Solar radiation pressure: Worked examples — Physics deep dive

Before we start, here are the only symbols we will use, each in plain words:

The two working formulas, restated: $P_{rad} (r) = \frac{S ( r )}{c}, S (r) = S_{\oplus} (\frac{1 AU}{r})^{2}, S_{\oplus} = 1361 W/m^{2} .$ $F = P_{rad} A (1 + η) cos θ, a_{SRP} = \frac{F}{m} .$

The $(\frac{1 AU}{r})^{2}$ shrink factor comes from the Inverse-square law — the same power spread over a bigger sphere.

The scenario matrix

Every SRP problem is one (or a mix) of these cells. Each example below is tagged with the cell(s) it fills.

Cell	What varies	Degenerate / limit to watch
A. Base pressure	just $P = S / c$ at one distance	—
B. Reflectivity dial	$η = 0, \frac{1}{2}, 1$	grey surface between the extremes
C. Angle sweep	$cos θ$ for $θ = 0, 6 0^{\circ}, 9 0^{\circ}$	edge-on $θ = 9 0^{\circ} \Rightarrow F = 0$
D. Distance scaling	$1/ r^{2}$ near ( $0.39$ AU) and far ( $5.2$ AU)	limit $r \to \infty \Rightarrow P \to 0$
E. Acceleration & $A / m$	dense probe vs featherweight sail	limit $A / m \to 0$
F. Force balance (word problem)	SRP vs Sun's gravity	ratio independent of $r$
G. Exam twist	solve backwards for a design number	infer $A / m$ from a required $a$

Cells covered: A→Ex1, B→Ex2, C→Ex3, D→Ex4, E→Ex5, F→Ex6, G→Ex7. Ex8 combines B+C.

Example 1 — Cell A: the base pressure at one distance

Worked example Pressure on a black panel at Venus (0.723 AU)

Q: A perfectly black solar panel sits at Venus's distance, facing the Sun. Find $P_{rad}$ .

Forecast: Venus is closer than Earth, so guess: bigger than the $4.54 \times 1 0^{- 6} Pa$ Earth value. By how much?

Step 1. Intensity at Venus: $S = 1361 \times (1/0.723)^{2}$ . Why this step? Intensity obeys the Inverse-square law; closer means more watts per square metre. $S = 1361 \times 1.913 = 2604 W/m^{2} .$

Step 2. Convert to pressure: $P = S / c = 2604/ (2.998 \times 1 0^{8})$ . Why this step? Each joule of light carries $1/ c$ of momentum (Photon momentum and relativity); momentum-per-second-per-area is pressure. Black surface ⇒ no reflection factor. $P_{rad} = 8.69 \times 1 0^{- 6} Pa .$

Verify: Ratio to Earth $= (1/0.723)^{2} = 1.91$ , and $4.54 \times 1 0^{- 6} \times 1.91 = 8.68 \times 1 0^{- 6}$ ✓. Units: $(W/m^{2}) / (m/s) = W\cdots / m^{3} = J/m^{3} = N/m^{2} = Pa$ ✓.

Example 2 — Cell B: the reflectivity dial ( $η = 0, \frac{1}{2}, 1$ )

Worked example Same panel, three coatings

Q: At 1 AU, face-on ( $θ = 0$ ), area $A = 10 m^{2}$ . Find the force for a black panel ( $η = 0$ ), a grey panel ( $η = 0.5$ ), and a mirror ( $η = 1$ ).

Forecast: Black is baseline; the mirror should be exactly double; grey lands in the middle at $\times 1.5$ .

Step 1. Base pressure at 1 AU: $P_{rad} = 4.54 \times 1 0^{- 6} Pa$ . Why this step? It's our reference number; every case just multiplies it.

Step 2. Force $= P_{rad} A (1 + η)$ , with $cos θ = 1$ . Why this step? Face-on kills the angle factor. The $(1 + η)$ counts recoil: absorb takes $p$ , reflect returns $- p$ for a change of $2 p$ . $F_{0} = 4.54 \times 1 0^{- 6} \times 10 \times 1 = 4.54 \times 1 0^{- 5} N,$ $F_{0.5} = 4.54 \times 1 0^{- 6} \times 10 \times 1.5 = 6.81 \times 1 0^{- 5} N,$ $F_{1} = 4.54 \times 1 0^{- 6} \times 10 \times 2 = 9.08 \times 1 0^{- 5} N .$

Verify: $F_{1} / F_{0} = 2$ exactly, $F_{0.5} / F_{0} = 1.5$ exactly ✓. The mirror doubling is the recoil rule of Section 2 in the parent note.

Example 3 — Cell C: angle sweep, including the edge-on limit

Step 1. Head-on force: $F = P_{rad} A (1 + η) cos 0^{\circ} = 4.54 \times 1 0^{- 6} \times 1000 \times 2 \times 1$ . Why this step? This is our maximum; every tilt scales it by $cos θ$ . $F (0^{\circ}) = 9.08 \times 1 0^{- 3} N \approx 9.1 mN .$

Step 2. At $6 0^{\circ}$ : multiply by $cos 6 0^{\circ} = 0.5$ . Why this step? $cos θ$ measures how much of the flat area the Sun actually "sees" — the projected area shrinks as you tilt. $F (6 0^{\circ}) = 9.08 \times 1 0^{- 3} \times 0.5 = 4.54 \times 1 0^{- 3} N .$

Step 3. At $9 0^{\circ}$ (edge-on): $cos 9 0^{\circ} = 0$ . Why this step? Edge-on, the projected area is a line — zero area, zero photons intercepted, zero force. This is the degenerate limit you must always name. $F (9 0^{\circ}) = 0 N .$

Verify: $F (6 0^{\circ}) / F (0^{\circ}) = cos 6 0^{\circ} = 0.5$ ✓, and $F (9 0^{\circ}) = 0$ matches "an edge catches nothing." Note this simple $cos θ$ model gives the along-Sun component only; the fuller mirror model gives a $cos^{2} θ$ normal force (see the parent's mistake box).

Example 4 — Cell D: distance scaling from Mercury to Jupiter

Worked example The

$1/ r^{2}$ span, plus the $r \to \infty$ limit Q: For the same spacecraft, compare $P_{rad}$ at Mercury (0.39 AU) and Jupiter (5.2 AU) to Earth. What happens as $r \to \infty$ ?

Forecast: Mercury much stronger (it's close), Jupiter much weaker (it's far). Deep space → nothing.

Step 1. Mercury factor $= (1/0.39)^{2}$ . Why this step? $P_{rad} \propto S \propto 1/ r^{2}$ ; square the distance ratio. $(1/0.39)^{2} = 6.57 \Rightarrow P = 4.54 \times 1 0^{- 6} \times 6.57 = 2.98 \times 1 0^{- 5} Pa .$

Step 2. Jupiter factor $= (1/5.2)^{2}$ . Why this step? Same law, far side. $(1/5.2)^{2} = 0.0370 \Rightarrow P = 4.54 \times 1 0^{- 6} \times 0.0370 = 1.68 \times 1 0^{- 7} Pa .$

Step 3 (limit). As $r \to \infty$ , $1/ r^{2} \to 0$ , so $P_{rad} \to 0$ . Why this step? You must state the limiting behaviour: SRP fades to nothing in the outer dark, which is why sails are useless past the outer planets.

Verify: Mercury/Jupiter pressure ratio $= (5.2/0.39)^{2} = 177.7$ , and $2.98 \times 1 0^{- 5} /1.68 \times 1 0^{- 7} = 177.4$ ✓ (rounding). Consistent with the Inverse-square law.

Example 5 — Cell E: acceleration and the $A / m$ extremes

Worked example Featherweight sail vs dense brick

Q: At 1 AU, mirror ( $η = 1$ ), face-on. Compare (a) a sail: $A = 1000 m^{2}$ , $m = 10 kg$ ; (b) a solid tungsten cube of side $0.1 m$ , $ρ = 19300 kg/m^{3}$ .

Forecast: The sail's $A / m$ is huge → real acceleration. The brick's $A / m$ is tiny → negligible.

Step 1. Sail acceleration: $a = P_{rad} (1 + η) (A / m)$ . Why this step? Orbits respond to acceleration, and mass cancels only in gravity, not here — SRP acceleration lives or dies by $A / m$ . $a_{sail} = 4.54 \times 1 0^{- 6} \times 2 \times (1000/10) = 9.08 \times 1 0^{- 4} m/s^{2} .$

Step 2. Brick numbers: mass $= ρ V = 19300 \times (0.1)^{3} = 19.3 kg$ ; face area $A = (0.1)^{2} = 0.01 m^{2}$ ; so $A / m = 0.01/19.3 = 5.18 \times 1 0^{- 4} m^{2} / kg$ . Why this step? We need $A / m$ for the brick — it's the whole story. $a_{brick} = 4.54 \times 1 0^{- 6} \times 2 \times 5.18 \times 1 0^{- 4} = 4.70 \times 1 0^{- 9} m/s^{2} .$

Step 3 (limit). As $A / m \to 0$ (a very dense, compact body), $a_{SRP} \to 0$ . Why this step? Name the degenerate case: a heavy dense probe barely notices SRP.

Verify: Ratio $a_{sail} / a_{brick} = (1000/10) / (5.18 \times 1 0^{- 4}) = 100/5.18 \times 1 0^{- 4} = 1.93 \times 1 0^{5}$ — the sail accelerates ~ $200, 000 \times$ harder ✓. This is exactly why Solar sails work and dense probes ignore SRP.

Example 6 — Cell F: word problem, SRP vs gravity balance

Worked example What area-to-mass makes SRP cancel the Sun's gravity?

Q: For a black spacecraft ( $η = 0$ ), what $A / m$ makes the outward SRP acceleration exactly equal the Sun's inward gravitational acceleration? (This defines a "solar-static" sail.) Sun mass $M_{⊙} = 1.989 \times 1 0^{30} kg$ , $G = 6.674 \times 1 0^{- 11}$ .

Forecast: Both accelerations go as $1/ r^{2}$ , so the answer should be a single number, independent of distance — this is the punchline from the parent's mistake box.

Step 1. SRP acceleration: $a_{SRP} = \frac{S _{\oplus} ( 1 AU / r ) ^{2}}{c} (1 + η) \frac{A}{m}$ . Why this step? Write both sides at general $r$ so we can watch $r$ cancel.

Step 2. Gravity: $a_{g} = \frac{G M _{⊙}}{r ^{2}}$ . Set $a_{SRP} = a_{g}$ : $\frac{S _{\oplus} ( 1 AU ) ^{2}}{c r ^{2}} (1 + η) \frac{A}{m} = \frac{G M _{⊙}}{r ^{2}} .$ Why this step? Both sides carry $1/ r^{2}$ — it divides out, proving distance-independence.

Step 3. Solve for $A / m$ with $η = 0$ : $\frac{A}{m} = \frac{G M _{⊙} c}{S _{\oplus} ( 1 AU ) ^{2} ( 1 + η )} .$ Plug numbers: numerator $= 6.674 \times 1 0^{- 11} \times 1.989 \times 1 0^{30} \times 2.998 \times 1 0^{8} = 3.979 \times 1 0^{28}$ . Denominator $= 1361 \times (1.496 \times 1 0^{11})^{2} \times 1 = 3.046 \times 1 0^{25}$ . $\frac{A}{m} = \frac{3.979 \times 1 0 ^{28}}{3.046 \times 1 0 ^{25}} = 1306 m^{2} / kg .$

Verify: $r$ vanished ✓ (the ratio is universal, as promised). Units: $\frac{( m ^{3} kg ^{- 1} s ^{- 2} ) ( kg ) ( m/s )}{( W/m ^{2} ) ( m ^{2} )} = \frac{m ^{4} s ^{- 3}}{W} = \frac{m ^{4} s ^{- 3}}{kg\cdotm ^{2} s ^{- 3}} = m^{2} / kg$ ✓. A required $A / m \approx 1300 m^{2} / kg$ is enormous — real sails only partly offset gravity. This links to Orbital perturbations.

Example 7 — Cell G: exam twist, solve backwards for a design number

Worked example Mission requires a set acceleration — find the sail area

Q: A mission needs $a_{SRP} = 1.0 \times 1 0^{- 4} m/s^{2}$ at 1 AU from a mirror sail ( $η = 1$ ) attached to a $50 kg$ payload. What sail area $A$ is required (ignore sail mass)?

Forecast: Rearrange the acceleration formula for $A$ ; expect hundreds of square metres.

Step 1. Start from $a_{SRP} = P_{rad} (1 + η) A / m$ and solve for $A$ : $A = \frac{a _{SRP} m}{P _{rad} ( 1 + η )} .$ Why this step? Exams often give the target performance and ask for the design — invert the formula.

Step 2. Plug in $P_{rad} = 4.54 \times 1 0^{- 6}$ , $η = 1$ , $m = 50$ : $A = \frac{1.0 \times 1 0 ^{- 4} \times 50}{4.54 \times 1 0 ^{- 6} \times 2} = \frac{5.0 \times 1 0 ^{- 3}}{9.08 \times 1 0 ^{- 6}} = 550.7 m^{2} .$ Why this step? Direct substitution.

Verify (plug back): $a = 4.54 \times 1 0^{- 6} \times 2 \times (550.7/50) = 4.54 \times 1 0^{- 6} \times 2 \times 11.01 = 1.0 \times 1 0^{- 4} m/s^{2}$ ✓ — recovers the target. Units: $\frac{( m/s ^{2} ) ( kg )}{Pa} = \frac{m\cdotkg\cdots ^{- 2}}{N/m ^{2}} = \frac{N}{N/m ^{2}} = m^{2}$ ✓.

Example 8 — Cells B + C combined: grey sail at an angle

Recall

Recall Test yourself (hide the answers)

Edge-on ( $θ = 9 0^{\circ}$ ) force on any flat plate? ::: exactly zero — projected area is a line As $r \to \infty$ , what happens to $P_{rad}$ ? ::: it goes to $0$ like $1/ r^{2}$ Why does SRP-vs-gravity give a distance-independent $A / m$ ? ::: both accelerations carry $1/ r^{2}$ , which cancels To hit a target acceleration, solve for area with which rearrangement? ::: $A = a m / [P_{rad} (1 + η)]$ A grey sail's two correction factors are? ::: $(1 + η)$ for recoil and $cos θ$ for projected area

Connections

Solar radiation pressure — parent note with the derivations.
Solar sails — Examples 3, 5, 7 are sail design in miniature.
Orbital perturbations — Example 6's balance sets the perturbation scale.
Inverse-square law — Examples 1 and 4 lean on it.
Atmospheric drag — the competing force at low altitude.
Yarkovsky effect — thermal cousin of these pushes on asteroids.
Photon momentum and relativity — the $p = E / c$ behind every $S / c$ .