Hohmann Δv calculation — both maneuvers: Exercises

3.2.20 · D4 · Physics › Orbital Mechanics & Astrodynamics › Hohmann Δv calculation — both maneuvers

Throughout, hum yeh toolkit reuse karte hain:

Circular speed (from Circular orbital velocity): $v_{circ} = μ / r$ .
Vis-viva equation: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ .
Transfer ellipse: $a_{t} = \frac{r _{1} + r _{2}}{2}$ .
$μ$ central body ka Standard gravitational parameter $= GM$ hai.

Jab tak problem mein alag na bataya jaye, Earth ke liye $μ_{\oplus} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ use karo.

Figure — Hohmann Δv calculation — both maneuvers

Woh picture har problem ka map hai: ek inner circle (radius $r_{1}$ , chalk blue), ek outer circle (radius $r_{2}$ , pale yellow), aur pink transfer ellipse jo dono ko touch karti hai. Burn 1 inner circle par pink dot par hoti hai; burn 2 outer circle par pink dot par.

Level 1 — Recognition

L1.1

$Δ v_{1}$ kaun si do speeds ko compare karta hai, aur kis radius par?

Recall Solution

KYA: $Δ v_{1} = v_{p} - v_{c 1}$ , dono radius $r_{1}$ par evaluate ki jaati hain. KYUN: Burn 1 inner circle par fire hoti hai. Burn se pehle tum circular speed $v_{c 1} = μ / r_{1}$ par move kar rahe ho; burn ke baad tum ellipse par ho, isliye tumhari speed ellipse ki periapsis speed $v_{p}$ hai. Dono " $r_{1}$ par" hain, isliye dono velocity vectors parallel (tangential) hain aur vector subtraction speeds ki subtraction mein collapse ho jaati hai. Answer: yeh circular speed $v_{c 1}$ aur transfer periapsis speed $v_{p}$ ko compare karta hai, dono $r_{1}$ par.

L1.2

$r_{2} > r_{1}$ ke liye, transfer ellipse apoapsis par outer circle se faster hai ya slower? Isliye, burn 2 prograde (speed up) hai ya retrograde (slow down)?

Recall Solution

KYA: $v_{a}$ (apoapsis par ellipse) ko $v_{c 2}$ (outer circle) se compare karo. KYUN: Apoapsis par spacecraft ellipse par apne slowest point par hoti hai (focus se sabse door, energy height mein trade ho gayi). Us same radius par circle ko circular rehne ke liye zyada speed chahiye. Toh $v_{a} < v_{c 2}$ . Answer: ellipse apoapsis par slower hai; burn 2 tumhe speed up karni chahiye (prograde), jo $Δ v_{2} = v_{c 2} - v_{a} > 0$ deta hai.

Level 2 — Application

L2.1

Earth transfer with $r_{1} = 7000$ km, $r_{2} = 20000$ km. $a_{t}$ , $v_{c 1}$ , $v_{p}$ , aur $Δ v_{1}$ compute karo.

Recall Solution

Step 1 — $a_{t}$ : $a_{t} = \frac{7000 + 20000}{2} = 13500$ km. (periapsis + apoapsis ka midpoint) Step 2 — $v_{c 1}$ : $v_{c 1} = 398600/7000 = 7.546$ km/s. Step 3 — $v_{p}$ (vis-viva at $r = r_{1}$ ): $v_{p} = 398600 (\frac{2}{7000} - \frac{1}{13500}) = 9.187 km/s .$ Step 4 — $Δ v_{1}$ : $9.187 - 7.546 = 1.641$ km/s.

L2.2

L2.1 jaisi hi orbits. $v_{c 2}$ , $v_{a}$ , aur $Δ v_{2}$ compute karo, phir total bhi.

Recall Solution

$v_{c 2}$ : $398600/20000 = 4.465$ km/s. $v_{a}$ (vis-viva at $r = r_{2}$ ): $398600 (\frac{2}{20000} - \frac{1}{13500}) = 3.213$ km/s. $Δ v_{2}$ : $4.465 - 3.213 = 1.252$ km/s. Total: $1.641 + 1.252 = 2.893$ km/s.

L2.3

Factored forms use karke L2.1 se $Δ v_{1}$ confirm karo, $v_{p}$ alag se compute kiye bina: $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) .$

Recall Solution

KYA: directly substitute karo. KYUN: factored form algebraically $v_{p} - v_{c 1}$ ke identical hai; bas vis-viva step hide ho jaata hai. $Δ v_{1} = 7.546 (\frac{2 \cdot 20000}{27000} - 1) = 7.546 (1.2172 - 1) = 1.641 km/s .$ L2.1 se exactly match karta hai. ✔

Level 3 — Analysis

L3.1

LEO→GEO ( $r_{1} = 6678$ km, $r_{2} = 42164$ km) ke liye parent note ne $Δ v_{1} \approx 2.42$ , $Δ v_{2} \approx 1.47$ km/s find kiya. Energy terms mein explain karo kyun pehla burn bada hota hai, aur batao yeh kis physical effect se connect karta hai.

Recall Solution

KYA: $Δ v_{1} > Δ v_{2}$ even though outer orbit slower hai. KYUN: Specific orbital energy $ε = - μ /2 a$ hai (dekho Specific orbital energy). Transfer ellipse ka $a_{t}$ , circle 1 se bahut zyada bada hai, isliye burn 1 ko ek bada energy jump inject karna padta hai. Kyunki tum $r_{1}$ par gravity well mein deep ho (speeds badi hain), ek given $Δ v$ wahan bahut energy buy karta hai — lekin tumhe bahut energy chahiye bhi, isliye required $Δ v$ bada hai. Burn 2 sirf already-high orbit ko circular tak tweak karta hai, jo ek smaller energy change hai. Connection: yeh exactly wahi reasoning hai jo Oberth effect ke peeche hai — burns deep in the well, jahan speed high hoti hai, energy-efficient hote hain, aur orbit raise karte waqt budget mein dominate bhi karte hain.

L3.2

Algebraically dikhao ki $Δ v_{1} = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ kisi bhi $r_{2} > r_{1}$ ke liye hamesha positive hai, aur yeh zero equal hota hai jab $r_{2} = r_{1}$ .

Recall Solution

KYA: bracket $B = \frac{2 r _{2}}{r _{1} + r _{2}} - 1$ ko analyse karo. KYUN $B > 0$ : $B > 0 ⟺ \frac{2 r _{2}}{r _{1} + r _{2}} > 1 ⟺ 2 r_{2} > r_{1} + r_{2} ⟺ r_{2} > r_{1}$ . Assumption se true hai, aur $v_{c 1} > 0$ , isliye $Δ v_{1} > 0$ . ✔ Degenerate case $r_{2} = r_{1}$ : phir $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2 r _{1}}{2 r _{1}} = 1$ , isliye $B = 0$ aur $Δ v_{1} = 0$ . Physical meaning: koi transfer needed nahi — tum already target circle par ho. Math sahi se self-cancel karta hai.

L3.3

Kya $r_{2}$ ko bada karna $Δ v_{1}$ ko hamesha badhata hai? $r_{2} \to \infty$ par $Δ v_{1}$ ki limit kya hai?

Recall Solution

KYA: $Δ v_{1} (r_{2})$ ka behaviour. Monotonic? Factor $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2}{r _{1} / r _{2} + 1}$ , $r_{2}$ badhne par increase karta hai (denominator shrink hota hai), isliye $Δ v_{1}$ , $r_{2}$ ke saath monotonically increase karta hai. Haan — hamesha zyada. Limit: jaise $r_{2} \to \infty$ , $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , isliye $Δ v_{1} \to v_{c 1} (2 - 1) .$ Physical meaning: yeh escape-like periapsis kick hai — inner circle se infinity tak push karne mein escape increment $(2 - 1) v_{c 1}$ lagta hai, kyunki escape speed $2 v_{c 1}$ hoti hai. Us ke baad, $Δ v_{1}$ aur nahi badhta; ek ever-larger orbit ka extra cost bounded hai.

Level 4 — Synthesis

L4.1

Ek Mars orbiter: $μ_{Mars} = 4.283 \times 1 0^{4} km^{3} / s^{2}$ , $r_{1} = 3690$ km (low Mars orbit), $r_{2} = 20428$ km (areostationary). Total Hohmann Δv find karo.

Recall Solution

$a_{t}$ : $(3690 + 20428) /2 = 12059$ km. $v_{c 1}$ : $42830/3690 = 3.407$ km/s. $v_{p}$ : $42830 (\frac{2}{3690} - \frac{1}{12059}) = 4.439$ km/s. $Δ v_{1}$ : $4.439 - 3.407 = 1.032$ km/s. $v_{c 2}$ : $42830/20428 = 1.448$ km/s. $v_{a}$ : $42830 (\frac{2}{20428} - \frac{1}{12059}) = 0.802$ km/s. $Δ v_{2}$ : $1.448 - 0.802 = 0.646$ km/s. Total: $1.678$ km/s. (Notice karo phir se $Δ v_{1} > Δ v_{2}$ — same deep-well logic jaise L3.1 mein, ab Mars ke around.)

L4.2

Orbit lowering: L2.1 reverse karo — $r_{1} = 20000$ km se $r_{2} = 7000$ km tak Earth ke around jao. Dono burns compute karo. Kya yeh prograde hain ya retrograde? Total ko L2.2 se compare karo.

Recall Solution

KYA: ab target andar hai. Transfer ellipse ka apoapsis starting (outer) radius par hai aur periapsis target (inner) radius par, isliye $a_{t}$ unchanged hai: $13500$ km. Burn 1 (outer radius $20000$ km par): tumhe ellipse par drop karna hai, jis ki speed wahan ( $v_{a} = 3.213$ km/s) wahan ki circular speed ( $4.465$ km/s) se kam hai. Isliye tum slow down karte ho: $Δ v_{1} = 4.465 - 3.213 = 1.252$ km/s — retrograde. Burn 2 (inner radius $7000$ km par): tum ellipse ke periapsis par pahunchte ho ( $v_{p} = 9.187$ km/s), inner circle ( $7.546$ km/s) se faster, isliye tum dobara slow down karte ho: $Δ v_{2} = 9.187 - 7.546 = 1.641$ km/s — retrograde. Total: $2.893$ km/s — L2.2 se identical. KYUN identical: Hohmann transfer ka total Δv budget symmetric hai — lowering mein exactly utna hi lagta hai jitna raising mein laga, bas dono burns prograde ki jagah retrograde hoti hain.

Level 5 — Mastery

L5.1

L4.2 ka symmetry result general mein prove karo: $r_{1} \to r_{2}$ jaane ka total Δv $r_{2} \to r_{1}$ jaane ke total ke barabar hota hai.

Recall Solution

KYA: dikhao ki $Δ v_{tot}$ tab unchanged rehta hai jab tum $r_{1} \leftrightarrow r_{2}$ swap karo. KYUN kaam karta hai: raising ke liye, dono burns $r_{1}$ aur $r_{2}$ par {circle, ellipse} compare karte hain. Lowering ke liye, same ellipse (same $a_{t}$ , kyunki $a_{t}$ , $r_{1}, r_{2}$ mein symmetric hai) ride hoti hai, aur dono burns fir se $r_{2}$ aur $r_{1}$ par {circle, ellipse} compare karte hain — same two speed gaps, bas retrograde execute hote hain. Formally raise total hai $∣ v_{p} - v_{c 1} ∣ + ∣ v_{c 2} - v_{a} ∣,$ aur lower total hai $∣ v_{a} - v_{c 2} ∣ + ∣ v_{c 1} - v_{p} ∣,$ jo term-by-term equal hain kyunki $∣ x ∣ = ∣ - x ∣$ . Hence equal totals. ∎

L5.2

Ek student burn 2 skip karne ka propose karta hai: sirf burn 1 fire karo $r_{2}$ par apoapsis reach karne ke liye, phir "ise ellipse rehne do." $r_{1} = 6678$ , $r_{2} = 42164$ km use karke, explain karo kaun si orbit result hoti hai, aur woh Δv compute karo jo tumhe circularize karne ke liye abhi bhi baad mein chahiye hoga.

Recall Solution

KYA sirf burn 1 se hota hai: craft transfer ellipse par hai — periapsis $r_{1}$ , apoapsis $r_{2}$ — unke beech hamesha coasting karta rehta hai (ek Impulsive maneuver approximation har burn ko instantaneous treat karta hai, isliye koi second burn nahi hone se ellipse simply repeat hoti hai). Yeh GEO par nahi hai. Apoapsis par uski speed $v_{a} = 398600 (2/42164 - 1/24421) = 1.609$ km/s hai, lekin ek GEO circle ko $v_{c 2} = 398600/42164 = 3.075$ km/s chahiye. Abhi bhi owed Δv: $Δ v_{2} = 3.075 - 1.609 = 1.467$ km/s — precisely the second burn. Ise skip karne se fuel nahi bachta; yeh bas tumhe ek ellipse par chhod deta hai, aur jab bhi tum finally circularize karo, tumhara wahi $1.467$ km/s owed hai.

L5.3

Limiting sanity check. LEO ( $r_{1} = 6678$ km, $v_{c 1} = 7.726$ km/s) ke liye, numerically verify karo ki $Δ v_{1} \to v_{c 1} (2 - 1)$ as $r_{2} \to \infty$ , aur interpret karo.

Recall Solution

Target value: $v_{c 1} (2 - 1) = 7.726 \times 0.41421 = 3.200$ km/s. Ek bahut bade $r_{2}$ se check karo, maan lo $r_{2} = 1 0^{10}$ km: $Δ v_{1} = 7.726 (\frac{2 \cdot 1 0 ^{10}}{6678 + 1 0 ^{10}} - 1) \approx 7.726 (2 - 1) = 3.200 km/s .$ Interpretation: LEO se escape speed $2 v_{c 1} = 10.926$ km/s hai, aur circular se $Δ v_{1}$ $(2 - 1) v_{c 1} = 3.200$ km/s hai. Infinitely far reach karne mein pehle burn ka koi zyada Δv nahi lagta jitna barely escape karne mein — ek beautiful bound jo motivate karta hai ki kyun, kuch radius ratio ke baad, Bi-elliptic transfer (jo $r_{2}$ se aage jaata hai aur wapas aata hai) Hohmann ko undercut kar sakta hai.

Recall Self-test summary (finish karne ke baad reveal karo)

Yahan har burn ek speed match tha un dono orbits ke beech jo actually ek radius par milte hain. Raising matlab dono burns prograde hain. ::: Lowering matlab dono burns retrograde hain, same total ke saath. $2 r / (r_{1} + r_{2})$ ke numerator mein kaun sa radius baithta hai? ::: Woh radius jis taraf tum reach kar rahe ho — near burn upar $r_{2}$ use karta hai, far burn upar $r_{1}$ use karta hai. $r_{2} \to \infty$ par $Δ v_{1}$ kya approach karta hai? ::: $v_{c 1} (2 - 1)$ , the escape-like periapsis kick.