Exercises — Sphere of influence — radius derivation

Aisha sahi hai: Laplace SOI mass ratio ka ==$2/5$ power== carry karta hai, jo ek tidal perturbation ki pehchaan hai. Ben ka $1/2$ power force-equality point $r_{eq}=R{q}^{1/2}$ hai — jahan planet ki raw pull Sun ki raw pull ke barabar hoti hai. Yeh ek alag (chhota) radius hai aur SOI nahi hai. Mnemonic: "Tides are two-fifths."

• $R$ = do bade bodies ke beech ki doori (jaise planet ↔ Sun).
• $m$ = chhote body ki mass — jiska bubble hum compute karte hain.
• $M$ = bade central body ki mass.
• Hamesha $m\ll M$, isliye $q=m/M<1$ aur $q^{2/5}<1$, jisse $r_{SOI}<R$ hota hai jaisa hona chahiye (bubble separation ke andar kafi deep hota hai).

Step 1 — mass ratio: $q=\frac{5.97\times 10^{24}}{1.989\times 10^{30}}=3.00\times 10^{-6}$. Step 2 — $2/5$ pe raise karo logs use karke (yaad karo: log power ko multiply mein badal deta hai). ${\log }_{10}q=-5.523$; times $0.4$ gives $-2.209$; toh $q^{0.4}=10^{-2.209}=6.18\times 10^{-3}$. Step 3 — $R$ se multiply karo: $r_{SOI}=1.496\times 10^{11}\times 6.18\times 10^{-3}\approx 9.2\times 10^8\text{ m}\approx 924,000\text{ km}.$ Yeh lagbhag 145 Earth radii hai.

$q=6.42\times 10^{23}/1.989\times 10^{30}=3.228\times 10^{-7}$. ${\log }_{10}q=-6.491$; $\times 0.4=-2.596$; $q^{0.4}=10^{-2.596}=2.535\times 10^{-3}$. $r_{SOI}=2.279\times 10^{11}\times 2.535\times 10^{-3}\approx 5.78\times 10^8\text{ m}\approx 578,000\text{ km}.$ Earth se chhota, Mars kaafi door hone ke bawajood, kyunki Mars bahut halka hai — mass ratio doori par bhaari padti hai.

Force-equality $1/2$ power use karta hai: $r_{eq}=R\sqrt{q}=2.279\times 10^{11}\times \sqrt{3.228\times 10^{-7}}.$ $\sqrt{3.228\times 10^{-7}}=5.681\times 10^{-4}$, toh $r_{eq}=2.279\times 10^{11}\times 5.681\times 10^{-4}=1.295\times 10^8$ m ≈ 129,500 km. Ratio: $r_{SOI}/r_{eq}=5.78\times 10^8/1.295\times 10^8\approx 4.46$. SOI force-balance point se lagbhag 4.5× bada hai — kyunki tidal criterion (chhote $q$ par $2/5$ power) raw-force criterion ($1/2$ power) se jeet jaata hai.

$r_{SOI}\propto q^{2/5}\propto m^{2/5}$. $m$ double karne se radius multiply hoti hai: $2^{2/5}=2^{0.4}=\mathrm{1.320.}$ Toh SOI lagbhag 32% badhta hai, 100% nahi. $2/5$ power mass change ko tame karta hai — mass par bada lever bubble par modest lever deta hai. Yahi softness hai jo wahi tidal $2/5$ hai jise hum upar picture kar chuke hain.

$(1/2{)}^{1/5}=2^{-0.2}=0.8706$. Ise L2.1 ke 924,000 km par apply karte hain: $924,000\times 0.8706\approx 804,000\text{ km}.$ Lagbhag ~13% reduction. Yeh tidal derivative $\frac{d}{dR}\left(\frac{GM}{R^2}\right)=-\frac{2GM}{R^3}$ mein factor 2 se aata hai. 1 ke kareeb hone ki wajah se, convention ise drop kar deta hai.

Hum ek Earth-centred orbit aur ek Moon-centred orbit ke beech patch kar rahe hain, isliye SOI Moon–Earth pair use karta hai. $q=7.35\times 10^{22}/5.97\times 10^{24}=0.01231$. ${\log }_{10}q=-1.9097$; $\times 0.4=-0.7639$; $q^{0.4}=10^{-0.7639}=0.1722$. $r_{SOI}=3.84\times 10^8\times 0.1722\approx 6.61\times 10^7\text{ m}\approx 66,000\text{ km}.$ Moon se ~66,000 km se kareeb spacecraft ko Moon ka orbit karte hua treat karna best hai; door ho toh Earth ka orbit karte hua. Yeh Patched Conic Approximation aur Gravity Assist / Flyby planning se directly connect karta hai.

$r_{SOI}=R{q}^{2/5}$ ko $q$ ke liye rearrange karo: $q={\left(\frac{r_{SOI}}{R}\right)}^{5/2}.$ $5/2$ power kyun? $q^{2/5}$ ko undo karne ka matlab hai reciprocal $5/2$ pe raise karna. $R=5\times 1.496\times 10^{11}=7.48\times 10^{11}$ m. Ratio $r_{SOI}/R=4.82\times 10^{10}/7.48\times 10^{11}=0.06444$. $q=0.06444^{2.5}$. ${\log }_{10}0.06444=-1.1908$; $\times 2.5=-2.977$; $q=10^{-2.977}=1.055\times 10^{-3}$. $m=qM=1.055\times 10^{-3}\times 1.989\times 10^{30}\approx 2.10\times 10^{27}\text{ kg}.$ Yeh roughly ek Jupiter-mass planet hai.

Neechey nested-sphere figure ise visual banata hai. Moon Earth se 384,000 km door hai; uska bubble ±66,000 km tak extend karta hai, toh yeh Earth se roughly 318,000 – 450,000 km tak span karta hai. Earth ka apna SOI 924,000 km tak pahunchta hai. Kyunki $450,000<924,000$, poora Moon SOI Earth ke SOI ke andar nested hai. Matlab: Moon ke paas ek craft do nested bubbles ke andar hai. Hum conics ko stages mein patch karte hain — Sun→Earth→Moon — ek hierarchy jise Restricted Three-Body Problem aur Hill Sphere treatments precise banati hain. Patched Conic Approximation har handoff ko ek fresh Two-Body Problem ki tarah handle karta hai.

Hill: $q/3=3.00\times 10^{-6}/3=1.00\times 10^{-6}$; cube root $=(10^{-6}{)}^{1/3}=10^{-2}=0.01$. $r_H=1.496\times 10^{11}\times 0.01=1.496\times 10^9\text{ m}\approx 1.50\times 10^6\text{ km}.$ Hill sphere (~1.5 million km) SOI (~924,000 km) se bada hai. Yeh alag sawaalon ke jawab dete hain: SOI poochhta hai "trajectory accuracy ke liye main kaun sa body orbit kar raha hoon switch kab karun?" (ek $2/5$ tidal-ratio law), jabki Hill Sphere poochhta hai "kahan tak ek moon Sun ke against gravitationally bound reh sakta hai?" (ek $1/3$ centrifugal-balance law Restricted Three-Body Problem mein). Alag physics ⇒ alag exponent ⇒ alag radius. Neechey figure teeno radii ko scale par stack karta hai.

SOI ke andar, Sun ke Tidal Forces construction se, Mars ke pull ke relative ek negligible perturbation hain — isliye sirf Mars ki gravity probe ko ek clean two-body hyperbola par bend karti hai. Bahar, Mars ka pull negligible perturbation hai aur Sun dominate karta hai, isliye probe ek heliocentric conic follow karta hai. SOI boundary exactly woh surface hai jahan yeh do "kaun-perturbation-hai" verdicts flip karte hain (Laplace ka equal-ratio condition). Do conics ko us surface par patch karna — position aur velocity match karna — flyby ka net velocity change minimal error ke saath deta hai. Yahi Patched Conic Approximation ka poora point hai.

Forward: $r_{SOI}=R{q}^{2/5}$. ::: Pehle $q=m/M$ compute karo, phir $q^{2/5}$, phir $\times R$. Invert for mass ratio: $q=(r_{SOI}/R{)}^{5/2}$. ::: Reciprocal power $5/2$, $2/5$ ko undo karta hai. Mass $\times 2\Rightarrow$ SOI $\times$? ::: $2^{2/5}\approx 1.32$ (32% bada). Force-equality radius? ::: $r_{eq}=R{q}^{1/2}$ — ek raw force balance, SOI se chhota. Earth ke liye SOI vs Hill: ::: SOI ≈ 0.92 Mkm, Hill ≈ 1.5 Mkm — alag exponents, alag sawaal. $r_{SOI}=R{q}^{2/5}$ kab break karta hai? ::: Jab $q\to 1$ ya orbit bahut eccentric ho — full three-body treatment use karo.