Hohmann Δv calculation — both maneuvers: Question bank

3.2.20 · D5 · Physics › Orbital Mechanics & Astrodynamics › Hohmann Δv calculation — both maneuvers

Vocabulary reminder (sab kuch parent parent note mein build kiya gaya hai):

$Δ v$ = velocity change ki magnitude jo ek burn supply karta hai.
$v_{c 1}, v_{c 2}$ = inner/outer circle par circular speeds.
$v_{p}, v_{a}$ = transfer-ellipse ki speeds periapsis (paas) aur apoapsis (door) par, vis-viva equation se.
$a_{t} = (r_{1} + r_{2}) /2$ = transfer semi-major axis; $μ$ = standard gravitational parameter.

True or false — justify

A Hohmann transfer kisi bhi do orbits ke beech kaam karta hai.

False. Yeh sirf do coplanar, circular orbits ke liye define kiya gaya hai; alag planes ya eccentric orbits ke liye extra maneuvers ya modified transfer ki zaroorat hoti hai.

Orbit raise karte waqt dono Hohmann burns prograde hote hain (speed badhti hai).

True. Periapsis par tum

v_{c 1}

se upar jaate ho tez

v_{p}

tak, aur apoapsis par dheemi

v_{a}

se upar jaate ho

v_{c 2}

tak — raise ke liye dono positive speed additions hain.

Total Δv do circular speeds ke difference ke barabar hota hai,

v_{c 1} - v_{c 2}

False. Tum kabhi seedha do circles ke beech nahi jaate; tum ek ellipse par jaate ho, isliye actual cost circle↔ellipse hai har end par:

Δ v_{1} = v_{p} - v_{c 1}

aur

Δ v_{2} = v_{c 2} - v_{a}

Orbit raise karne ke liye, pehla (periapsis) burn hamesha dono mein se bada hota hai.

Classic LEO→GEO-scale jump ke liye True hai, kyunki yeh energy inject karta hai gravity well ki gehraai mein ek door apoapsis tak pahunchne ke liye; yeh modest ratios ke liye flip ho sakta hai, isliye yeh ek tendency hai, law nahi.

Hohmann transfer sabhi radius ratios ke liye circular orbits ke beech possible cheapest two-impulse transfer hai.

False. Bahut bade ratios ke liye (

r_{2} / r_{1} ≳ 11.94

) teen burns wala bi-elliptic transfer total Δv mein isse beat kar sakta hai.

Target orbit ko baar baar bada karne se total Δv bina kisi limit ke badhta rehta hai.

False. Jab

r_{2} \to \infty

Δ v_{1} \to v_{c 1} (2 - 1)

(ek escape-jaisi kick) aur

Δ v_{2} \to 0

, isliye total ek finite limit ke paas pahunchta hai jo escape budget ke kareeb hai.

Transfer ellipse dono circles ke tangent hota hai.

True. Tangency (har end par velocity vectors parallel) exactly yahi allow karta hai ki har burn ek pure scalar speed change ho bina kisi bekar sideways thrust ke.

Hohmann transfer mein exactly transfer ellipse ki half period lagti hai.

True. Tum periapsis se apoapsis tak coast karte ho — ek half-orbit — isliye transfer time hai

t = π a_{t}^{3} / μ

Δv ek vector hai, isliye tum simply speeds subtract nahi kar sakte.

Yahan False hai. Δv generally ek vector change hai, lekin kyunki Hohmann burns tangential hain, pehle aur baad ki velocities parallel hain, isliye iska magnitude ek plain speed subtraction tak reduce ho jaata hai.

Spot the error

"Apoapsis par main transfer par aa raha hoon, isliye

Δ v_{2} = v_{a} - v_{c 2}

Sign ulta hai: apoapsis par ellipse outer circle se dheema hai, isliye tumhe speed add karni hogi —

Δ v_{2} = v_{c 2} - v_{a} > 0

"Main

v_{p}

calculate karoonga

a = r_{1}

use karke kyunki burn

r_{1}

par hai."

Galat

a

: periapsis par

r = r_{1}

hai lekin orbit ellipse hai, isliye vis-viva mein

a = a_{t} = (r_{1} + r_{2}) /2

use hota hai,

r_{1}

nahi.

"Δv spacecraft ki mass par depend karta hai, isliye ek bhaari ship ko zyada Δv chahiye."

Nahi: Δv ek per-unit-mass velocity budget hai;

m

har orbital-speed formula mein cancel ho jaata hai. Mass fuel ko affect karta hai (rocket equation ke through), Δv ko nahi.

μ = G

hai, isliye main

6.674 \times 1 0^{- 11}

plug in karoonga."

μ = GM

hai —

G

aur central body ki mass ka product. Sirf

G

use karne se planet poori tarah chhoot jaata hai aur nonsense aata hai.

"Transfer ki semi-major axis

r_{2} - r_{1}

hai."

Woh axis length bhi nahi hai. Major axis focus se periapsis+apoapsis tak span karta hai:

2 a_{t} = r_{1} + r_{2}

, isliye

a_{t} = (r_{1} + r_{2}) /2

"Orbit lower karne ke liye (

r_{2} < r_{1}

), formulas break ho jaate hain aur dono burns negative ho jaate hain."

Woh break nahi hote; geometry bas reverse ho jaati hai — dono burns retrograde ho jaate hain (speed kam hoti hai), isliye tum magnitudes lete ho aur total Δv usi raise ke jaisa hota hai jo reverse direction mein ho.

"Ek burn kai minutes leta hai, isliye mujhe burn ke dauraan gravity ko model karna chahiye."

Textbook Hohmann ke liye hum impulsive maneuver approximation use karte hain: burn ko ek point par instantaneous treat kiya jaata hai, jo valid hai jab burn time

≪

orbital period ho.

Why questions

Exactly do burns kyun hone chahiye, na zyada na kam?

Periapsis par ek burn apoapsis ko

r_{2}

tak raise karta hai; apoapsis par doosra periapsis ko

r_{2}

tak raise karta hai circularize karne ke liye. Ek burn sirf do circle conditions mein se ek ko fix kar sakta hai.

Tangential (prograde) burns yahan sabse efficient kyun hain?

Kyunki velocity already desired thrust ke parallel hai, Δv ka har bit directly speed mein add hota hai — kuch bhi velocity vector ko sideways turn karne mein kharch nahi hota.

Deep (periapsis) burn energy-wise costly kyun hai?

Well ki gehraai mein orbit sabse tez hoti hai, aur jahan tum already sabse tez move kar rahe ho wahan speed add karna sabse zyada energy khareedata hai (Oberth effect) — lekin ek door apoapsis tak pahunchna simply wahan ek badi energy injection demand karta hai.

Outer (apoapsis) burn chhota kyun hota hai?

Outer circle dheemi hai, aur transfer already lagbhag us speed tak coast kar chuka hai, isliye circularize karne ke liye sirf ek chhoti top-up ki zaroorat hai.

Specific orbital energy sirf

a

par kyun depend karta hai, orbit mein position par nahi?

Kinetic aur potential energy ellipse ke along trade off karti hain lekin unka sum conserved rehta hai; woh constant

- μ /2 a

ke barabar hai, isliye yeh sirf size se fixed hota hai (dekho Specific orbital energy).

Vis-viva circular-speed formula mein reduce kyun ho jaata hai jab

a = r

ho?

a = r

set karo:

v^{2} = μ (2/ r - 1/ r) = μ / r

, exactly

v_{c} = μ / r

— ek circle bas ek aisa ellipse hai jiska semi-major axis uski radius ke barabar hai.

Tum simply inner circle se outer circle par "jump" kyun nahi kar sakte?

Ek single instantaneous position-jump orbital mechanics ko violate karta hai; tumhe ek continuous conic path (ellipse) follow karna hoga jo physically do radii ko connect kare.

Edge cases

Agar

r_{1} = r_{2}

ho toh kya hoga?

Transfer ellipse same circle mein degenerate ho jaata hai,

a_{t} = r_{1}

, aur dono Δv's zero hain — koi maneuver ki zaroorat nahi kyunki tum already target orbit par ho.

Agar target lower hai,

r_{2} < r_{1}

Same machinery, reversed: dono burns retrograde hain (tum har end par slow down karte ho). Δv magnitudes corresponding raise se match karte hain, isliye lowering ki cost same hai jitni same do radii ke beech raising ki.

Jab

r_{2} \to \infty

Δ v_{1}

kya approach karta hai aur yeh kya represent karta hai?

Δ v_{1} \to v_{c 1} (2 - 1)

, woh extra speed jo

r_{1}

se just barely escape karne ke liye chahiye — periapsis kick ek escape-jaisa burn ban jaata hai jabki

Δ v_{2} \to 0

Agar

r_{2}

sirf thoda

r_{1}

se bada hai, toh total Δv roughly kaise behave karta hai?

Yeh smoothly zero ki taraf shrink karta hai, kyunki transfer ellipse almost starting circle hi hai; tiny radius changes tiny, near-linear Δv cost karti hain.

Agar tum different planes mein do circles ke beech Hohmann try karo toh kya hoga?

Tangency aur pure-scalar-subtraction reasoning fail ho jaata hai; tumhe ek added plane-change Δv (ek vector addition) ki zaroorat hogi, jo ise simple two-scalar-burn problem nahi rehne deta.

Agar tum same radii use karo lekin heavier planet (bada

μ

) use karo toh transfer time badlega kya?

Haan:

t = π a_{t}^{3} / μ

, isliye bada

μ

coast time ko short kar deta hai chahe geometry (

a_{t}

) unchanged ho.

Recall One-line self-test

Woh single fact jo har sign error fix karta hai ::: Raise mein, dono burns tumhari speed badhate hain ( $v_{p} > v_{c 1}$ periapsis par, $v_{c 2} > v_{a}$ apoapsis par), isliye dono Δv's positive prograde additions hain.