Hohmann transfer — derivation, minimum energy transfer

WHAT is a Hohmann transfer?

WHY tangent burns? Because a burn changes speed most efficiently when the thrust is parallel to velocity. At the tangent points the transfer ellipse and the circle share the same velocity direction, so the burn only has to change the magnitude of $v$ — no wasted sideways thrust.

The physics you build everything from

We only need three first-principles facts about the two-body problem ( $\mu = GM$ ):

1. Vis-viva equation (energy conservation for orbits): $v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$

Formula Deriving vis-viva from scratch

Specific orbital energy = kinetic + potential per unit mass: $\varepsilon = \frac{1}{2}v^2 - \frac{\mu}{r}$ Why? Newtonian gravity gives potential energy $-\mu/r$ per unit mass; energy is conserved. For any closed orbit the total energy depends only on the semi-major axis: $\varepsilon = -\frac{\mu}{2a}$ Why this form? Evaluate $\varepsilon$ at periapsis and apoapsis and use $r_p+r_a=2a$ plus angular-momentum conservation — the two eliminate to leave energy as a function of $a$ alone. Setting the two expressions for $\varepsilon$ equal: $\tfrac{1}{2}v^2 - \frac{\mu}{r} = -\frac{\mu}{2a}\;\Longrightarrow\; v^2 = \mu\!\left(\frac{2}{r}-\frac{1}{a}\right)$

2. Circular speed (special case, $r=a$ ): $v^2 = \mu\left(\frac{2}{r}-\frac{1}{r}\right)=\frac{\mu}{r}\;\Rightarrow\; v_\text{circ}=\sqrt{\frac{\mu}{r}}$

3. Semi-major axis of the transfer ellipse. Its longest diameter spans from inner to outer orbit: $2a_t = r_1 + r_2 \;\Rightarrow\; \boxed{a_t = \frac{r_1+r_2}{2}}$ Why? Periapsis $=r_1$ , apoapsis $=r_2$ , and by definition $r_p+r_a=2a$ .

HOW to get the two burns (the derivation)

Figure — Hohmann transfer — derivation, minimum energy transfer

Step 1 — Speed on the inner circle

$v_{c1}=\sqrt{\frac{\mu}{r_1}}$ Why this step? This is the speed you actually have before you touch the throttle.

Step 2 — Speed needed at periapsis of the transfer ellipse

Use vis-viva at $r=r_1$ with $a=a_t$ : $v_{p}=\sqrt{\mu\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=\sqrt{\mu\left(\frac{2}{r_1}-\frac{2}{r_1+r_2}\right)}$ Why this step? At periapsis you're closest to the planet on the ellipse, so you must be moving faster than the circle to have enough energy to swing out to $r_2$ .

Step 3 — First burn

$\boxed{\Delta v_1 = v_p - v_{c1}=\sqrt{\frac{\mu}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)}$ Why this step? Both velocities are tangent (same direction), so $\Delta v$ is just the difference of magnitudes.

Step 4 — Speed at apoapsis of the transfer ellipse

Vis-viva at $r=r_2$ : $v_{a}=\sqrt{\mu\left(\frac{2}{r_2}-\frac{2}{r_1+r_2}\right)}$ This is slower than the outer circular speed — you arrive at $r_2$ having "run out of steam."

Step 5 — Second burn to circularize

$v_{c2}=\sqrt{\frac{\mu}{r_2}}$ $\boxed{\Delta v_2 = v_{c2}-v_a=\sqrt{\frac{\mu}{r_2}}\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right)}$ Why this step? You need to speed up to hold the larger circle.

Total cost

$\Delta v_\text{total}=\Delta v_1+\Delta v_2$

WHY is it minimum energy? (Forecast-then-verify)

The Oberth insight (WHY tangential + low-altitude burns win): kinetic energy is $\frac12 v^2$ . A fixed $\Delta v$ added where you're already fast (deep in the gravity well, at periapsis) buys more orbital energy: $d(\frac12 v^2)=v\,dv$ grows with $v$ . That's why the first burn is placed at $r_1$ .

Worked Examples

Worked example LEO → GEO (canonical)

$\mu_\oplus = 3.986\times10^{5}\,\text{km}^3/\text{s}^2$ , $r_1=7000$ km, $r_2=42164$ km.

Step 1 $v_{c1}=\sqrt{398600/7000}=7.546$ km/s. Why? current speed. Step 2 $a_t=(7000+42164)/2=24582$ km. Step 3 $v_p=\sqrt{398600(2/7000-1/24582)}=10.10$ km/s → $\Delta v_1=10.10-7.55=\mathbf{2.55}$ km/s. Step 4 $v_a=\sqrt{398600(2/42164-1/24582)}=1.677$ km/s. Step 5 $v_{c2}=\sqrt{398600/42164}=3.075$ km/s → $\Delta v_2=3.075-1.677=\mathbf{1.40}$ km/s. Total $\Delta v\approx\mathbf{3.95}$ km/s. Transfer time $=\pi\sqrt{24582^3/398600}\approx 19100$ s ≈ 5.3 h.

Common Mistakes

Recall

Recall Explain to a 12-year-old (Feynman)

Imagine you're on a merry-go-round and want to jump to a bigger merry-go-round outside you. You can't step straight out — you'd fly off. So you give one hard push to swing out on a big loopy path that just reaches the outer ring, and when you get there you're moving too slowly to stay on it, so you give one more push to catch up to its speed. Two pushes, cheapest ride. That looping path is a squished circle (an ellipse), and it's the lazy, fuel-saving way to change orbits.

Recall Active recall — cover the answers

Why two burns and not one? → Because one burn only stretches the circle into an ellipse; a second is needed to re-circularize at the new radius.
Where are the burns applied? → At periapsis ( $r_1$ ) and apoapsis ( $r_2$ ) of the transfer ellipse, tangentially.
What is $a_t$ ? → $(r_1+r_2)/2$ .
When does bi-elliptic beat Hohmann? → When $r_2/r_1 \gtrsim 11.94$ .

Flashcards

What is a Hohmann transfer?

A two-impulse, minimum-energy maneuver between two coplanar circular orbits via an ellipse tangent to both (periapsis at

r_1

, apoapsis at

r_2

Vis-viva equation?

v^2=\mu(2/r-1/a)

, derived from

\varepsilon=\tfrac12 v^2-\mu/r=-\mu/2a

Semi-major axis of the transfer ellipse?

a_t=(r_1+r_2)/2

Formula for the first burn

\Delta v_1

\sqrt{\mu/r_1}\left(\sqrt{2r_2/(r_1+r_2)}-1\right)

Formula for the second burn

\Delta v_2

\sqrt{\mu/r_2}\left(1-\sqrt{2r_1/(r_1+r_2)}\right)

Why tangential burns?

Velocities are collinear, so

\Delta v

is a pure magnitude change — no thrust wasted turning.

Transfer time?

Half the ellipse period:

t=\pi\sqrt{a_t^3/\mu}

At apoapsis do you speed up or slow down to circularize when raising?

Speed up (prograde), since

v_a<v_{c2}

When does bi-elliptic beat Hohmann?

For radius ratios

r_2/r_1\gtrsim 11.94

Why is periapsis the best place for the big burn (Oberth)?

Energy gain

=v\,dv

; adding

\Delta v

where

v

is large yields more orbital energy per fuel.

Connections

Vis-viva equation
Kepler's Third Law
Orbital energy & semi-major axis
Oberth effect
Bi-elliptic transfer
Delta-v budget
Two-body problem

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Hohmann transfer ka basic idea simple hai: tumhe ek chhoti circular orbit se badi circular orbit me jana hai (jaise LEO se GEO), aur tum minimum fuel me jana chahte ho. Straight bahar nahi jaa sakte kyunki gravity ek curved track hai. Toh trick yeh hai — do baar engine fire karo. Pehla burn periapsis pe, jisse tumhari circle stretch hoke ek ellipse ban jaati hai jiska door wala point ( $r_2$ ) exactly badi orbit ko chhuta hai. Coast karke wahan pahuncho, phir doosra burn karke wapas circle bana lo.

Sab kuch do formulas se nikalta hai. Ek hai vis-viva: $v^2=\mu(2/r-1/a)$ — yeh energy conservation hai, kuch yaad karne ki zarurat nahi, derive ho jata hai from $\varepsilon=\frac12 v^2-\mu/r=-\mu/2a$ . Doosra: transfer ellipse ka semi-major axis $a_t=(r_1+r_2)/2$ , kyunki uska periapsis $r_1$ aur apoapsis $r_2$ hai. Bas vis-viva me $r_1$ aur $a_t$ daal ke $v_p$ nikalo, circular speed se subtract karo, mil gaya $\Delta v_1$ . Same cheez $r_2$ pe karo for $\Delta v_2$ .

Ek important galti yaad rakhna: apoapsis pe log sochte hain "upar aaye hain toh brake lagana hoga." Galat! Apoapsis pe tum circle ke liye zaroorat se kam speed pe ho ( $v_a<v_{c2}$ ), toh tumhe prograde fire karke speed badhani padegi. Ulta tabhi hoga jab badi se chhoti orbit me aa rahe ho.

Yeh matter kyun karta hai? Kyunki har satellite mission me fuel = paisa aur weight. Hohmann tumhe do sabse chhote burns deta hai (tangential burns, kyunki velocity same direction me hoti hai, koi thrust turning me waste nahi hota). Sirf jab ratio $r_2/r_1$ bahut bada ho (~11.94 se zyada), tab bi-elliptic transfer thoda better hota hai. Practical missions ke liye Hohmann hi king hai.

Hohmann transfer — derivation, minimum energy transfer

WHAT is a Hohmann transfer?

The physics you build everything from

HOW to get the two burns (the derivation)

Step 1 — Speed on the inner circle

Step 2 — Speed needed at periapsis of the transfer ellipse

Step 3 — First burn

Step 4 — Speed at apoapsis of the transfer ellipse

Step 5 — Second burn to circularize

Total cost

WHY is it minimum energy? (Forecast-then-verify)

Worked Examples

Common Mistakes

Recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Orbital Mechanics & Astrodynamics

Connections