Solar radiation pressure: Worked examples — Hinglish deep dive

3.2.35 · D3 · Physics › Orbital Mechanics & Astrodynamics › Solar radiation pressure

Shuru karne se pehle, ye hain wo sirf wahi symbols jo hum use karenge, har ek seedhi baat mein:

Definition Toolkit (har symbol earn kiya hua)

$S (r)$ — intensity: kitne watts ki sunlight ek square metre se cross karti hai distance $r$ pe. Units $W/m^{2}$ .
$c$ — speed of light $= 2.998 \times 1 0^{8} m/s$ . Energy se momentum ka conversion factor.
$P_{rad} = S / c$ — base radiation pressure ek black (fully absorbing) surface pe jo Sun ki taraf face kar rahi ho. Units $Pa = N/m^{2}$ .
$η$ — reflectivity: ek dial $0$ (black, sab absorb karta hai) se $1$ (perfect mirror, sab bounce karta hai) tak. Koi units nahi.
$θ$ — incidence angle: surface ke normal (surface se seedha bahar nikla hua arrow) aur Sun ki taraf wapas jaane wali direction ke beech ka angle. Jab surface seedha Sun ki taraf face kare, $θ = 0$ .
$A$ — plate ka flat area. $m$ — uska mass. $A / m$ — area-to-mass ratio.
$AU$ — astronomical unit, Earth–Sun distance $\approx 1.496 \times 1 0^{11} m$ .

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

$(\frac{1 AU}{r})^{2}$ shrink factor Inverse-square law se aata hai — same power ek bade sphere par spread ho jaati hai.

Scenario matrix

Har SRP problem in cells mein se ek (ya mix) hoti hai. Neeche har example ko us cell(s) ke saath tag kiya gaya hai jo wo fill karta hai.

Cell	Kya vary karta hai	Degenerate / limit dhyaan rakhne wali
A. Base pressure	sirf $P = S / c$ ek distance pe	—
B. Reflectivity dial	$η = 0, \frac{1}{2}, 1$	extremes ke beech grey surface
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) aur 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 ki gravity	ratio $r$ se independent
G. Exam twist	ulta solve karo ek design number ke liye	$A / m$ infer karo ek required $a$ se

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

Example 1 — Cell A: ek distance pe base pressure

Worked example Venus pe ek black panel par pressure (0.723 AU)

Q: Ek perfectly black solar panel Venus ki distance pe Sun ki taraf face karke baitha hai. $P_{rad}$ nikalo.

Forecast: Venus Earth se zyada close hai, toh guess karo: Earth ki value $4.54 \times 1 0^{- 6} Pa$ se bada hoga. Kitna zyada?

Step 1. Venus pe intensity: $S = 1361 \times (1/0.723)^{2}$ . Ye step kyun? Intensity Inverse-square law follow karti hai; closer matlab zyada watts per square metre. $S = 1361 \times 1.913 = 2604 W/m^{2} .$

Step 2. Pressure mein convert karo: $P = S / c = 2604/ (2.998 \times 1 0^{8})$ . Ye step kyun? Light ka har joule $1/ c$ momentum carry karta hai (Photon momentum and relativity); momentum-per-second-per-area hi pressure hai. Black surface ⇒ koi reflection factor nahi. $P_{rad} = 8.69 \times 1 0^{- 6} Pa .$

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

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

Worked example Same panel, teen coatings

Q: 1 AU pe, face-on ( $θ = 0$ ), area $A = 10 m^{2}$ . Teen cases ke liye force nikalo: black panel ( $η = 0$ ), grey panel ( $η = 0.5$ ), aur mirror ( $η = 1$ ).

Forecast: Black baseline hai; mirror bilkul double hona chahiye; grey beech mein $\times 1.5$ pe aayega.

Step 1. 1 AU pe base pressure: $P_{rad} = 4.54 \times 1 0^{- 6} Pa$ . Ye step kyun? Ye hamara reference number hai; har case bas ise multiply karta hai.

Step 2. Force $= P_{rad} A (1 + η)$ , with $cos θ = 1$ . Ye step kyun? Face-on angle factor ko khatam kar deta hai. $(1 + η)$ recoil count karta hai: absorb $p$ leta hai, reflect $- p$ wapas karta hai $2 p$ ka change hota hai. $F_{0} = 4.54 \times 1 0^{- 6} \times 10 \times 1 = 4.54 \times 1 0^{- 5} N,$ $F_{0.5} = 4.54 \times 1 0^{- 6} \times 10 \times 1.5 = 6.81 \times 1 0^{- 5} N,$ $F_{1} = 4.54 \times 1 0^{- 6} \times 10 \times 2 = 9.08 \times 1 0^{- 5} N .$

Verify: $F_{1} / F_{0} = 2$ exactly, $F_{0.5} / F_{0} = 1.5$ exactly ✓. Mirror doubling parent note ke Section 2 ka recoil rule hai.

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

Step 1. Head-on force: $F = P_{rad} A (1 + η) cos 0^{\circ} = 4.54 \times 1 0^{- 6} \times 1000 \times 2 \times 1$ . Ye step kyun? Ye hamara maximum hai; har tilt ise $cos θ$ se scale karta hai. $F (0^{\circ}) = 9.08 \times 1 0^{- 3} N \approx 9.1 mN .$

Step 2. $6 0^{\circ}$ pe: $cos 6 0^{\circ} = 0.5$ se multiply karo. Ye step kyun? $cos θ$ measure karta hai ki flat area ka kitna hissa Sun actually "dekhta" hai — tilt karte waqt projected area shrink hota hai. $F (6 0^{\circ}) = 9.08 \times 1 0^{- 3} \times 0.5 = 4.54 \times 1 0^{- 3} N .$

Step 3. $9 0^{\circ}$ pe (edge-on): $cos 9 0^{\circ} = 0$ . Ye step kyun? Edge-on, projected area ek line hai — zero area, zero photons intercept hote hain, zero force. Ye degenerate limit hai jo tumhe hamesha name karni chahiye. $F (9 0^{\circ}) = 0 N .$

Verify: $F (6 0^{\circ}) / F (0^{\circ}) = cos 6 0^{\circ} = 0.5$ ✓, aur $F (9 0^{\circ}) = 0$ "ek edge kuch nahi pakadta" se match karta hai. Note karo ki ye simple $cos θ$ model sirf along-Sun component deta hai; fuller mirror model ek $cos^{2} θ$ normal force deta hai (parent ke mistake box mein dekho).

Example 4 — Cell D: Mercury se Jupiter tak distance scaling

Worked example

$1/ r^{2}$ span, plus $r \to \infty$ limit Q: Same spacecraft ke liye, Mercury (0.39 AU) aur Jupiter (5.2 AU) pe $P_{rad}$ ko Earth se compare karo. $r \to \infty$ hone par kya hota hai?

Forecast: Mercury bahut strong (close hai), Jupiter bahut weak (door hai). Deep space → kuch nahi.

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

Step 2. Jupiter factor $= (1/5.2)^{2}$ . Ye step kyun? 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). Jab $r \to \infty$ , $1/ r^{2} \to 0$ , toh $P_{rad} \to 0$ . Ye step kyun? Tumhe limiting behaviour state karni chahiye: SRP outer dark mein kuch nahi ho jaata, isliye sails outer planets se aage useless hain.

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

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

Worked example Featherweight sail vs dense brick

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

Forecast: Sail ka $A / m$ bahut bada hai → real acceleration. Brick ka $A / m$ bahut chhota hai → negligible.

Step 1. Sail acceleration: $a = P_{rad} (1 + η) (A / m)$ . Ye step kyun? Orbits acceleration pe respond karti hain, aur mass sirf gravity mein cancel hota hai, yahan nahi — SRP acceleration $A / m$ se jeeta ya marta hai. $a_{sail} = 4.54 \times 1 0^{- 6} \times 2 \times (1000/10) = 9.08 \times 1 0^{- 4} m/s^{2} .$

Step 2. Brick numbers: mass $= ρ V = 19300 \times (0.1)^{3} = 19.3 kg$ ; face area $A = (0.1)^{2} = 0.01 m^{2}$ ; toh $A / m = 0.01/19.3 = 5.18 \times 1 0^{- 4} m^{2} / kg$ . Ye step kyun? Brick ke liye $A / m$ chahiye — yahi puri kahaani hai. $a_{brick} = 4.54 \times 1 0^{- 6} \times 2 \times 5.18 \times 1 0^{- 4} = 4.70 \times 1 0^{- 9} m/s^{2} .$

Step 3 (limit). Jab $A / m \to 0$ (bahut dense, compact body), $a_{SRP} \to 0$ . Ye step kyun? Degenerate case name karo: ek heavy dense probe SRP ko barely notice karta hai.

Verify: Ratio $a_{sail} / a_{brick} = (1000/10) / (5.18 \times 1 0^{- 4}) = 100/5.18 \times 1 0^{- 4} = 1.93 \times 1 0^{5}$ — sail ~ $200, 000 \times$ zyada hard accelerate karta hai ✓. Isliye hi Solar sails kaam karte hain aur dense probes SRP ignore karte hain.

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

Worked example Kaun sa area-to-mass SRP ko Sun ki gravity cancel kara deta hai?

Q: Ek black spacecraft ( $η = 0$ ) ke liye, kaun sa $A / m$ outward SRP acceleration ko Sun ke inward gravitational acceleration ke exactly equal banata hai? (Ye ek "solar-static" sail define karta hai.) Sun mass $M_{⊙} = 1.989 \times 1 0^{30} kg$ , $G = 6.674 \times 1 0^{- 11}$ .

Forecast: Dono accelerations $1/ r^{2}$ ki tarah jaati hain, toh jawab ek single number hona chahiye, distance se independent — ye parent ke mistake box ka punchline hai.

Step 1. SRP acceleration: $a_{SRP} = \frac{S _{\oplus} ( 1 AU / r ) ^{2}}{c} (1 + η) \frac{A}{m}$ . Ye step kyun? Dono sides ko general $r$ pe likho taaki hum dekh sakein ki $r$ cancel hota hai.

Step 2. Gravity: $a_{g} = \frac{G M _{⊙}}{r ^{2}}$ . Set $a_{SRP} = a_{g}$ : $\frac{S _{\oplus} ( 1 AU ) ^{2}}{c r ^{2}} (1 + η) \frac{A}{m} = \frac{G M _{⊙}}{r ^{2}} .$ Ye step kyun? Dono sides mein $1/ r^{2}$ hai — wo divide out ho jaata hai, distance-independence prove hoti hai.

Step 3. $η = 0$ ke saath $A / m$ ke liye solve karo: $\frac{A}{m} = \frac{G M _{⊙} c}{S _{\oplus} ( 1 AU ) ^{2} ( 1 + η )} .$ Numbers plug karo: 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: $r$ vanish ho gaya ✓ (ratio universal hai, jaise promise kiya). Units: $\frac{( m ^{3} kg ^{- 1} s ^{- 2} ) ( kg ) ( m/s )}{( W/m ^{2} ) ( m ^{2} )} = \frac{m ^{4} s ^{- 3}}{W} = \frac{m ^{4} s ^{- 3}}{kg\cdotm ^{2} s ^{- 3}} = m^{2} / kg$ ✓. Required $A / m \approx 1300 m^{2} / kg$ bahut enormous hai — real sails sirf partly gravity offset karte hain. Ye Orbital perturbations se link karta hai.

Example 7 — Cell G: exam twist, ulta solve karo ek design number ke liye

Worked example Mission ko ek set acceleration chahiye — sail area nikalo

Q: Ek mission ko $a_{SRP} = 1.0 \times 1 0^{- 4} m/s^{2}$ chahiye 1 AU pe ek mirror sail ( $η = 1$ ) se jo ek $50 kg$ payload se attached hai. Kaun sa sail area $A$ required hai (sail mass ignore karo)?

Forecast: Acceleration formula ko $A$ ke liye rearrange karo; hundreds of square metres expect karo.

Step 1. $a_{SRP} = P_{rad} (1 + η) A / m$ se shuru karo aur $A$ ke liye solve karo: $A = \frac{a _{SRP} m}{P _{rad} ( 1 + η )} .$ Ye step kyun? Exams aksar target performance dete hain aur design maangte hain — formula invert karo.

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

Verify (plug back): $a = 4.54 \times 1 0^{- 6} \times 2 \times (550.7/50) = 4.54 \times 1 0^{- 6} \times 2 \times 11.01 = 1.0 \times 1 0^{- 4} m/s^{2}$ ✓ — target recover ho gaya. 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 ek angle pe

Worked example Grey sail (

$η = 0.6$ ) $4 5^{\circ}$ pe tilted Q: $A = 200 m^{2}$ , $η = 0.6$ , $θ = 4 5^{\circ}$ , 1 AU pe. Force?

Forecast: Reflectivity factor $1.6$ , angle factor $cos 4 5^{\circ} \approx 0.707$ — roughly $1.6 \times 0.71 \approx 1.13$ times black-face-on value expect karo.

Step 1. Black face-on baseline: $F_{0} = P_{rad} A = 4.54 \times 1 0^{- 6} \times 200 = 9.08 \times 1 0^{- 4} N$ . Ye step kyun? Reference isolate karo taaki do correction factors transparent hon.

Step 2. Dono factors apply karo: $F = F_{0} (1 + η) cos θ = 9.08 \times 1 0^{- 4} \times 1.6 \times 0.7071$ . Ye step kyun? $(1 + η)$ partial recoil handle karta hai; $cos θ$ projected area handle karta hai. $F = 9.08 \times 1 0^{- 4} \times 1.1314 = 1.027 \times 1 0^{- 3} N .$

Verify: Combined factor $1.6 \times 0.7071 = 1.131$ ; $9.08 \times 1 0^{- 4} \times 1.131 = 1.027 \times 1 0^{- 3}$ ✓. Sanity: black-face-on $9.08 \times 1 0^{- 4}$ se bada hai (reflectivity tilt loss pe haavi hai) — forecast se match karta hai.

Recall

Recall Apne aap test karo (answers chhupao)

Kisi bhi flat plate pe edge-on ( $θ = 9 0^{\circ}$ ) force? ::: bilkul zero — projected area ek line hai $r \to \infty$ hone par $P_{rad}$ ka kya hota hai? ::: $1/ r^{2}$ ki tarah $0$ ho jaata hai SRP-vs-gravity distance-independent $A / m$ kyun deta hai? ::: dono accelerations mein $1/ r^{2}$ hota hai, jo cancel ho jaata hai Target acceleration hit karne ke liye, area kaunse rearrangement se nikalo? ::: $A = a m / [P_{rad} (1 + η)]$ Grey sail ke do correction factors kya hain? ::: $(1 + η)$ recoil ke liye aur $cos θ$ projected area ke liye

Connections

Solar radiation pressure — parent note with derivations.
Solar sails — Examples 3, 5, 7 miniature mein sail design hain.
Orbital perturbations — Example 6 ka balance perturbation scale set karta hai.
Inverse-square law — Examples 1 aur 4 iske upar lean karte hain.
Atmospheric drag — low altitude pe competing force.
Yarkovsky effect — asteroids pe in pushes ka thermal cousin.
Photon momentum and relativity — har $S / c$ ke peeche $p = E / c$ .