = −ω²x

a

accelerationm s⁻²

ω

angular frequencyrad s⁻¹

x

displacement from equilibriumm

The minus sign is the whole idea: the acceleration is proportional to the displacement and always points back towards equilibrium.

T = 1 / f

T

periods

f

frequencyHz

ω = 2π / T = 2πf

ω

angular frequencyrad s⁻¹

T

periods

f

frequencyHz

T = 2π√(m / k)

T

periods

m

masskg

k

spring constantN m⁻¹

Independent of the amplitude — which is what makes the motion isochronous.

T = 2π√(l / g)

T

periods

l

length of the pendulumm

g

gravitational field strengthm s⁻²

For small amplitudes only, where sin θ ≈ θ. The mass of the bob does not appear.

x = x₀ sin(ωt + φ) · x = x₀ cos(ωt + φ)

x₀

amplitudem

φ

phase anglerad

t

times

Higher level only. Which of the two you use is a choice of where t = 0 sits — the sine form starts at the equilibrium point, the cosine form at maximum displacement.

v = ωx₀ cos(ωt + φ)

v

velocitym s⁻¹

ω

angular frequencyrad s⁻¹

x₀

amplitudem

Higher level only. The maximum speed is ωx₀, reached at the equilibrium point.

v = ±ω√(x₀² − x²)

v

velocitym s⁻¹

x

displacementm

x₀

amplitudem

Higher level only. The form to use when the question gives a position rather than a time.

Ek = ½mω²(x₀² − x²)

Ek

kinetic energyJ

m

masskg

x₀

amplitudem

x

displacementm

Higher level only.

Ep = ½mω²x²

Ep

potential energyJ

m

masskg

x

displacementm

Higher level only.

ET = ½mω²x₀²

ET

total energyJ

m

masskg

ω

angular frequencyrad s⁻¹

x₀

amplitudem

Higher level only. Constant while the oscillator is undamped, and proportional to the square of the amplitude.

v = fλ

Rearrangedf = v / λ · λ = v / f

v

wave speedm s⁻¹

f

frequencyHz

λ

wavelengthm

When a wave crosses into a new medium the frequency is fixed by the source, so the speed and the wavelength change together.

T = 1 / f

T

periods

f

frequencyHz

One wavelength passes a fixed point in one period, which is where v = fλ comes from.

I ∝ A²

I

intensityW m⁻²

A

amplitudem

Intensity also falls as 1/d² from a point source — the same inverse square law as B.1.

n = c / v

n

refractive index of the medium (a ratio, no unit)

c

speed of light in a vacuumm s⁻¹

v

speed of light in the mediumm s⁻¹

n₁ sin θ₁ = n₂ sin θ₂ · sin θ₁ / sin θ₂ = v₁ / v₂ = n₂ / n₁

n₁, n₂

refractive indices of the two media

θ₁, θ₂

angles to the normal in each medium°

v₁, v₂

wave speeds in each mediumm s⁻¹

Angles are measured from the normal, not the surface.

sin θc = n₂ / n₁

θc

critical angle°

n₁

refractive index of the denser medium

n₂

refractive index of the less dense medium

Total internal reflection needs n₁ > n₂ and an angle of incidence beyond θc.

path difference = nλ (constructive) · (n + ½)λ (destructive)

n

any whole number

λ

wavelengthm

For sources in phase. Two sources must be coherent — same frequency, constant phase difference — for a steady pattern.

s = λD / d

s

fringe separationm

λ

wavelengthm

D

slit-to-screen distancem

d

slit separationm

Valid where D is far greater than d.

θ = λ / b

θ

angle to the first minimumrad

λ

wavelengthm

b

slit widthm

Higher level only. A narrower slit spreads the pattern wider. This envelope also modulates the double-slit fringes.

nλ = d sin θ

n

order of the maximum

λ

wavelengthm

d

spacing between adjacent slitsm

θ

angle of the maximum from the straight-through direction°

Higher level only. More slits make the maxima sharper, which is why a grating separates spectral lines better than a double slit.

θ = 1.22 λ / b

θ

smallest resolvable angular separationrad

λ

wavelengthm

b

diameter of the circular aperturem

Higher level only. Two point sources are just resolved when the central maximum of one falls on the first minimum of the other.

λn = 2L / n

λn

wavelength of the nth harmonicm

L

length of the string or pipem

n

harmonic number, 1, 2, 3, …

A string fixed at both ends has a node at each end; a pipe open at both ends has an antinode at each end. The wavelengths come out the same.

λn = 4L / n, with n odd

λn

wavelength of the nth harmonicm

L

length of the pipem

n

1, 3, 5, … only

Node at the closed end, antinode at the open end. Only the odd harmonics exist.

fn = v / λn

fn

frequency of the nth harmonicHz

v

wave speed in the string or the airm s⁻¹

Not a separate booklet equation: it is v = fλ applied to the standing wave. For two fixed ends this gives fn = nv/2L.

driving frequency = natural frequency at resonance

f₀

natural frequency of the systemHz

No equation in the course: resonance and the three degrees of damping — light, critical, heavy — are examined qualitatively and from resonance curves. Heavier damping lowers and broadens the peak.

Δf / f ≈ Δλ / λ ≈ v / c

Δf

change in observed frequencyHz

Δλ

change in observed wavelengthm

v

relative speed of source and observerm s⁻¹

c

wave speed — the speed of light for lightm s⁻¹

An approximation, good only for v much less than c. It is the form used for the redshift of light from a receding source.

f′ = f v / (v ∓ us)

f′

observed frequencyHz

f

frequency emitted by the sourceHz

v

speed of the wave in the mediumm s⁻¹

us

speed of the sourcem s⁻¹

Higher level only. Minus when the source approaches — a smaller denominator, so a higher pitch — and plus when it recedes.

f′ = f (v ± uo) / v

f′

observed frequencyHz

f

frequency emitted by the sourceHz

v

speed of the wave in the mediumm s⁻¹

uo

speed of the observerm s⁻¹

Higher level only. Plus when the observer moves towards the source. The two cases are not symmetric, because the medium fixes the wave speed.

F = GMm / r²

F

gravitational force between the two massesN

G

gravitational constantN m² kg⁻²

M, m

the two masseskg

r

separation of their centresm

For point masses, and for uniform spheres measured centre to centre.

g = F / m = GM / r²

g

gravitational field strengthN kg⁻¹, m s⁻²

M

mass producing the fieldkg

r

distance from its centrem

The force per unit mass, and numerically the same as the acceleration of free fall at that point.

v = √(GM / r)

v

orbital speedm s⁻¹

M

mass being orbitedkg

r

orbital radiusm

Not printed in the data booklet: it comes from setting the gravitational force equal to the centripetal force, GMm/r² = mv²/r. The orbiting mass cancels.

T² = 4π²r³ / (GM)

T

orbital periods

r

orbital radiusm

M

mass being orbitedkg

Not marked given here because the booklet's coverage of it has not been confirmed. It follows from GMm/r² = 4π²mr/T², and the useful form is that T² is proportional to r³ for everything orbiting the same central mass.

Vg = −GM / r

Vg

gravitational potentialJ kg⁻¹

M

mass producing the fieldkg

r

distance from its centrem

Higher level only. Negative because the zero is taken at infinity and gravity only attracts.

Ep = −GMm / r

Ep

potential energy of the pairJ

M, m

the two masseskg

r

their separationm

Higher level only. This replaces ΔEp = mgΔh once the field can no longer be treated as uniform.

g = −ΔVg / Δr

g

gravitational field strengthN kg⁻¹

ΔVg/Δr

potential gradientJ kg⁻¹ m⁻¹

Higher level only. The field is the gradient of the potential graph, and equipotential surfaces are everywhere perpendicular to the field lines.

W = mΔVg

W

work doneJ

m

mass movedkg

ΔVg

change in gravitational potentialJ kg⁻¹

Higher level only. No work is done moving along an equipotential.

vesc = √(2GM / r)

vesc

escape speedm s⁻¹

M

mass of the planet or starkg

r

distance from its centrem

Higher level only. The speed at which kinetic energy just cancels the negative potential energy, so the total energy reaches zero.

ET = −GMm / (2r)

ET

total energy of the orbiting bodyJ

M

mass being orbitedkg

m

mass in orbitkg

r

orbital radiusm

Higher level only. Kinetic plus potential. It is negative for a bound orbit, and a satellite losing energy to drag falls to a smaller r and speeds up.

F = kQ₁Q₂ / r², with k = 1 / (4πε₀)

F

force between the chargesN

k

Coulomb constantN m² C⁻²

Q₁, Q₂

the two point chargesC

ε₀

permittivity of free spaceC² N⁻¹ m⁻²

r

separationm

Same inverse square shape as gravitation, but this one can repel as well as attract.

E = F / q = kQ / r²

E

electric field strengthN C⁻¹, V m⁻¹

F

force on the test chargeN

q

test chargeC

Q

charge producing the fieldC

Force per unit positive charge, so the field direction is the direction of the force on a positive charge.

E = V / d

E

field strengthV m⁻¹

V

potential difference between the platesV

d

plate separationm

Ve = kQ / r

Ve

electric potentialV, J C⁻¹

Q

charge producing the fieldC

r

distance from itm

Higher level only. Unlike gravitational potential this can be positive or negative, following the sign of Q.

Ep = kQq / r

Ep

potential energy of the pair of chargesJ

Q, q

the two chargesC

r

their separationm

Higher level only. Positive for two like charges, which is why they fly apart when released.

E = −ΔVe / Δr

E

electric field strengthV m⁻¹

ΔVe/Δr

potential gradientV m⁻¹

Higher level only. In a uniform field this is exactly E = V/d.

W = qΔVe

W

work doneJ

q

charge movedC

ΔVe

change in electric potentialV

Higher level only.

ΔEk = qV

ΔEk

kinetic energy gainedJ, eV

q

chargeC

V

accelerating potential differenceV

This is what the electronvolt is: the energy an electron gains across one volt.

F = qvB sin θ

F

magnetic forceN

q

chargeC

v

speedm s⁻¹

B

magnetic flux densityT

θ

angle between the velocity and the field°

The force is perpendicular to both v and B, so it does no work and cannot change the speed — only the direction.

F = BIL sin θ

F

force on the conductorN

B

magnetic flux densityT

I

currentA

L

length of conductor in the fieldm

θ

angle between the conductor and the field°

Direction from the left-hand rule for conventional current.

r = mv / (qB)

r

radius of the pathm

m

mass of the particlekg

v

speedm s⁻¹

q

chargeC

B

magnetic flux densityT

Not printed in the data booklet: it is qvB = mv²/r rearranged, for a particle moving at right angles to the field.

F / L = μ₀ I₁ I₂ / (2πr)

F/L

force per unit lengthN m⁻¹

μ₀

permeability of free spaceT m A⁻¹

I₁, I₂

the two currentsA

r

separation of the wiresm

Currents in the same direction attract; opposite directions repel.

Φ = BA cos θ

Φ

magnetic fluxWb

B

magnetic flux densityT

A

area of the loopm²

θ

angle between the field and the normal to the area°

Higher level only. Note that θ is measured from the normal, so the flux is greatest when the field is perpendicular to the plane of the loop.

flux linkage = NΦ

N

number of turns in the coil

Φ

flux through one turnWb

Higher level only.

ε = −N ΔΦ / Δt

ε

induced electromotive forceV

N

number of turns

ΔΦ/Δt

rate of change of fluxWb s⁻¹

Higher level only. The magnitude is the rate of change of flux linkage; the minus sign is Lenz's law, which is conservation of energy — the induced current opposes the change that made it.

ε = BvL

ε

induced e.m.f.V

B

magnetic flux densityT

v

speed of the conductorm s⁻¹

L

length of the conductor in the fieldm

Higher level only. For a rod moving at right angles to a uniform field, sweeping out area as it goes.

ᴬZX · A = Z + N

A

nucleon number

Z

proton number

N

number of neutrons

X

chemical symbol of the element

Not printed in the data booklet — it is notation rather than an equation. Isotopes share Z and differ in N.

E = hf = hc / λ

E

difference between the two energy levelsJ, eV

h

Planck constantJ s

f

frequency of the photonHz

λ

wavelength of the photonm

Discrete energy levels are why emission and absorption spectra are lines rather than a continuum, and why the two are complementary.

En = −13.6 / n² eV

En

energy of the nth leveleV

n

principal quantum number, 1, 2, 3, …

Higher level only. Hydrogen only — a hydrogen-like ion of proton number Z has En = −13.6Z² / n² eV. The levels crowd together as n grows and reach zero at ionisation.

R = R₀ A^(1/3)

R

radius of the nucleusm

R₀

Fermi radiusm

A

nucleon number

Higher level only. R³ is proportional to A, so nuclear density is roughly the same for every nuclide.

qV = Ek, giving d = kQq / Ek

d

closest approach of the alpha particle to the nucleusm

Ek

initial kinetic energy of the alpha particleJ

Q, q

charges of the nucleus and the alpha particleC

Higher level only, and not printed in the data booklet: it is the kinetic energy converted entirely to electric potential energy. This is how Rutherford scattering puts an upper bound on nuclear size.

E = hf · p = h / λ

E

photon energyJ, eV

h

Planck constantJ s

f

frequencyHz

p

photon momentumkg m s⁻¹

Higher level only. A photon has momentum without having mass.

Emax = hf − Φ

Emax

maximum kinetic energy of an emitted electronJ, eV

hf

energy of the incident photonJ, eV

Φ

work function of the metalJ, eV

Higher level only. One photon, one electron — which is why intensity changes the number emitted and never their maximum energy.

Φ = hf₀

Φ

work functionJ, eV

f₀

threshold frequencyHz

Higher level only. Below f₀ nothing is emitted however bright the light is.

eVs = Emax

e

elementary chargeC

Vs

stopping voltageV

Emax

maximum kinetic energy of the photoelectronsJ

Higher level only. The p.d. that just stops the fastest electron, and the way Emax is measured.

λ = h / p

λ

de Broglie wavelengthm

h

Planck constantJ s

p

momentum of the particlekg m s⁻¹

Higher level only. Electron diffraction is the evidence: matter shows interference where the wavelength is comparable to the spacing it meets.

λf − λi = (h / (me c))(1 − cos θ)

λi, λf

wavelength of the photon before and after scatteringm

me

rest mass of the electronkg

θ

angle through which the photon is scattered°

Higher level only. The photon loses energy to the electron, so the scattered wavelength is always the longer one.

A decreases by 4 · Z decreases by 2

α

an alpha particle is a helium nucleus, ⁴₂He

Not printed in the data booklet: it is bookkeeping you write out as a nuclear equation, with nucleon number and proton number balancing on both sides.

β⁻: n → p + e⁻ + ν̄e · β⁺: p → n + e⁺ + νe

ν̄e

electron antineutrino

νe

electron neutrino

Not printed in the data booklet. The neutrino was proposed to account for the continuous beta energy spectrum — without it energy and momentum do not balance.

A unchanged · Z unchanged

γ

a photon emitted as the nucleus drops to a lower energy state

Not printed in the data booklet. It usually follows an alpha or beta decay that leaves the nucleus excited.

fraction remaining = (½)ⁿ

n

number of half-lives elapsed

Not printed in the data booklet: it is the definition of half-life applied repeatedly, and it answers most SL half-life questions without any logarithms.

Δm = Zmp + Nmn − mnucleus · E = Δmc²

Δm

mass defectkg, u

mp, mn

rest masses of a free proton and a free neutronkg, u

E

binding energy releasedJ, MeV

A nucleus weighs less than its parts; the shortfall is the energy that had to be supplied to pull it apart. Working in u and MeV c⁻² is faster — 1 u is 931.5 MeV c⁻².

binding energy per nucleon = binding energy / A

A

nucleon number

binding energy per nucleon

how tightly bound the nuclide isMeV

Not printed in the data booklet. The curve peaks around iron-56, which is why fusion releases energy below the peak and fission above it.

A = λN

A

activityBq

λ

decay constants⁻¹

N

number of nuclei not yet decayed

Higher level only. λ is the probability per unit time that any one nucleus decays.

N = N₀ e^(−λt) · A = A₀ e^(−λt)

N

nuclei remaining after time t

N₀

number present at t = 0

A₀

initial activityBq

t

times

Higher level only. Activity and count rate fall on the same exponential, since both are proportional to N.

λ = ln 2 / t½

λ

decay constants⁻¹

t½

half-lifes

Higher level only. ln 2 is 0.693.

E = Δmc² = c² × (total mass before − total mass after)

E

energy releasedJ, MeV

Δm

loss of mass in the reactionkg, u

c

speed of light in a vacuumm s⁻¹

The same mass–energy relation as E.3, read off the binding-energy-per-nucleon curve: splitting a heavy nucleus moves the fragments up towards the peak, and the difference comes out as energy. The rest of E.4 — chain reactions, moderators, control rods, enrichment — is qualitative.

E = Δmc²

E

energy releasedJ, MeV

Δm

loss of mass in the reactionkg, u

Light nuclei fusing move up the binding-energy curve. Fusion needs enormous temperature and pressure because the nuclei have to overcome their mutual electrostatic repulsion.

outward radiation pressure = inward gravitational pressure

equilibrium

what keeps a main sequence star at a stable radius

Not an equation in the course. A star leaves the main sequence when the fuel for the outward pressure runs out.

L = σAT⁴

L

luminosity, the total power radiatedW

σ

Stefan–Boltzmann constantW m⁻² K⁻⁴

A

surface area, 4πR² for a star of radius Rm²

T

surface temperatureK

The same law as B.1, applied to a star treated as a black body.

b = L / (4πd²)

b

apparent brightnessW m⁻²

L

luminosityW

d

distance to the starm

Measure b, know L, and you have the distance — the whole basis of the standard candle method.

λmax = 2.90 × 10⁻³ / T

λmax

wavelength of peak emissionm

T

surface temperatureK

Wien's law again, and the axis a Hertzsprung–Russell diagram is plotted against.

L ∝ M^3.5

L

luminosityW

M

mass of the starkg

For main sequence stars only. The steep power is why massive stars burn out fastest despite having more fuel.

d / parsec = 1 / (p / arcsecond)

d

distance to the starpc

p

parallax anglearcsecond

Measured against the Earth's orbit six months apart. The angles are tiny, so the method runs out for distant stars.