Exercises — Solar radiation pressure

WHAT: A photon carries momentum $p=E/c$. WHY: Relativity gives $E^2=(pc{)}^2+(m{c}^2{)}^2$. A photon has mass $m=0$, so the second term vanishes and $E=pc$, i.e. $p=E/c$. The old rule $p=mv$ is only the low-speed rule; it is not the definition of momentum. Massless light still carries momentum because momentum is fundamentally about energy and direction of flow, not about mass. See Photon momentum and relativity.

WHAT: Spread the power over a sphere of radius $r=1\text{AU}$. WHY the sphere? Energy is conserved, so all the Sun's power passes through every sphere centred on the Sun (see Inverse-square law). Sphere area is $4\pi r^2$. $S=\frac{L_{\odot }}{4\pi r^2}=\frac{3.828\times 10^{26}}{4\pi (1.496\times 10^{11}{)}^2}.$ Denominator: $4\pi (1.496\times 10^{11}{)}^2=2.812\times 10^{23}{\text{m}}^2$. $S=\frac{3.828\times 10^{26}}{2.812\times 10^{23}}\approx 1361{\text{W/m}}^2.✓$

WHAT: Convert energy flux to momentum flux by dividing by $c$. WHY: Each joule of light carries $1/c$ of momentum, so momentum-per-second-per-area (= pressure) is $S/c$. $P_{\text{rad}}=\frac{S}{c}=\frac{1361}{2.998\times 10^8}\approx 4.54\times 10^{-6}\text{Pa}.$

WHAT: Use $F=P_{\text{rad}}A(1+\eta )\cos \theta$ with $\eta =1$ (mirror), $\theta =0$ so $\cos \theta =1$. WHY the $(1+\eta )$: An absorber takes momentum $p$; a mirror sends the photon back with momentum $-p$, so its momentum change is $p-(-p)=2p$. With $\eta =1$ that is the factor $2$. $F=(4.54\times 10^{-6})(1000)(2)(1)\approx 9.1\times 10^{-3}\text{N}\approx 9\text{mN}.$ Tiny, but never switches off — over months this builds huge $\Delta v$ (see Solar sails).

WHAT: $a_{\text{SRP}}=\frac{P_{\text{rad}}(1+\eta )A}{m}$ with $\eta =0$ (absorber), so $(1+\eta )=1$. WHY divide by $m$: orbits respond to acceleration, and $a=F/m$. $a=\frac{(4.54\times 10^{-6})(1)(0.20)}{0.50}=1.82\times 10^{-6}{\text{m/s}}^2.$

WHAT: $a_{\text{SRP}}\propto S\propto 1/r^2$ (the $A/m$ and $(1+\eta )$ factors are unchanged for the same craft). WHY the square: intensity spreads over $4\pi r^2$, so it falls as $1/r^2$ (Inverse-square law). $\frac{a_{\text{Mars}}}{a_{\text{Earth}}}={\left(\frac{r_{\text{Earth}}}{r_{\text{Mars}}}\right)}^2={\left(\frac{1}{1.52}\right)}^2=0.433\approx 43\%.$

WHAT: Write both accelerations. $a_{\text{SRP}}=\frac{S(r)}{c}\frac{(1+\eta )A}{m}=\frac{L_{\odot }}{4\pi r^{2}c}\frac{(1+\eta )A}{m},a_{\text{grav}}=\frac{G{M}_{\odot }}{r^2}.$ WHY the $r^2$ cancels: both carry a lone $1/r^2$. Divide: $\frac{a_{\text{SRP}}}{a_{\text{grav}}}=\frac{L_{\odot }(1+\eta )}{4\pi cG{M}_{\odot }}\cdot \frac{A}{m}.$ Every symbol except $A/m$ (and the fixed reflectivity $\eta$) is a universal constant. So the ratio is distance-independent but proportional to $A/m$ — a feather-light solar sail feels SRP strongly, a dense probe barely. This is why SRP is not "just extra gravity"; see Orbital perturbations.

WHAT: Set the ratio $=1$ and solve for $A/m$. $\frac{A}{m}=\frac{4\pi cG{M}_{\odot }}{L_{\odot }(1+\eta )}.$ Numerator: $4\pi (2.998\times 10^8)(1.327\times 10^{20})=5.000\times 10^{29}$. Denominator ($\eta =0$): $3.828\times 10^{26}\times 1=3.828\times 10^{26}$. $\frac{A}{m}=\frac{5.000\times 10^{29}}{3.828\times 10^{26}}\approx 1.306\times 10^3{\text{m}}^2/\text{kg}\approx 1300{\text{m}}^2/\text{kg}.$ A staggering $1300$ square metres per kilogram — far beyond any real spacecraft, which is why SRP perturbs orbits but never fully cancels the Sun's pull for ordinary craft.

WHAT: $a=F/m$, then $\Delta v=at$ (constant acceleration). WHY constant-a is fine here: over one year at 1 AU the intensity barely changes, so treating $a$ as constant is a good first estimate. $a=\frac{9.08\times 10^{-3}}{100}=9.08\times 10^{-5}{\text{m/s}}^2.$ $\Delta v=at=(9.08\times 10^{-5})(3.156\times 10^7)\approx 2.87\times 10^3\text{m/s}\approx 2.9\text{km/s}.$ A 9 mN whisper becomes nearly 3 km/s of velocity change in a year — that is the whole point of Solar sails: small force, no fuel, enormous integrated effect.

(a) WHAT: Compare magnitudes. $a_{\text{drag}}/a_{\text{SRP}}=2.0\times 10^{-5}/4.5\times 10^{-6}\approx 4.4$. Drag is about $4.4\times$ stronger, so drag dominates in LEO. (b) WHY altitude flips it: Drag depends on air density, which falls off exponentially with altitude, while SRP is essentially constant with altitude (the Sun's distance barely changes). So climb high enough and drag collapses while SRP holds steady — SRP overtakes drag somewhere in the upper thousands of km. See Atmospheric drag for the density model.

(a) WHAT: $p=E/c=(hc/\lambda )/c=h/\lambda$. $p=\frac{6.626\times 10^{-34}}{500\times 10^{-9}}=1.325\times 10^{-27}\text{kg⋅m/s}.$ (b) WHY: Force = momentum delivered per second = (rate $N$) $\times p$. Set $=1\text{N}$: $N=\frac{F}{p}=\frac{1}{1.325\times 10^{-27}}\approx 7.5\times 10^{26}\text{photons/s}.$ Nearly a trillion trillion photons per second for one newton — one photon really is nothing; the flood is everything.

• First $\cos \theta$ — projected area. The tilted plate intercepts only $A\cos \theta$ of the beam (Problem 2.2 logic).
• Second $\cos \theta$ — direction of momentum change. For a mirror the reflected photons leave symmetrically about the normal, so the net momentum kick points along the normal. Only the component of the incoming momentum along the normal reverses, and that component is the full momentum times $\cos \theta$. Multiply the two effects → ${\cos }^2\theta$.

Look at the figure: the blue arrow (incoming momentum) splits into a normal part (length $\propto \cos \theta$) and a tangential part; only the normal part reverses on reflection, giving the green net force $\propto {\cos }^2\theta$.

(b) Numbers (drop the common factor $2P_{\text{rad}}A$; compare ${\cos }^2\theta$ vs $\cos \theta$):

At $0^{\circ }$ they agree; as you tilt, the rigorous model gives a smaller normal force. A sail designer who used the simple model at $60^{\circ }$ would over-predict thrust by a factor of two.

Photon momentum ::: $p=E/c=h/\lambda$ Base pressure at 1 AU ::: $P_0=S_0/c\approx 4.54\times 10^{-6}\text{Pa}$ Simple flat-plate force ::: $F=P_{\text{rad}}A(1+\eta )\cos \theta$ Rigorous mirror normal force ::: $F_n=2P_{\text{rad}}A{\cos }^2\theta$ Why SRP/gravity is distance-independent ::: both $\propto 1/r^2$; ratio $\propto A/m$ Optimal solar-sail tack angle ::: $\theta =\arctan (1/\sqrt{2})\approx 35.3^{\circ }$