Hohmann Δv calculation — both maneuvers: Worked examples

3.2.20 · D3 · Physics › Orbital Mechanics & Astrodynamics › Hohmann Δv calculation — both maneuvers

Kuch bhi shuru karne se pehle, wo symbols jo hum baar baar use karte hain:

Definition Recurring cast

$μ$ (mu) = standard gravitational parameter $= GM$ central body ka. Units $km^{3} / s^{2}$ . Dekho Standard gravitational parameter.
$r_{1}$ = starting circular orbit ka radius (planet ke center se measure kiya, surface se nahi).
$r_{2}$ = target circular orbit ka radius.
$a_{t} = \frac{r _{1} + r _{2}}{2}$ = transfer ellipse ka semi-major axis (periapsis-to-apoapsis span ka aadha).
$v_{c 1} = μ / r_{1}, v_{c 2} = μ / r_{2}$ = orbit 1 aur orbit 2 par circular speeds.
$v_{p}$ = periapsis speed = transfer ellipse ki speed uske near point par (radius $r_{1}$ ). "Periapsis" = orbit ka wo point jo planet ke sabse qareeb hai. Yeh ellipse ka fastest point hota hai.
$v_{a}$ = apoapsis speed = transfer ellipse ki speed uske far point par (radius $r_{2}$ ). "Apoapsis" = orbit ka wo point jo planet se sabse door hai. Yeh ellipse ka slowest point hota hai.
$Δ v_{1}, Δ v_{2}$ = burn 1 (at $r_{1}$ ) aur burn 2 (at $r_{2}$ ) ka speed change.
$Δ v_{tot} \equiv Δ v_{1} + Δ v_{2}$ = total speed budget. Hum ise simple sum define karte hain kyunki dono burns alag, independent impulses hain (dekho Impulsive maneuver approximation); koi vector cancellation nahi hai, isliye scalar magnitudes simply add ho jaati hain.

Saari chaar speeds sirf do tools se aati hain, jinpar hum pura page rely karte hain:

The scenario matrix

Har Hohmann problem in cells mein se ek (ya blend) hai. Neeche ke examples sab ko hit karte hain. Diagram shared geometry dikhata hai — har cell apna start circle (blue), target circle (green), transfer ellipse (dashed orange) aur do red burn points pick karta hai, toh iske paas refer karte rehna.

Cell	Kya special hai	Example
A · Standard raise	$r_{2} > r_{1}$ , dono burns prograde, ordinary numbers	Ex 1
B · Small raise	$r_{2}$ sirf thoda $> r_{1}$ ; dono $Δ v$ tiny aur almost equal	Ex 2
C · Large raise / "which burn bigger" flip	bahut bada $r_{2}$ ; check karo burn 1 ya burn 2 dominate karta hai	Ex 3
D · Lowering (going inward)	$r_{2} < r_{1}$ ; burns retrograde ho jaate hain (speed decreases)	Ex 4
E · Degenerate $r_{1} = r_{2}$	koi transfer needed nahi; $Δ v = 0$ dena zaroori	Ex 5
F · Limit $r_{2} \to \infty$	escape-like; $Δ v_{1} \to (2 - 1) v_{c 1}$ , $Δ v_{2} \to 0$	Ex 6
G · Real-world word problem	radii altitudes ke form mein di hain, pehle planet radius add karo	Ex 7
H · Exam twist	transfer time (Ex 8a), ya given $Δ v$ aur backward solve karo (Ex 8b)	Ex 8

Pehle yeh figure padho. Blue circle orbit 1 hai, green circle orbit 2 hai, dashed orange oval transfer ellipse hai jo blue ko uske near point (periapsis, right, jahan $v_{p}$ rehta hai) par touch karta hai aur green ko uske far point (apoapsis, left, jahan $v_{a}$ rehta hai) par. Do red dots exactly wahan hain jahan engine fire karta hai — aur kahin nahi. Blue aur green arrows $r_{1}$ aur $r_{2}$ ko planet se focus par measure karte hain. Neeche har cell bas yahi picture hai alag-alag radii ke saath.

Figure — Hohmann Δv calculation — both maneuvers

Cell A — the standard raise

Neeche ka figure dekho — yahi exact case hai. Right wala red dot (periapsis, $r_{1}$ par) jahan $v_{c 1}$ jump karta hai $v_{p}$ tak; left wala red dot (apoapsis, $r_{2}$ par) jahan $v_{a}$ jump karta hai $v_{c 2}$ tak. Labelled speeds dikhate hain ki burn 1 bada kyun hai.

Step 1 — transfer ka semi-major axis. $a_{t} = \frac{6678 + 20000}{2} = 13339 km$ Yeh step kyun? Vis-viva ko $a$ chahiye; ellipse $r_{1}$ (periapsis) se $r_{2}$ (apoapsis) tak span karta hai, isliye $a_{t}$ unka average hai.

Step 2 — $r_{1}$ par circular speed. $v_{c 1} = \frac{398600}{6678} = 7.726 km/s$ Kyun? Yeh wo speed hai jis par hum start karte hain, circle 1 par ride karte hue.

Step 3 — periapsis par transfer speed ( $r = r_{1}$ , $a = a_{t}$ ). $v_{p} = 398600 (\frac{2}{6678} - \frac{1}{13339}) = 9.462 km/s$ Kyun? Yeh wo speed hai jo humein burn 1 ke baad ellipse par hone ke liye chahiye. Kyunki $a_{t} > r_{1}$ , ellipse yahan circle se fast hai.

Step 4 — burn 1. $Δ v_{1} = v_{p} - v_{c 1} = 9.462 - 7.726 = 1.736 km/s$ Is tarah subtract kyun? Dono velocities same direction (prograde) point karti hain, aur $v_{p} > v_{c 1}$ , isliye $Δ v = ∣ v_{p} - v_{c 1} ∣ = v_{p} - v_{c 1}$ — badi minus chhoti.

Step 5 — $r_{2}$ par circular speed. $v_{c 2} = \frac{398600}{20000} = 4.464 km/s$ Kyun? Yeh wo speed hai jis par humein end up karna hai circle 2 par ride karne ke liye, isliye burn 2 mein ellipse ki apoapsis speed se compare karna zaroori hai.

Step 6 — apoapsis par transfer speed ( $r = r_{2}$ , $a = a_{t}$ ). $v_{a} = 398600 (\frac{2}{20000} - \frac{1}{13339}) = 3.159 km/s$ Kyun? Ellipse apne far point par slower hota hai, isliye yeh $v_{c 2}$ se kam hai — humein speed up karna hoga.

Step 7 — burn 2 aur total. $Δ v_{2} = v_{c 2} - v_{a} = 4.464 - 3.159 = 1.305 km/s, Δ v_{tot} = 1.736 + 1.305 = 3.041 km/s$ $v_{c 2} - v_{a}$ kyun aur add kyun? Apoapsis par $v_{a} < v_{c 2}$ , isliye $Δ v_{2} = ∣ v_{c 2} - v_{a} ∣ = v_{c 2} - v_{a}$ (badi minus chhoti, ek prograde speed-up). Hum dono add karte hain kyunki wo alag points par alag impulses hain, $Δ v_{tot}$ ki definition ke mutabiq.

Verify: Dono burns positive ✔ (raise ke liye expected prograde). $Δ v_{1} > Δ v_{2}$ — forecast payoff: gehri burn zyada kaam karti hai (yeh Oberth effect ka seed hai). Units: $km^{3} s^{- 2} / km = km/s$ ✔.

Cell B — the small raise

Step 1 — $a_{t}$ . $a_{t} = (7000 + 7100) /2 = 7050$ km. Kyun? same averaging rule — transfer ellipse periapsis $r_{1}$ se apoapsis $r_{2}$ tak span karta hai.

Step 2 — chaar speeds. $v_{c 1} = 398600/7000 = 7.546 km/s — circular speed jis par hum start karte hain (v_{circ} at r_{1})$ $v_{p} = 398600 (2/7000 - 1/7050) = 7.573 km/s — vis-viva at r = r_{1} on the ellipse (a = a_{t})$ $v_{c 2} = 398600/7100 = 7.493 km/s — circular speed jo humein reach karni hai (v_{circ} at r_{2})$ $v_{a} = 398600 (2/7100 - 1/7050) = 7.466 km/s — vis-viva at r = r_{2} on the ellipse$ Yeh formulae yahan kyun? Har burn ek circle speed ko same radius par ellipse speed se compare karta hai: circular speed $v_{circ} (r)$ use karta hai, ellipse speed $a = a_{t}$ ke saath vis-viva use karta hai. Isliye $v_{c}$ mein $μ / r$ hai aur $v_{p}, v_{a}$ mein $μ (2/ r - 1/ a_{t})$ hai.

Step 3 — dono burns. $Δ v_{1} = v_{p} - v_{c 1} = 7.573 - 7.546 = 0.0267 km/s, Δ v_{2} = v_{c 2} - v_{a} = 7.493 - 7.466 = 0.0264 km/s$ Itne close kyun? Jab $r_{2} \approx r_{1}$ , ellipse barely eccentric hota hai, isliye periapsis aur apoapsis excesses almost mirror images hain.

Verify: $Δ v_{1} \approx Δ v_{2}$ (<2% ka fark), 50/50 forecast confirm karta hai. Sanity: total $\approx 53 m/s$ , 100 km station-keeping raise ke liye sahi ballpark ✔.

Cell C — the large raise & the "which burn is bigger" flip

Step 1 — $a_{t}$ . $a_{t} = (6678 + 400000) /2 = 203339$ km. Average kyun? Transfer ellipse ab bhi periapsis $r_{1}$ aur apoapsis $r_{2}$ ko touch karta hai, isliye uska semi-major axis unka midpoint hai — rule kabhi nahi badlta, sirf numbers bade hote hain.

Step 2 — burn 1 (near-escape kick). $v_{c 1} = 7.726 km/s, v_{p} = 398600 (2/6678 - 1/203339) = 10.847 km/s$ $Δ v_{1} = v_{p} - v_{c 1} = 10.847 - 7.726 = 3.121 km/s$ Itna bada kyun, aur yeh subtraction kyun? $r_{2}$ huge hone par, periapsis speed escape speed $2 v_{c 1} = 10.926$ km/s ke paas pahunch jaati hai. Kyunki $v_{p} > v_{c 1}$ , $Δ v_{1} = ∣ v_{p} - v_{c 1} ∣ = v_{p} - v_{c 1}$ .

Step 3 — burn 2 (deep-space circularize). $v_{c 2} = 398600/400000 = 0.998 km/s, v_{a} = 398600 (2/400000 - 1/203339) = 0.181 km/s$ $Δ v_{2} = v_{c 2} - v_{a} = 0.998 - 0.181 = 0.817 km/s$ $v_{c 2} - v_{a}$ kyun? Door apoapsis par ellipse crawl karta hai ( $v_{a} ≪ v_{c 2}$ ), isliye humein speed up karna hoga: $Δ v_{2} = ∣ v_{c 2} - v_{a} ∣ = v_{c 2} - v_{a}$ , badi minus chhoti.

Verify: $Δ v_{1} ≫ Δ v_{2}$ — burn 1 ab bhi jeetta hai, aur Ex 1 se bhi zyada lopsidedly. Lesson: raise ke liye, periapsis burn hamesha dominate karta hai jab $r_{2}$ badhta hai. (Yahi reason hai ki Bi-elliptic transfer bahut bade ratios ke liye Hohmann ko beat kar sakta hai — woh expensive kick ko split karta hai.) Units ✔.

Cell D — orbit lower karna (andar jaana)

Step 1 — $a_{t}$ . $a_{t} = (42164 + 6678) /2 = 24421$ km. Kyun? transfer ellipse same shape ka hai jaise LEO→GEO — Hohmann symmetric hai; do radii ka average karna ab bhi semi-major axis deta hai.

Step 2 — speeds at (ab outer) start radius $r_{1} = 42164$ par. $v_{c 1} = 398600/42164 = 3.075 km/s — circular speed jis par hum start karte hain$ $v_{a} = 398600 (2/42164 - 1/24421) = 1.607 km/s — ellipse speed yahan$ Yeh ellipse ki apoapsis speed kyun hai, aur yeh formulae kyun? Upar se start karke, hum ellipse par uske far point par drop karte hain, isliye yahan transfer speed slow apoapsis speed hai — vis-viva use karo $r = r_{1}$ , $a = a_{t}$ ke saath. Circular speed ab bhi $v_{circ} (r_{1})$ use karta hai. Note: $v_{a} < v_{c 1}$ .

Step 3 — burn 1 magnitude. $Δ v_{1} = v_{c 1} - v_{a} = 3.075 - 1.607 = 1.468 km/s$ $v_{c 1} - v_{a}$ kyun? Yahan before speed $v_{c 1}$ hai aur after speed $v_{a}$ hai, $v_{c 1} > v_{a}$ ke saath. Definition $Δ v = ∣ v_{after} - v_{before} ∣ = ∣ v_{a} - v_{c 1} ∣ = v_{c 1} - v_{a}$ se — phir badi minus chhoti, positive. Physically hum slow down karte hain (retrograde) andar drop karne ke liye.

Step 4 — low target radius $r_{2} = 6678$ par speeds. $v_{p} = 398600 (2/6678 - 1/24421) = 10.15 km/s — ellipse periapsis speed$ $v_{c 2} = 398600/6678 = 7.726 km/s — circular speed jo humein reach karni hai$ Kyun? Neeche pahunchte hue, hum ellipse ke near point (periapsis) par pahunchte hain, sabse fast, vis-viva ke through $r = r_{2}$ par; target circle speed $v_{circ} (r_{2})$ hai. Note: $v_{p} > v_{c 2}$ .

Step 5 — burn 2 magnitude. $Δ v_{2} = v_{p} - v_{c 2} = 10.15 - 7.726 = 2.421 km/s$ $v_{p} - v_{c 2}$ kyun? Ab before speed $v_{p}$ (ellipse par) hai aur after $v_{c 2}$ (circle) hai, $v_{p} > v_{c 2}$ ke saath. Toh $Δ v = ∣ v_{after} - v_{before} ∣ = ∣ v_{c 2} - v_{p} ∣ = v_{p} - v_{c 2}$ — badi minus chhoti, positive. Physically ek aur retrograde brake.

Verify: Total $= 1.468 + 2.421 = 3.889$ km/s — parent note ke LEO→GEO total (3.89 km/s) se identical ✔. Hohmann reversible hai: lower karna utna hi kharcha karta hai jitna raise karna, bas retrograde burns ke saath. Dono $Δ v$ 's positive rahe kyunki humne hamesha badi-minus-chhoti likha, exactly jo $∣ v_{after} - v_{before} ∣$ demand karta hai. Burn sizes ki ordering bhi flip ho jaati hai (badi burn ab doosri hai).

Cell E — degenerate case $r_{1} = r_{2}$

Pehle humein woh closed-form formula earn karni hogi jo yeh cell test karta hai. Burn 1 lo aur $v_{p}$ ke liye vis-viva expression substitute karo $a_{t} = (r_{1} + r_{2}) /2$ ke saath:

$v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = μ (\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}}) = \frac{μ}{r _{1}} \frac{2 r _{2}}{r _{1} + r _{2}}$

Yeh simplify kyun hota hai? $\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}} = \frac{2}{r _{1}} \cdot \frac{r _{2}}{r _{1} + r _{2}}$ factor karo, phir $μ / r_{1}$ bahar nikalo. $v_{c 1} = μ / r_{1}$ subtract karne par burn 1 ki factored form milti hai.

Ab $Δ v_{2}$ same tarike se derive karo, step by step. Apoapsis speed $v_{a}$ (vis-viva at $r = r_{2}$ ) se shuru karo: $v_{a} = μ (\frac{2}{r _{2}} - \frac{1}{a _{t}}) = μ (\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}})$ Kyun? same $a_{t} = (r_{1} + r_{2}) /2$ substitution. Andar factor karo: $\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}} = \frac{2}{r _{2}} \cdot \frac{r _{1}}{r _{1} + r _{2}}$ , toh $v_{a} = \frac{μ}{r _{2}} \frac{2 r _{1}}{r _{1} + r _{2}}$ Phir $Δ v_{2} = v_{c 2} - v_{a} = \frac{μ}{r _{2}} - \frac{μ}{r _{2}} \frac{2 r _{1}}{r _{1} + r _{2}}$ , aur $μ / r_{2}$ bahar nikalne par neeche boxed form milta hai.

Step 1 — $a_{t}$ . $a_{t} = (10000 + 10000) /2 = 10000 = r_{1} = r_{2}$ . Check kyun? "Ellipse" circle mein hi degenerate ho jaata hai.

Step 2 — abhi derive ki gayi factored formula mein plug karo. $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) = \frac{μ}{r _{1}} (\frac{2 r _{1}}{2 r _{1}} - 1) = \frac{μ}{r _{1}} (1 - 1) = 0$ $Δ v_{2} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{2 r _{1}}) = 0$

Verify: $Δ v_{tot} = 0$ ✔. Kahin bhi division-by-zero nahi ( $a_{t}$ finite aur nonzero hai). Formula degenerate boundary par well-behaved hai — ek acha correctness check.

Cell F — limit $r_{2} \to \infty$

Step 1 — burn 1 limit. $Δ v_{1} = μ / r_{1} (2 r_{2} / (r_{1} + r_{2}) - 1)$ mein, $r_{2} \to \infty$ hone par ratio $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ . Toh $Δ v_{1} \to \frac{μ}{r _{1}} (2 - 1) = (2 - 1) v_{c 1}$ Kyun? $2 v_{c 1}$ escape speed hai; circular ke upar extra push $(2 - 1) v_{c 1}$ hai.

Numerically: $(2 - 1) \times 7.726 = 3.200$ km/s.

Step 2 — burn 2 limit. $Δ v_{2} = μ / r_{2} (1 - 2 r_{1} / (r_{1} + r_{2}))$ mein, leading $μ / r_{2} \to 0$ jabki bracket $\to 1$ . Toh $Δ v_{2} \to 0$ Kyun? Infinite radius par circular speed zero hai, aur ellipse ki apoapsis speed bhi zero hai, isliye kuch bhi burn karne ko nahi bacha.

Verify: $Δ v_{1} \to 3.200$ km/s, $Δ v_{2} \to 0$ ✔. Parent ke forecast box se exactly match karta hai. Isliye interplanetary escape burn essentially sirf "burn 1 to infinity" hai.

Cell G — real-world word problem (altitudes, radii nahi)

Step 1 — altitudes ko orbital radii mein convert karo. $r_{1} = 6371 + 500 = 6871 km, r_{2} = 6371 + 800 = 7171 km$ Yeh step kyun? Gravity mass ke center se distance par depend karta hai, isliye vis-viva mein har $r$ center-relative hai. $+ R_{\oplus}$ bhool jaana #1 real-world error hai.

Step 2 — $a_{t}$ aur speeds. $a_{t} = (6871 + 7171) /2 = 7021 km$ $v_{c 1} = 398600/6871 = 7.617 km/s, v_{p} = 398600 (2/6871 - 1/7021) = 7.696 km/s$ $v_{c 2} = 398600/7171 = 7.456 km/s, v_{a} = 398600 (2/7171 - 1/7021) = 7.375 km/s$ Yeh formulae kyun? Circular speeds $v_{circ} (r)$ use karte hain; ellipse speeds vis-viva use karte hain $a = a_{t}$ ke saath — exactly jaise har raise cell mein.

Step 3 — burns. $Δ v_{1} = v_{p} - v_{c 1} = 7.696 - 7.617 = 0.0791 km/s, Δ v_{2} = v_{c 2} - v_{a} = 7.456 - 7.375 = 0.0805 km/s$ $Δ v_{tot} = 0.1596 km/s \approx 160 m/s$

Verify: Small raise ⇒ tiny, nearly-equal burns (Cell B behavior se consistent) ✔. Agar tumne galti se altitudes (500 aur 800) use kiye hote, $v_{c 1} = 398600/500 = 28.2$ km/s — absurd, kisi bhi real orbit se fast — ek instant red flag ki tumne $R_{\oplus}$ bhool gaye.

Cell H — the exam twists

Ex 8a — transfer time

Step 1 — transfer semi-major axis. $a_{t} = (6678 + 42164) /2 = 24421$ km. Kyun? Period ko $a$ chahiye, aur transfer is ellipse ka aadha hai.

Step 2 — transfer ellipse ka full period (Kepler III). $T = 2 π \frac{a _{t}^{3}}{μ} = 2 π \frac{2442 1 ^{3}}{398600} s$ Yeh tool kyun? Specific orbital energy speed fix karta hai lekin timing nahi; Kepler's third law $T = 2 π a^{3} / μ$ orbit size ko elapsed time mein convert karta hai.

Step 3 — full period evaluate karo. $T = 2 π \frac{1.4566 \times 1 0 ^{13}}{398600} = 2 π 3.654 \times 1 0^{7} = 3.798 \times 1 0^{4} s$

Step 4 — Hohmann sirf half-ellipse use karta hai (periapsis → apoapsis). $t_{transfer} = \frac{T}{2} = π \frac{a _{t}^{3}}{μ} = 1.899 \times 1 0^{4} s \approx 5.28 hours$ Half kyun, aur isliye $π a_{t}^{3} / μ$ form? Tum periapsis se apoapsis tak ride karte ho — exactly ellipse ka ek aadha loop — isliye coast time $T /2$ hai, jo $2 π$ ko $π$ mein badal deta hai: standard transfer-time formula $t_{transfer} = π a_{t}^{3} / μ$ .

Verify: $\approx 5.3$ h textbook LEO→GEO transfer time hai ✔. Units: $km^{3} / (km^{3} s^{- 2}) = s$ ✔.

Ex 8b — given Δv se backward solve karo

Step 1 — $Δ v_{1}$ factored form mein likho aur square root isolate karo. $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) \Rightarrow \frac{2 r _{2}}{r _{1} + r _{2}} = 1 + \frac{Δ v _{1}}{μ / r _{1}}$ Kyun? $r_{2}$ sirf us ek square root ke andar appear karta hai, isliye hum pehle baaki sab kuch peel off karte hain.

Step 2 — numbers plug karo. $v_{c 1} = 398600/6678 = 7.726$ km/s, toh $\frac{2 r _{2}}{6678 + r _{2}} = 1 + \frac{2.42}{7.726} = 1.3133 \Rightarrow \frac{2 r _{2}}{6678 + r _{2}} = 1.7248$ Square kyun? Root hatane ke liye aur $r_{2}$ mein linear equation paane ke liye.

Step 3 — linear equation solve karo. $2 r_{2} = 1.7248 (6678 + r_{2}) \Rightarrow (2 - 1.7248) r_{2} = 1.7248 \times 6678 \Rightarrow r_{2} = \frac{11518}{0.2752} = 41857 km$ Linear kyun? Denominator clear karne ke baad dono sides $r_{2}$ mein first-degree hain.

Verify: $r_{2} \approx 41900$ km — essentially GEO (42 164 km), forecast se match ✔. Forward-check: $Δ v_{1} = 7.726 (2 \cdot 41857/ (6678 + 41857) - 1) = 2.42$ km/s ✔ (round-trip consistent).

Recall Kaunsa cell kaunsa hai? (self-test)

Sirf altitudes diye hone par, pehla step kya? ::: Planet radius add karo center-relative $r_{1}, r_{2}$ paane ke liye (Cell G). Orbit lower karna — burns prograde hain ya retrograde? ::: Retrograde (tum slow down karte ho), lekin $Δ v$ magnitudes phir bhi positive hain (Cell D). $r_{2} \to \infty$ par $Δ v_{1}$ ? ::: $(2 - 1) v_{c 1}$ , near-escape kick (Cell F). $r_{1} = r_{2}$ hone par $Δ v_{tot}$ ? ::: Exactly zero (Cell E). Raise ke liye kaunsa burn bada hai aur kya woh $r_{2}$ ke saath change hota hai? ::: Burn 1 (periapsis), aur woh aur zyada dominate karta hai jaise $r_{2}$ badhta hai (Cells A, C). Transfer time formula? ::: $t_{transfer} = π a_{t}^{3} / μ$ (ellipse period ka aadha) (Cell H). Given $Δ v_{1}$ se $r_{2}$ nikalne ke liye? ::: Square root isolate karo, square karo, resulting linear equation solve karo (Cell H, Ex 8b).