Hohmann transfer — derivation, minimum energy transfer: Question bank

3.2.19 · D5 · Physics › Orbital Mechanics & Astrodynamics › Hohmann transfer — derivation, minimum energy transfer

Neeche diye har item ke liye teen tools pehle se trusted hone chahiye: Vis-viva equation ( $v^{2} = μ (2/ r - 1/ a)$ ), Kepler's Third Law (period sirf $a$ par depend karta hai), aur Orbital energy & semi-major axis ka link ( $ε = - μ /2 a$ ). Agar koi bhi symbol unfamiliar lage, pehle woh notes padho — yeh page unhe use karta hai, rebuild nahi karta.

True or false — justify

Answers cover karo. Justification bolein, sirf verdict nahi.

Transfer ellipse ke apoapsis par aap outer circular speed se zyada fast chal rahe hote hain.

False. Apoapsis par aap ellipse ke slow end par hote hain aur us radius ki circular speed se neeche hote hain (

v_{a} < v_{c 2}

); yahi gap exact reason hai ki ek second prograde burn ki zaroorat padti hai.

Apna orbit raise karna matlab aapko slow karna hoga kyunki aap "climb" kar rahe hain.

False. Orbit raise karne ke liye energy add karni padti hai, yaani prograde burn se speed up karna hota hai; climb karte waqt coasting slow-down apne aap ho jaata hai, lekin raise mein har intentional burn ek acceleration hi hota hai.

Ek chhoti se badi orbit mein jaate waqt dono Hohmann burns prograde hoti hain.

True. Periapsis burn energy add karke circle ko ellipse mein stretch karta hai, aur apoapsis burn phir energy add karke circularize karta hai oopar — dono velocity ke saath direction mein hoti hain.

Hohmann transfer kisi bhi do circular orbits ke beech sabse sasta possible tarika hai.

Aam taur par False. Yeh sabse sasta two-impulse tangential transfer hai, aur radius ratio

r_{2} / r_{1} \approx 11.94

tak yahi jeetta hai; usse zyada ratio par ek Bi-elliptic transfer kam cost mein ho sakta hai.

Transfer ellipse par periapsis par vis-viva mein

a = r_{1}

plug karna chahiye.

False. Ellipse par

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

har jagah hota hai; sirf radius

r

change hota hai

r_{1}

aur

r_{2}

ke beech.

a = r

use karne par aap khaamoshi se apni ellipse ko wapas circle bana dete hain.

Ek Hohmann transfer alag-alag planes mein hone wali do orbits ke beech kaam karta hai.

Jaise likha hai False. Classic derivation coplanar orbits assume karta hai; plane change ke liye ek extra out-of-plane

Δ v

chahiye jo Delta-v budget mein add hoti hai.

Kyunki dono burns par velocities tangent hain, aap speeds ko plain numbers ki tarah subtract kar sakte hain.

True. Tangency ka matlab hai old aur new velocity vectors collinear hain, isliye

Δ v = ∣ v_{new} ∣ - ∣ v_{old} ∣

exact hai — yahi reason hai ki tangent burns choose ki jaati hain.

Transfer time launch site ya spacecraft ki mass par depend karta hai.

False. Half-period

π a_{t}^{3} / μ

hai: yeh sirf

a_{t}

aur central body ke

μ

par depend karta hai, vehicle ki mass par nahi (Two-body problem chhoti mass ko cancel kar deta hai).

Ek Hohmann transfer orbit ko lower bhi kar sakti hai.

True. Ise reverse mein chalao: outer circle par retrograde burn karo taaki ek ellipse mein aao jiska apoapsis

r_{2}

ho aur periapsis

r_{1}

, phir periapsis par retrograde burn karo neeche circularize karne ke liye. Tab dono burns retrograde hoti hain.

Spot the error

Har line mein ek flawed statement hai; reveal mein flaw ka naam hai.

"Aap apoapsis par brake karte hain taaki badi circle mein slow ho jao."

Error brake word mein hai. Apoapsis par

v_{a} < v_{c 2}

hota hai, isliye aapko prograde fire karna hota hai speed up karne ke liye; braking se aap wapas periapsis ki taraf gir jaate.

"Transfer ellipse ka semi-major axis

r_{2}

hai kyunki apoapsis

r_{2}

tak pahunchta hai."

Apoapsis radius ko semi-major axis se confuse kar raha hai. Full major axis

r_{1} + r_{2}

hai, isliye

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

, jo hamesha

r_{2}

se kam hota hai.

"Pehla burn inner circle par kahin bhi karo; geometry symmetric hai."

Burn point periapsis ban jaata hai, isliye ellipse ka low point wahan fix ho jaata hai. Yeh theek hai — lekin iska high point (apoapsis)

18 0^{\circ}

door padta hai, isliye departure timing taaki apoapsis target ke saath coincide kare does matter karta hai (phasing).

"Inner circle se ek single burn directly outer circle par le ja sakta hai."

Ek burn sirf inner circle se tangent ek ellipse produce karta hai;

r_{2}

par second burn ke bina aap seedha

r_{1}

par wapas swing ho jaate. Circles ko matched speed aur radius dono chahiye.

"Kyunki kinetic energy

\frac{1}{2} v^{2}

hai, gravity well ke andar ek burn aur ek door ka burn same

Δ v

ke liye same energy gain denge."

Oberth effect ko ignore kar raha hai. Gain

d (\frac{1}{2} v^{2}) = v d v

hai, jo current speed

v

ke saath badhta hai; wohi

Δ v

wahan zyada energy kharidta hai jahan aap already fast hain (neeche, periapsis par).

"Vis-viva kehta hai

v

sirf

r

par depend karta hai, isliye ek radius pick karne par speed fix ho jaati hai."

v^{2} = μ (2/ r - 1/ a)

a

par bhi depend karta hai. Usi

r_{1}

par, circular orbit (

a = r_{1}

) aur transfer ellipse (

a = a_{t}

) ki speeds alag hoti hain — yahi difference

Δ v_{1}

hai.

Why questions

Hohmann mein ek ki jagah do impulses kyun lagti hain?

Ek burn energy change karta hai lekin aapko ek ellipse par chhod deta hai jo

r_{1}

par wapas cross karti hai;

r_{2}

par second burn periapsis ko

r_{2}

tak uthane aur orbit ko phir se circular banane ke liye chahiye.

Burns ko velocity se tangent kyun rakhte hain, kisi angle par kyun nahi?

Tangential burn apna saara

Δ v

speed change karne mein kharach karta hai (hence energy), koi bhi velocity redirect karne mein nahi, isliye turning mein koi fuel waste nahi hota — yahi har burn ko minimal banata hai.

Pehla burn inner (low) orbit par kyun kiya jaata hai, outward coast karne ke baad kyun nahi?

Oberth effect: aap gravity well mein deepest point par sabse fast chal rahe hote hain, isliye wahan ek diya hua

Δ v

orbital energy

ε

mein sabse bada gain yield karta hai.

Transfer time

r_{1}

aur

r_{2}

par alag-alag kyun depend nahi karta, sirf unke sum par kyun?

Kepler's Third Law period ko

a^{3}

se jodata hai, aur

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

hai, isliye half period

π a_{t}^{3} / μ

sirf combination

r_{1} + r_{2}

dekhta hai.

Bade raise ke liye

Δ v_{2}

(circularizing burn) usually

Δ v_{1}

se chhhota kyun hota hai?

r_{2}

par door hone par har speed chhoti hoti hai, isliye do chhhoti speeds ka difference

v_{c 2} - v_{a}

bhi chhota hai;

r_{1}

ke paas speeds badi hoti hain isliye unka difference

v_{p} - v_{c 1}

bada hota hai.

Bi-elliptic sirf bahut bade ratios par Hohmann ko beat karta hai, aisa kyun?

Pehle bahut door jaane se plane-raising/lowering burns wahan ho sakti hain jahan speeds tiny hain (extreme Oberth savings), lekin us door point tak pahunchne ki extra cost bhi aati hai; sirf jab

r_{2} / r_{1} ≳ 11.94

ho tab saving cost se zyada hoti hai.

Har speed formula mein spacecraft ki mass irrelevant kyun hai?

Two-body problem mein chhoti mass cancel ho jaati hai; equations specific quantities use karti hain (energy aur

μ = GM

per unit spacecraft mass), isliye trajectories mass-independent hoti hain.

Edge cases

Jab

r_{2} \to r_{1}

ho to

Δ v_{total}

aur ellipse ka kya hoga?

Ellipse original circle mein degenerate ho jaata hai,

a_{t} \to r_{1}

, aur dono burns zero tak shrink ho jaate hain — aap already wahan hain, koi maneuver nahi chahiye.

Jab

r_{2} \to \infty

(escape-like) ho to transfer kaisi dikhti hai?

Periapsis speed

v_{p} \to 2 μ / r_{1}

, local escape speed, ban jaati hai, isliye

Δ v_{1}

escape burn ke paas pahunch jaata hai; ellipse ek ever-flatter, ever-longer path ban jaata hai aur transfer time

\to \infty

ho jaata hai.

Agar

r_{2} < r_{1}

ho (target start orbit ke andar hai), kya tab bhi Hohmann hai?

Haan — yahi maneuver reverse mein hai. Periapsis ab outer burn point ka opposite hai; dono burns retrograde hain aur aap andar girne ke liye energy khoते hain.

Kya Hohmann do elliptical (non-circular) orbits ke beech defined hai?

Classic Hohmann do circular coplanar orbits ke beech hoti hai. Ellipses ke beech transfer ek zyada general problem hai; clean two-tangent-burn result circular endpoints assume karta hai.

Exactly

r_{2} / r_{1} = 11.94

par kaun sa transfer sasta hai?

Dono essentially tied hain — yahi ratio woh crossover hai jahan Hohmann aur (limiting) bi-elliptic ki cost same hoti hai; usse neeche Hohmann jeetta hai, usse oopar bi-elliptic kar sakta hai.

Agar aapka engine sirf lambe time tak burn kar sakta hai (instant nahi), kya ideal Hohmann tab bhi apply hota hai?

Exactly nahi. Derivation impulsive (instantaneous) burns assume karta hai; ek lamba finite burn thrust ko changing positions par spread karta hai, gravity losses incur karta hai jo real Delta-v budget ko ideal se oopar le jaata hai.

Recall Ek line ka self-test jaane se pehle

Yeh cover karo: woh ek reason batao kyun tangent burns scalar (vector nahi) subtraction use karne deti hain. ::: Har burn par old aur new velocity vectors collinear hote hain, isliye sirf unke magnitudes alag hote hain.