Exercises — Solar radiation pressure

WHAT: Ek photon momentum $p=E/c$ carry karta hai. WHY: Relativity deta hai $E^2=(pc{)}^2+(m{c}^2{)}^2$. Photon ka mass $m=0$ hota hai, toh doosra term vanish ho jaata hai aur $E=pc$ ho jaata hai, yaani $p=E/c$. Purana rule $p=mv$ sirf low-speed ke liye hai; yeh momentum ki definition nahi hai. Massless light phir bhi momentum carry karti hai kyunki momentum fundamentally energy aur flow ki direction ke baare mein hai, mass ke baare mein nahi. Dekho Photon momentum and relativity.

WHAT: Power ko radius $r=1\text{AU}$ wale sphere par spread karo. WHY sphere? Energy conserved hoti hai, isliye Sun ki saari power har us sphere se pass karti hai jo Sun ke centre par centred ho (dekho Inverse-square law). Sphere ka area $4\pi r^2$ hota hai. $S=\frac{L_{\odot }}{4\pi r^2}=\frac{3.828\times 10^{26}}{4\pi (1.496\times 10^{11}{)}^2}.$ 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: Energy flux ko $c$ se divide karke momentum flux mein convert karo. WHY: Light ka har joule $1/c$ momentum carry karta hai, toh momentum-per-second-per-area (= pressure) $S/c$ hoti hai. $P_{\text{rad}}=\frac{S}{c}=\frac{1361}{2.998\times 10^8}\approx 4.54\times 10^{-6}\text{Pa}.$

WHAT: $F=P_{\text{rad}}A(1+\eta )\cos \theta$ use karo jahan $\eta =1$ (mirror), $\theta =0$ toh $\cos \theta =1$. WHY $(1+\eta )$: Absorber momentum $p$ leta hai; mirror photon ko wapas bhejta hai momentum $-p$ ke saath, toh uska momentum change $p-(-p)=2p$ hota hai. $\eta =1$ ke saath yeh factor $2$ banta hai. $F=(4.54\times 10^{-6})(1000)(2)(1)\approx 9.1\times 10^{-3}\text{N}\approx 9\text{mN}.$ Chhota hai, lekin kabhi band nahi hota — mahino mein yeh bahut bada $\Delta v$ build kar leta hai (dekho Solar sails).

WHAT: $a_{\text{SRP}}=\frac{P_{\text{rad}}(1+\eta )A}{m}$ jahan $\eta =0$ (absorber), toh $(1+\eta )=1$. WHY $m$ se divide karo: orbits acceleration par respond karti hain, aur $a=F/m$. $a=\frac{(4.54\times 10^{-6})(1)(0.20)}{0.50}=1.82\times 10^{-6}{\text{m/s}}^2.$

WHAT: $a_{\text{SRP}}\propto S\propto 1/r^2$ (usi craft ke liye $A/m$ aur $(1+\eta )$ factors unchanged hain). WHY square: intensity $4\pi r^2$ par spread hoti hai, toh yeh $1/r^2$ ki tarah girta hai (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: Dono accelerations likho. $a_{\text{SRP}}=\frac{S(r)}{c}\frac{(1+\eta )A}{m}=\frac{L_{\odot }}{4\pi r^{2}c}\frac{(1+\eta )A}{m},a_{\text{grav}}=\frac{G{M}_{\odot }}{r^2}.$ WHY $r^2$ cancel ho jaata hai: dono mein ek aakela $1/r^2$ hai. Divide karo: $\frac{a_{\text{SRP}}}{a_{\text{grav}}}=\frac{L_{\odot }(1+\eta )}{4\pi cG{M}_{\odot }}\cdot \frac{A}{m}.$ $A/m$ (aur fixed reflectivity $\eta$) ke alaawa har symbol ek universal constant hai. Toh ratio distance-independent hai lekin $A/m$ ke proportional hai — ek feather-light solar sail SRP strongly feel karta hai, ek dense probe barely. Isliye SRP "sirf extra gravity" nahi hai; dekho Orbital perturbations.

WHAT: Ratio $=1$ set karo aur $A/m$ solve karo. $\frac{A}{m}=\frac{4\pi cG{M}_{\odot }}{L_{\odot }(1+\eta )}.$ Numerator: $4\pi (2.998\times 10^8)(1.327\times 10^{20})=5.000\times 10^{29}$. Denominator ($\eta =0$): $3.828\times 10^{26}\times 1=3.828\times 10^{26}$. $\frac{A}{m}=\frac{5.000\times 10^{29}}{3.828\times 10^{26}}\approx 1.306\times 10^3{\text{m}}^2/\text{kg}\approx 1300{\text{m}}^2/\text{kg}.$ Ek shocking $1300$ square metres per kilogram — kisi bhi real spacecraft se bahut zyada, isliye SRP orbits ko perturb karta hai lekin ordinary craft ke liye Sun ki pull ko kabhi puri tarah cancel nahi kar sakta.

WHAT: $a=F/m$, phir $\Delta v=at$ (constant acceleration). WHY constant-a theek hai: 1 AU par ek year mein intensity barely badlti hai, toh $a$ ko constant treat karna ek achha first estimate hai. $a=\frac{9.08\times 10^{-3}}{100}=9.08\times 10^{-5}{\text{m/s}}^2.$ $\Delta v=at=(9.08\times 10^{-5})(3.156\times 10^7)\approx 2.87\times 10^3\text{m/s}\approx 2.9\text{km/s}.$ 9 mN ki whisper ek saal mein almost 3 km/s velocity change ban jaati hai — Solar sails ka yahi point hai: chhoti force, koi fuel nahi, enormous integrated effect.

(a) WHAT: Magnitudes compare karo. $a_{\text{drag}}/a_{\text{SRP}}=2.0\times 10^{-5}/4.5\times 10^{-6}\approx 4.4$. Drag almost $4.4\times$ zyada strong hai, toh LEO mein drag dominate karta hai. (b) WHY altitude flip karta hai: Drag air density par depend karta hai, jo altitude ke saath exponentially fall karti hai, jabki SRP altitude ke saath essentially constant hai (Sun ki distance barely badlti hai). Toh kaafi upar jao aur drag collapse kar jaata hai jabki SRP steady rehta hai — SRP kahin upper thousands of km mein drag ko overtake kar leta hai. Density model ke liye dekho Atmospheric drag.

(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 $N$) $\times p$. $=1\text{N}$ set karo: $N=\frac{F}{p}=\frac{1}{1.325\times 10^{-27}}\approx 7.5\times 10^{26}\text{photons/s}.$ Ek newton ke liye almost ek trillion trillion photons per second — ek photon sach mein kuch nahi hai; barsat sab kuch hai.

• Pehla $\cos \theta$ — projected area. Tilted plate beam ka sirf $A\cos \theta$ intercept karti hai (Problem 2.2 logic).
• Doosra $\cos \theta$ — momentum change ki direction. Mirror ke liye reflected photons normal ke baare mein symmetrically leave karte hain, toh net momentum kick normal ke along point karta hai. Incoming momentum ka sirf woh component normal ke along reverse hota hai, aur woh component full momentum times $\cos \theta$ hai. Dono effects multiply karo → ${\cos }^2\theta$.

Figure dekho: blue arrow (incoming momentum) normal part (length $\propto \cos \theta$) aur tangential part mein split hota hai; sirf normal part reflection par reverse hota hai, green net force $\propto {\cos }^2\theta$ deta hai.

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

$0^{\circ }$ par dono agree karte hain; jaise tum tilt karo, rigorous model chhota normal force deta hai. Ek sail designer jo $60^{\circ }$ par simple model use kare, thrust do baar over-predict karta.

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 ::: $F_n=2P_{\text{rad}}A{\cos }^2\theta$ SRP/gravity distance-independent kyun hai ::: dono $\propto 1/r^2$; ratio $\propto A/m$ Optimal solar-sail tack angle ::: $\theta =\arctan (1/\sqrt{2})\approx 35.3^{\circ }$