Hohmann transfer — derivation, minimum energy transfer: Worked examples

3.2.19 · D3 · Physics › Orbital Mechanics & Astrodynamics › Hohmann transfer — derivation, minimum energy transfer

Yeh page Hohmann parent note ka drill hall hai. Humne jo formulas wahan banaye the, unhe hum har tarah ke case mein use karte hain jo yeh topic de sakta hai: orbit raise karna, orbit lower karna, degenerate "same orbit" case, ek bade jump ki limit, ek real spacecraft word problem, aur ek exam twist jahan Hohmann quietly ek bi-elliptic se haar jaata hai.

Kisi bhi number se pehle, hum vocabulary fix karte hain taaki koi bhi symbol unexplained na rahe.

Definition Woh symbols jo hum har jagah reuse karte hain

$μ$ (Greek "mu") — gravitational parameter $μ = GM$ , central body ki pull-strength. Units: $km^{3} / s^{2}$ .
$r_{1}$ — starting circular orbit ka radius (planet centre se distance).
$r_{2}$ — target circular orbit ka radius.
$a_{t}$ — transfer ellipse ka semi-major axis, uska half-longest-diameter, $a_{t} = (r_{1} + r_{2}) /2$ .
$v_{c 1}, v_{c 2}$ — start aur target orbits par circular speeds, $μ / r_{1}$ aur $μ / r_{2}$ .
$v_{p}$ — transfer ellipse par periapsis speed: ship kitni fast move kar rahi hai ellipse ke us point par jo planet ke sabse paas hai (radius $= r_{1}$ ek raise ke liye). Vis-viva se compute hoti hai $r = r_{1}, a = a_{t}$ ke saath.
$v_{a}$ — transfer ellipse par apoapsis speed: ship kitni fast move kar rahi hai planet se sabse door wale point par (radius $= r_{2}$ ek raise ke liye). Vis-viva se compute hoti hai $r = r_{2}, a = a_{t}$ ke saath.
$Δ v$ (padho "delta-vee") — fuel se kharida gaya speed mein badlaav. Positive matlab speed up, negative matlab slow down.

Do engines hain jinka hum sahara lete hain: Vis-viva equation $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ aur circular speed $v_{circ} = μ / r$ . Neeche sab kuch sirf yahi do hain, sahi jagah evaluate kiye gaye.

The scenario matrix

Har Hohmann problem inhi cells mein se ek hai. Jo worked examples follow karte hain, wo un cells ke saath tagged hain jo woh cover karte hain.

Cell	Situation	Kya khaas hai	$Δ v_{1}, Δ v_{2}$ ka sign	Example
A	Raise: $r_{2} > r_{1}$	canonical, dono burns prograde	$+, +$	Ex 1
B	Lower: $r_{2} < r_{1}$	mirror image, dono burns retrograde	$-, -$	Ex 2
C	Degenerate: $r_{2} = r_{1}$	koi transfer needed nahi	$0, 0$	Ex 3
D	Small hop: $r_{2} \approx r_{1}$	limiting behaviour, tiny burns	$\to 0$	Ex 4
E	Huge jump: $r_{2} ≫ r_{1}$	$Δ v_{1}$ ki escape-like limit	large $+$	Ex 5
F	Real-world word problem	Mars-mission style, Sun as centre	$+, +$	Ex 6
G	Exam twist: $r_{2} / r_{1}$ bahut bada	Hohmann vs bi-elliptic showdown	totals compare karo	Ex 7
H	Transfer-time question	Kepler + half-ellipse	—	Ex 8

Hum do constants reuse karte hain:

Earth: $μ_{\oplus} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ .
Sun: $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ .

Figure — Hohmann transfer — derivation, minimum energy transfer

Cell A — Orbit raise karo (canonical)

Worked example Ex 1 · LEO → GEO raise (

$r_{2} > r_{1}$ ) Earth. $r_{1} = 7000$ km, $r_{2} = 42164$ km. $Δ v_{1}$ , $Δ v_{2}$ , total nikalo.

Forecast: Dono burns tumhe speed up karenge (prograde). Guess karo: kya $Δ v_{1}$ ya $Δ v_{2}$ zyada costly hai? (Padhne se pehle guess karo.)

Step 1 — starting speed. $v_{c 1} = μ / r_{1} = 398600/7000 = 7.546$ km/s. Yeh step kyun? Yeh woh speed hai jo tumhare paas already hai; burn sirf difference ke liye pay karta hai.

Step 2 — transfer semi-major axis. $a_{t} = (7000 + 42164) /2 = 24582$ km. Yeh step kyun? Ellipse par Vis-viva ko $a = a_{t}$ chahiye, $r$ nahi.

Step 3 — ellipse par periapsis speed. $v_{p} = 398600 (\frac{2}{7000} - \frac{1}{24582}) = 9.879$ km/s. Yeh step kyun? Periapsis par ( $r = r_{1}$ ) tumhe circle se faster hona padega taaki $r_{2}$ tak swing out karne ki energy ho.

Step 4 — first burn. $Δ v_{1} = v_{p} - v_{c 1} = 9.879 - 7.546 = 2.333$ km/s. Yeh step kyun? Velocities collinear hain (tangent), isliye yeh plain subtraction hai.

Step 5 — apoapsis speed. $v_{a} = 398600 (\frac{2}{42164} - \frac{1}{24582}) = 1.640$ km/s. Yeh step kyun? Tum $r_{2}$ par bahut slow arrive karte ho wahan ek circle ke liye.

Step 6 — second burn. $v_{c 2} = 398600/42164 = 3.075$ km/s, isliye $Δ v_{2} = 3.075 - 1.640 = 1.435$ km/s.

Total: $Δ v_{total} = 2.333 + 1.435 = 3.768$ km/s.

Verify: Dono positive → dono prograde ✓ (Cell A signs se match). Pehla burn bada hai kyunki yeh well mein deep hota hai jahan speeds high hain — Oberth effect ne ise "worth more" per fuel banaya, lekin yahan yeh sirf bada required jump hai. Units: sab $(km^{3} / s^{2}) / km = km/s$ ✓.

Cell B — Orbit lower karo (mirror image)

Worked example Ex 2 · GEO → LEO lower (

$r_{2} < r_{1}$ ) Same radii, reversed: $r_{1} = 42164$ km se start, target $r_{2} = 7000$ km.

Forecast: Ab tum drop karna chahte ho. Kya tum brake karte ho, coast karte ho, phir brake — ya kahin speed up? Signs guess karo.

Step 1 — starting speed. $v_{c 1} = 398600/42164 = 3.075$ km/s. Yeh step kyun? High orbit par current speed.

Step 2 — $a_{t}$ . $(42164 + 7000) /2 = 24582$ km — Ex 1 waali wahi ellipse, doosre taraf fly ki gayi. Yeh step kyun? Transfer ellipse sirf pair ${r_{1}, r_{2}}$ par depend karta hai, travel direction par nahi, isliye yeh Ex 1 ki tarah identical hai.

Step 3 — first burn (at $r_{1} = 42164$ , ab apoapsis). Wahan ellipse par speed $v_{a} = 1.640$ km/s hai (Ex 1 ke apoapsis jaisi). $Δ v_{1} = 1.640 - 3.075 = - 1.435$ km/s. Negative kyun? Andar fall karne ke liye tumhe slow down karna padega taaki orbit ek lower periapsis par sag jaye.

Step 4 — second burn (at $r_{2} = 7000$ , ab periapsis). Wahan ellipse speed $v_{p} = 9.879$ km/s hai; target circle speed $v_{c 2} = 7.546$ km/s. $Δ v_{2} = 7.546 - 9.879 = - 2.333$ km/s. Negative kyun? Tum chhote circle ke liye bahut fast arrive karte ho aur brake karna padta hai.

Total magnitude: $∣Δ v ∣ = 1.435 + 2.333 = 3.768$ km/s.

Verify: Dono signs negative (retrograde) — exactly Cell A ka mirror ✓. Total cost Ex 1 ke barabar (same ellipse, same speed differences) ✓. Yeh general rule confirm karta hai: raising = do baar accelerate, lowering = do baar decelerate, aur cost symmetric hai.

Cell C — Degenerate: same orbit

Worked example Ex 3 ·

$r_{2} = r_{1}$ Earth, $r_{1} = r_{2} = 10000$ km. Cost nikalo.

Forecast: Koi movement needed nahi — zero expect karo. Lekin dekhte hain formulas blow up to nahi karte.

Step 1 — $a_{t}$ . $(10000 + 10000) /2 = 10000$ km $= r_{1} = r_{2}$ . Yeh step kyun? "Ellipse" degenerate hokar ek circle ban gayi hai.

Step 2 — periapsis speed. $v_{p} = μ (2/ r_{1} - 1/ a_{t}) = μ / r_{1} = v_{c 1}$ . Yeh step kyun? $a_{t} = r_{1}$ ke saath, vis-viva circular-speed formula mein reduce ho jaati hai.

Step 3 — burns. $Δ v_{1} = v_{p} - v_{c 1} = 0$ , aur identically $Δ v_{2} = 0$ .

Total: $0$ km/s.

Verify: $r_{1} = r_{2} = r$ set karo first-burn ratio $2 r_{2} / (r_{1} + r_{2}) - 1 = 2 r /2 r - 1 = 1 - 1 = 0$ mein ✓. Formula degenerate point par continuous hai — koi division-by-zero nahi, koi NaN nahi. Yahi anchor hai jis taraf "small hop" limit approach karti hai aage.

Cell D — Small hop (limiting behaviour)

Worked example Ex 4 ·

$r_{2}$ just above $r_{1}$ Earth, $r_{1} = 7000$ km, $r_{2} = 7100$ km (100 km ki raise).

Forecast: Bahut chhote burns. Guess karo: kya $Δ v_{1}$ aur $Δ v_{2}$ roughly equal hain yahan? (Tiny-hop limit mein unhe hona chahiye.)

Step 1 — $a_{t}$ . $(7000 + 7100) /2 = 7050$ km. Yeh step kyun? Vis-viva evaluations ke liye transfer ellipse set karta hai.

Step 2 — first burn. $v_{c 1} = 398600/7000 = 7.5464$ km/s; $v_{p} = 398600 (2/7000 - 1/7050) = 7.5732$ km/s. $Δ v_{1} = 0.0268$ km/s. Yeh step kyun? Ex 1 jaisi hi periapsis logic, bas ek tiny stretch.

Step 3 — second burn. $v_{c 2} = 398600/7100 = 7.4931$ km/s; $v_{a} = 398600 (2/7100 - 1/7050) = 7.4664$ km/s. $Δ v_{2} = 0.0267$ km/s. Yeh step kyun? Thodi si higher orbit par circularize karo.

Total: $\approx 0.0535$ km/s $= 53.5$ m/s.

Verify: $Δ v_{1} \approx Δ v_{2}$ $0.4%$ ke andar ✓ — small-hop limit mein split symmetric hai, jaise forecast kiya tha.

$\frac{1}{4}$ rule kahan se aata hai (ek derivation, guess nahi). Likho $r_{2} = r_{1} + Δ r$ jahan $Δ r$ tiny hai. First-burn factor $\frac{2 r _{2}}{r _{1} + r _{2}} - 1$ ko approximation $1 + x \approx 1 + \frac{x}{2}$ use karke expand karo jo small $x$ ke liye valid hai (yeh binomial expansion ka pehla term hai — yeh tool kyun? kyunki yeh ek awkward square root ko simple arithmetic mein turn karta hai $x = 0$ ke paas). Yahan $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2 ( r _{1} + Δ r )}{2 r _{1} + Δ r} = 1 + \frac{Δ r}{2 r _{1} + Δ r} \approx 1 + \frac{Δ r}{2 r _{1}}$ , isliye $x \approx \frac{Δ r}{2 r _{1}}$ aur factor $\approx \frac{x}{2} = \frac{Δ r}{4 r _{1}}$ . $v_{c 1}$ se multiply karo: $Δ v_{1} \approx v_{c 1} \cdot \frac{1}{4} \cdot \frac{Δ r}{r _{1}}$ . Yahi hai jahan $\frac{1}{4}$ rehta hai. Numerically $\frac{1}{4} (7.546) (100/7000) = 0.0270$ km/s per burn — exact $0.0268$ se match karta hai ✓. Burns smoothly zero hoti jaati hain jaise $r_{2} \to r_{1}$ , Cell C se connect karta hai.

Cell E — Huge jump (escape-like limit)

Worked example Ex 5 ·

$r_{2} ≫ r_{1}$ (near-escape) Earth, $r_{1} = 7000$ km, $r_{2} = 7 \times 1 0^{6}$ km ( $r_{2} / r_{1} = 1000$ ).

Forecast: Jaise $r_{2} \to \infty$ , $Δ v_{1}$ ko us extra speed ke paas approach karna chahiye jo escape ke liye chahiye (ek parabolic kick), kyunki ellipse almost ek parabola ban jaati hai. Dekhte hain number us ceiling ki taraf creep karta hai.

Step 1 — first-burn formula ko factor karo. $Δ v_{1} = v_{p} - v_{c 1}$ se start karo. $v_{c 1} = μ / r_{1}$ ko vis-viva expression se bahar nikalo: $v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = \frac{μ}{r _{1}} 2 - \frac{r _{1}}{a _{t}} = v_{c 1} 2 - \frac{r _{1}}{a _{t}} .$ Kyunki $a_{t} = (r_{1} + r_{2}) /2$ , humhare paas $\frac{r _{1}}{a _{t}} = \frac{2 r _{1}}{r _{1} + r _{2}}$ hai, isliye $2 - \frac{r _{1}}{a _{t}} = \frac{2 r _{2}}{r _{1} + r _{2}}$ . Therefore $Δ v_{1} = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) .$ Yeh step kyun? $v_{c 1}$ ko factor out karne se poori $r_{2}$ -dependence ek bracket ke andar isolate ho jaati hai, aur yahi exactly woh hai jiska hum $r_{2} \to \infty$ limit lena chahte hain.

Step 2 — numbers plug in karo. $r_{2} / r_{1} = 1000$ ke saath: $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2000}{1001} = 1.998$ , isliye bracket $= 1.998 - 1 = 0.4135$ . $Δ v_{1} = 7.5464 \times 0.4135 = 3.120$ km/s. Yeh step kyun? Poori $r_{2}$ dependence us ek bracket mein hai.

Step 3 — the ceiling. Jaise $r_{2} \to \infty$ , $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , bracket $\to 2 - 1 = 0.4142$ . Isliye $Δ v_{1}^{m a x} = v_{c 1} (2 - 1) = 7.5464 \times 0.4142 = 3.126$ km/s. Yeh step kyun? $2 v_{c 1}$ exactly escape (parabolic) speed hai; $Δ v_{1}$ us burn se zyada kabhi nahi ho sakti jo barely escape kare.

Step 4 — second burn. $r_{2} = 7 \times 1 0^{6}$ km par, $v_{c 2} = 398600/7 \times 1 0^{6} = 0.2387$ km/s. Transfer ellipse par apoapsis speed $v_{a} = 398600 (2/7 \times 1 0^{6} - 1/3503500) = 0.2385$ km/s hai (yahan almost circular). $Δ v_{2} = 0.2387 - 0.2385 = 0.00024$ km/s. Yeh step kyun? Near-parabolic ellipse ke apoapsis par ship itna fast ya slow barely hi move kar rahi hoti hai huge outer circle se compare mein, isliye second burn tiny hai.

Verify: $Δ v_{1} = 3.120$ km/s escape ceiling $3.126$ km/s ke $0.2%$ ke andar hai ✓ — "almost-parabola" picture confirm karta hai. Doosra burn $Δ v_{2} \approx 0.24$ m/s negligible hai — almost saari cost pehle burn ki hai. Yahi exactly woh regime hai jahan ek Bi-elliptic transfer matter karna shuru karta hai (Cell G).

Cell F — Real-world word problem

Worked example Ex 6 · Earth → Mars heliocentric transfer

Ek probe Sun ke around Earth ki orbit ( $r_{1} = 1.496 \times 1 0^{8}$ km) se Mars ki orbit ( $r_{2} = 2.279 \times 1 0^{8}$ km) ke liye nikalta hai. $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ use karo. Do heliocentric burns aur total nikalo.

Forecast: Yeh classic "≈2.9 + ≈2.6" km/s heliocentric numbers hain. Guess karo is baar pehla ya doosra burn bada hai.

Step 1 — Earth ki orbital speed. $v_{c 1} = 1.327 \times 1 0^{11} /1.496 \times 1 0^{8} = 29.78$ km/s. Yeh step kyun? Yoh speed hai jo probe already Sun ke around Earth ke saath share karta hai.

Step 2 — $a_{t}$ . $(1.496 + 2.279) \times 1 0^{8} /2 = 1.8875 \times 1 0^{8}$ km. Yeh step kyun? Do vis-viva evaluations ke liye transfer ellipse fix karta hai.

Step 3 — first burn. $v_{p} = 1.327 \times 1 0^{11} (2/1.496 \times 1 0^{8} - 1/1.8875 \times 1 0^{8}) = 32.73$ km/s. $Δ v_{1} = 32.73 - 29.78 = 2.94$ km/s. Yeh step kyun? Speed up karo taaki probe ka aphelion Mars tak reach kare.

Step 4 — Mars circular & apoapsis speeds. $v_{c 2} = 1.327 \times 1 0^{11} /2.279 \times 1 0^{8} = 24.13$ km/s; $v_{a} = 1.327 \times 1 0^{11} (2/2.279 \times 1 0^{8} - 1/1.8875 \times 1 0^{8}) = 21.48$ km/s. $Δ v_{2} = 24.13 - 21.48 = 2.65$ km/s. Yeh step kyun? Mars ki orbit mein circularize karo; tum slow arrive karte ho aur speed up karna padta hai.

Total (heliocentric): $5.59$ km/s.

Verify: Dono prograde (Cell A pattern) ✓; pehla burn bada ✓ — textbook Earth→Mars values (~2.9 aur ~2.6 km/s) se consistent. Note yeh Earth/Mars gravity wells ignore karta hai; ek poora Delta-v budget unhe bhi add karta hai. Units check: km/s throughout ✓.

Cell G — Exam twist: Hohmann vs bi-elliptic

Worked example Ex 7 · Hohmann kab haarta hai? (

$r_{2} / r_{1}$ bada) Earth. $r_{1} = 7000$ km, $r_{2} = 112000$ km (ratio $R = 16$ ). Hohmann ko ek bi-elliptic se compare karo jo $r_{2}$ par drop karne se pehle $r_{b} = 560000$ km tak fly karta hai.

Forecast: Threshold $R \approx 11.94$ hai. Kyunki $16 > 11.94$ , guess karo: bi-elliptic should win. Numbers se confirm karte hain — aur ek twist ke liye watch karo.

Step 1 — Hohmann total. $v_{c 1} = 398600/7000 = 7.546$ km/s. $Δ v_{1} = 7.546 (2 \cdot 112000/119000 - 1) = 7.546 (1.8824 - 1) = 2.808$ km/s. $v_{c 2} = 398600/112000 = 1.887$ km/s; $Δ v_{2} = 1.887 (1 - 2 \cdot 7000/119000) = 1.887 (1 - 0.3429) = 1.240$ km/s. Hohmann total $= 4.048$ km/s. Yeh step kyun? Baseline two-burn cost jo beat karni hai.

Step 2 — Bi-elliptic (teen burns via $r_{b} = 560000$ km). Ek bi-elliptic do bade ellipses use karta hai: pehle ek far point $r_{b}$ tak, phir wapas $r_{2}$ mein.

Burn 1 ( $r_{1}$ par, outbound ellipse par jahan $a_{A} = (r_{1} + r_{b}) /2 = 283500$ km): $v = 398600 (2/7000 - 1/283500) = 10.578$ km/s; $Δ v_{A} = 10.578 - 7.546 = 3.032$ km/s. Yeh sub-step kyun? Well mein deep ek fast burn (Oberth effect) tumhe $r_{b}$ tak fling kar deta hai — yeh mehnga wala hai.
Burn 2 ( $r_{b}$ par, doosre end ko $r_{1}$ se $r_{2}$ tak raise karna; second ellipse $a_{B} = (r_{b} + r_{2}) /2 = 336000$ km): $r_{b}$ par outbound ellipse par speed $398600 (2/560000 - 1/283500) = 0.4676$ km/s hai; inbound ellipse par needed $398600 (2/560000 - 1/336000) = 0.9926$ km/s; $Δ v_{B} = 0.9926 - 0.4676 = 0.5250$ km/s. Yeh sub-step kyun? Way out $r_{b}$ par tum barely move kar rahe ho, isliye ek chhoti nudge saste mein orbit ke near-side ko $r_{1}$ se $r_{2}$ tak lift kar deti hai — yahi woh jagah hai jahan bi-elliptic fuel bachata hai.
Burn 3 ( $r_{2}$ par, circularize — ek brake): $r_{2} = 112000$ par inbound ellipse par speed $398600 (2/112000 - 1/336000) = 2.398$ km/s; $v_{c 2} = 1.887$ km/s; $Δ v_{C} = 2.398 - 1.887 = 0.511$ km/s (retrograde). Yeh sub-step kyun? $r_{b}$ se wapas fall karte waqt tum $r_{2}$ circle ke liye bahut fast arrive karte ho, isliye brake karna padta hai — teen mein se yeh akela retrograde burn hai.

Bi-elliptic total $= 3.032 + 0.525 + 0.511 = 4.068$ km/s.

Step 3 — result dhyan se padho. Hohmann $= 4.048$ km/s, yeh bi-elliptic $= 4.068$ km/s. $R = 16 > 11.94$ hone ke bawajood, yeh particular bi-elliptic thoda bura hai. Twist kyun? " $R > 11.94$ " rule sirf guarantee karta hai ki ek sufficiently distant $r_{b}$ exist karta hai jo Hohmann ko beat kare — yeh nahi kehta ki $r_{b}$ ka har choice win kare. Hamara $r_{b} = 560000$ km ( $= 5 r_{2}$ ) kaafi door nahi hai. $r_{b} \to \infty$ push karo aur bi-elliptic total Hohmann se neeche gir jaata hai.

Verify: $r_{b} = 560000$ km ke saath totals $4.048$ vs $4.068$ km/s hain — Hohmann abhi bhi $20$ m/s se jeet raha hai ✓ (machine check dekho). $r_{b} = 1 0^{9}$ km ke saath doosra check ek bi-elliptic total Hohmann se below deta hai ✓, threshold rule confirm karta hai ki yeh best bi-elliptic ke baare mein hai, ek vastly longer flight time ki keemat par. Exam trap dodged: " $R > 11.94$ " ⇒ ek behtar bi-elliptic exist karta hai, yeh nahi ki koi bhi given wala sasta hai.

Cell H — Transfer-time question

Worked example Ex 8 · LEO→GEO trip kitna time leta hai?

Ex 1 ka transfer ellipse use karo: $a_{t} = 24582$ km, $μ_{\oplus} = 398600 km^{3} / s^{2}$ .

Forecast: Tum ellipse ka aadha fly karte ho, isliye time $\frac{1}{2}$ full orbital period hona chahiye. Guess karo: kuch ghante.

Step 1 — Kepler's Third Law se period. $T = 2 π a_{t}^{3} / μ = 2 π 2458 2^{3} /398600$ . Yeh step kyun? Kepler's Third Law ek orbit ki period ko sirf uske semi-major axis se jodta hai — koi doosri orbit shape info needed nahi. Yeh sahi tool hai kyunki hum sirf $a_{t}$ jaante hain.

Step 2 — uska aadha lo. Ek Hohmann arc periapsis se apoapsis tak run karta hai — exactly ellipse ka aadha — isliye $t_{transfer} = \frac{T}{2} = π \frac{a _{t}^{3}}{μ} = π \frac{2458 2 ^{3}}{398600} = π 3.727 \times 1 0^{7} = π (6105) = 19180 s .$ Yeh step kyun? Tum poori ellipse fly nahi karte — tum doosre burn ke saath apoapsis par isko chhhod dete ho, isliye sirf aadhi period elapse hoti hai.

Step 3 — ghanton mein convert karo. $19180 s \div 3600 = 5.33$ h. Yeh step kyun? Mission planner ke liye human-readable units.

Verify: $19180$ s $= 5.33$ h ✓ — GEO ke liye ek reasonable half-day-ish transfer, published GTO coast times se match karta hai. Units: $km^{3} / (km^{3} / s^{2}) = s^{2} = s$ ✓.

Recall Answers cover karo — kya tumne har cell hit ki?

Raise vs lower: signs? ::: Raise = dono burns positive (prograde); lower = dono negative (retrograde); magnitudes identical hain.
$r_{2} = r_{1}$ par kya hota hai? ::: $a_{t} = r_{1}$ , dono $Δ v = 0$ ; formulas finite rehte hain.
$r_{2} \to \infty$ ke saath $Δ v_{1}$ par ceiling? ::: $v_{c 1} (2 - 1)$ — escape (parabolic) kick.
Threshold jahan bi-elliptic Hohmann ko beat karta hai? ::: $r_{2} / r_{1} ≳ 11.94$ — lekin sirf ek sufficiently distant $r_{b}$ hi saving realise karta hai.
Transfer time formula? ::: $t = π a_{t}^{3} / μ$ (period ka aadha).