Hohmann transfer — derivation, minimum energy transfer: Worked examples

This page is the drill hall for the Hohmann parent note. We take the formulas it built and push them through every kind of case the topic can hand you: raising an orbit, lowering an orbit, the degenerate "same orbit" case, the limit of a huge jump, a real spacecraft word problem, and an exam twist where Hohmann quietly loses to a bi-elliptic one.

Before any numbers, we fix the vocabulary so no symbol appears unexplained.

Definition The symbols we reuse everywhere

$μ$ (Greek "mu") — the gravitational parameter $μ = GM$ , the pull-strength of the central body. Units: $km^{3} / s^{2}$ .
$r_{1}$ — radius of the starting circular orbit (distance from planet centre).
$r_{2}$ — radius of the target circular orbit.
$a_{t}$ — semi-major axis of the transfer ellipse, its half-longest-diameter, $a_{t} = (r_{1} + r_{2}) /2$ .
$v_{c 1}, v_{c 2}$ — the circular speeds on the start and target orbits, $μ / r_{1}$ and $μ / r_{2}$ .
$v_{p}$ — the periapsis speed on the transfer ellipse: how fast the ship is moving at the point of the ellipse closest to the planet (radius $= r_{1}$ for a raise). Computed from vis-viva with $r = r_{1}, a = a_{t}$ .
$v_{a}$ — the apoapsis speed on the transfer ellipse: how fast the ship is moving at the point farthest from the planet (radius $= r_{2}$ for a raise). Computed from vis-viva with $r = r_{2}, a = a_{t}$ .
$Δ v$ (read "delta-vee") — a change in speed you buy with fuel. Positive means speed up, negative means slow down.

The two engines we lean on are the Vis-viva equation $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ and the circular speed $v_{circ} = μ / r$ . Everything below is just these two, evaluated at the right place.

The scenario matrix

Every Hohmann problem is one of these cells. The worked examples that follow are tagged with the cell they cover.

Cell	Situation	What's special	Sign of $Δ v_{1}, Δ v_{2}$	Example
A	Raise: $r_{2} > r_{1}$	canonical, both burns prograde	$+, +$	Ex 1
B	Lower: $r_{2} < r_{1}$	mirror image, both burns retrograde	$-, -$	Ex 2
C	Degenerate: $r_{2} = r_{1}$	no transfer needed	$0, 0$	Ex 3
D	Small hop: $r_{2} \approx r_{1}$	limiting behaviour, tiny burns	$\to 0$	Ex 4
E	Huge jump: $r_{2} ≫ r_{1}$	escape-like limit of $Δ v_{1}$	large $+$	Ex 5
F	Real-world word problem	Mars-mission style, Sun as centre	$+, +$	Ex 6
G	Exam twist: $r_{2} / r_{1}$ very large	Hohmann vs bi-elliptic showdown	compare totals	Ex 7
H	Transfer-time question	Kepler + half-ellipse	—	Ex 8

We reuse two constants:

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 — Raise an orbit (canonical)

Worked example Ex 1 · LEO → GEO raise (

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

Forecast: Both burns speed you up (prograde). Guess: does $Δ v_{1}$ or $Δ v_{2}$ cost more? (Guess before reading.)

Step 1 — starting speed. $v_{c 1} = μ / r_{1} = 398600/7000 = 7.546$ km/s. Why this step? This is the speed you already have; the burn only pays for the difference.

Step 2 — transfer semi-major axis. $a_{t} = (7000 + 42164) /2 = 24582$ km. Why this step? Vis-viva on the ellipse needs $a = a_{t}$ , not $r$ .

Step 3 — periapsis speed on the ellipse. $v_{p} = 398600 (\frac{2}{7000} - \frac{1}{24582}) = 9.879$ km/s. Why this step? At periapsis ( $r = r_{1}$ ) you must be faster than the circle to have the energy to swing out to $r_{2}$ .

Step 4 — first burn. $Δ v_{1} = v_{p} - v_{c 1} = 9.879 - 7.546 = 2.333$ km/s. Why this step? Velocities are collinear (tangent), so it's plain subtraction.

Step 5 — apoapsis speed. $v_{a} = 398600 (\frac{2}{42164} - \frac{1}{24582}) = 1.640$ km/s. Why this step? You arrive at $r_{2}$ too slow for a circle there.

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

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

Verify: Both positive → both prograde ✓ (matches Cell A signs). The first burn is larger because it happens deep in the well where speeds are high — the Oberth effect made it "worth more" per fuel, but here it's simply the bigger required jump. Units: all $(km^{3} / s^{2}) / km = km/s$ ✓.

Cell B — Lower an orbit (mirror image)

Worked example Ex 2 · GEO → LEO lower (

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

Forecast: Now you want to drop. Do you brake, coast, brake again — or speed up somewhere? Guess the signs.

Step 1 — starting speed. $v_{c 1} = 398600/42164 = 3.075$ km/s. Why this step? Current speed on the high orbit.

Step 2 — $a_{t}$ . $(42164 + 7000) /2 = 24582$ km — same ellipse as Ex 1, flown the other way. Why this step? The transfer ellipse depends only on the pair ${r_{1}, r_{2}}$ , not on the travel direction, so it's identical to Ex 1's.

Step 3 — first burn (at $r_{1} = 42164$ , now the apoapsis). Speed there on the ellipse is $v_{a} = 1.640$ km/s (same as Ex 1's apoapsis). $Δ v_{1} = 1.640 - 3.075 = - 1.435$ km/s. Why negative? To fall inward you must slow down so the orbit sags to a lower periapsis.

Step 4 — second burn (at $r_{2} = 7000$ , now periapsis). Ellipse speed there is $v_{p} = 9.879$ km/s; target circle speed $v_{c 2} = 7.546$ km/s. $Δ v_{2} = 7.546 - 9.879 = - 2.333$ km/s. Why negative? You arrive too fast for the small circle and must brake.

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

Verify: Both signs negative (retrograde) — exactly the mirror of Cell A ✓. Total cost equals Ex 1 (same ellipse, same speed differences) ✓. This confirms the general rule: raising = accelerate twice, lowering = decelerate twice, and cost is symmetric.

Cell C — Degenerate: same orbit

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 (a 100 km raise).

Forecast: Very small burns. Guess: are $Δ v_{1}$ and $Δ v_{2}$ roughly equal here? (In the tiny-hop limit they should be.)

Step 1 — $a_{t}$ . $(7000 + 7100) /2 = 7050$ km. Why this step? Sets the transfer ellipse for the vis-viva evaluations.

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. Why this step? Same periapsis logic as Ex 1, just a 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. Why this step? Circularize at the slightly higher orbit.

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

Verify: $Δ v_{1} \approx Δ v_{2}$ to within $0.4%$ ✓ — in the small-hop limit the split is symmetric, as forecast.

Where the $\frac{1}{4}$ rule comes from (a derivation, not a guess). Write $r_{2} = r_{1} + Δ r$ with $Δ r$ tiny. Expand the first-burn factor $\frac{2 r _{2}}{r _{1} + r _{2}} - 1$ using the approximation $1 + x \approx 1 + \frac{x}{2}$ valid for small $x$ (this is the first term of the binomial expansion — why this tool? because it turns an awkward square root into simple arithmetic near $x = 0$ ). Here $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2 ( r _{1} + Δ r )}{2 r _{1} + Δ r} = 1 + \frac{Δ r}{2 r _{1} + Δ r} \approx 1 + \frac{Δ r}{2 r _{1}}$ , so $x \approx \frac{Δ r}{2 r _{1}}$ and the factor $\approx \frac{x}{2} = \frac{Δ r}{4 r _{1}}$ . Multiply by $v_{c 1}$ : $Δ v_{1} \approx v_{c 1} \cdot \frac{1}{4} \cdot \frac{Δ r}{r _{1}}$ . That is where the $\frac{1}{4}$ lives. Numerically $\frac{1}{4} (7.546) (100/7000) = 0.0270$ km/s per burn — matches the exact $0.0268$ ✓. The burns shrink to zero smoothly as $r_{2} \to r_{1}$ , connecting to Cell C.

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: As $r_{2} \to \infty$ , $Δ v_{1}$ should approach the extra speed needed to escape (a parabolic kick), because the ellipse becomes almost a parabola. Let's see the number creep toward that ceiling.

Step 1 — factor the first-burn formula. Start from $Δ v_{1} = v_{p} - v_{c 1}$ . Pull $v_{c 1} = μ / r_{1}$ out of the vis-viva expression: $v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = \frac{μ}{r _{1}} 2 - \frac{r _{1}}{a _{t}} = v_{c 1} 2 - \frac{r _{1}}{a _{t}} .$ Since $a_{t} = (r_{1} + r_{2}) /2$ , we have $\frac{r _{1}}{a _{t}} = \frac{2 r _{1}}{r _{1} + r _{2}}$ , so $2 - \frac{r _{1}}{a _{t}} = \frac{2 r _{2}}{r _{1} + r _{2}}$ . Therefore $Δ v_{1} = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) .$ Why this step? Factoring $v_{c 1}$ out isolates the entire $r_{2}$ -dependence inside one bracket, which is exactly what we want to take the $r_{2} \to \infty$ limit of.

Step 2 — plug in the numbers. With $r_{2} / r_{1} = 1000$ : $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2000}{1001} = 1.998$ , so the bracket $= 1.998 - 1 = 0.4135$ . $Δ v_{1} = 7.5464 \times 0.4135 = 3.120$ km/s. Why this step? The whole $r_{2}$ dependence is in that one bracket.

Step 3 — the ceiling. As $r_{2} \to \infty$ , $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , bracket $\to 2 - 1 = 0.4142$ . So $Δ v_{1}^{m a x} = v_{c 1} (2 - 1) = 7.5464 \times 0.4142 = 3.126$ km/s. Why this step? $2 v_{c 1}$ is exactly escape (parabolic) speed; $Δ v_{1}$ can never exceed the burn that just barely escapes.

Step 4 — second burn. At $r_{2} = 7 \times 1 0^{6}$ km, $v_{c 2} = 398600/7 \times 1 0^{6} = 0.2387$ km/s. The apoapsis speed on the transfer ellipse is $v_{a} = 398600 (2/7 \times 1 0^{6} - 1/3503500) = 0.2385$ km/s (nearly circular out here). $Δ v_{2} = 0.2387 - 0.2385 = 0.00024$ km/s. Why this step? At apoapsis of a near-parabolic ellipse the ship is barely moving faster or slower than the huge outer circle, so the second burn is tiny.

Verify: $Δ v_{1} = 3.120$ km/s is within $0.2%$ of the escape ceiling $3.126$ km/s ✓ — confirming the "almost-parabola" picture. The second burn $Δ v_{2} \approx 0.24$ m/s is negligible — nearly all the cost is the first burn. This is exactly the regime where a Bi-elliptic transfer starts to matter (Cell G).

Cell F — Real-world word problem

Worked example Ex 6 · Earth → Mars heliocentric transfer

A probe leaves Earth's orbit around the Sun ( $r_{1} = 1.496 \times 1 0^{8}$ km) for Mars' orbit ( $r_{2} = 2.279 \times 1 0^{8}$ km). Use $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ . Find the two heliocentric burns and total.

Forecast: These are the classic "≈2.9 + ≈2.6" km/s heliocentric numbers. Guess whether the first or second burn is bigger this time.

Step 1 — Earth's orbital speed. $v_{c 1} = 1.327 \times 1 0^{11} /1.496 \times 1 0^{8} = 29.78$ km/s. Why this step? That's the speed the probe already shares with Earth around the Sun.

Step 2 — $a_{t}$ . $(1.496 + 2.279) \times 1 0^{8} /2 = 1.8875 \times 1 0^{8}$ km. Why this step? Fixes the transfer ellipse for the two vis-viva evaluations.

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. Why this step? Speed up so the probe's aphelion reaches Mars.

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. Why this step? Circularize into Mars' orbit; you arrive slow and must speed up.

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

Verify: Both prograde (Cell A pattern) ✓; first burn larger ✓ — consistent with textbook Earth→Mars values (~2.9 and ~2.6 km/s). Note this ignores Earth/Mars gravity wells; a full Delta-v budget adds those. Units check: km/s throughout ✓.

Cell G — Exam twist: Hohmann vs bi-elliptic

Worked example Ex 7 · When does Hohmann lose? (

$r_{2} / r_{1}$ large) Earth. $r_{1} = 7000$ km, $r_{2} = 112000$ km (ratio $R = 16$ ). Compare Hohmann to a bi-elliptic that flies out to $r_{b} = 560000$ km before dropping to $r_{2}$ .

Forecast: The threshold is $R \approx 11.94$ . Since $16 > 11.94$ , guess: bi-elliptic should win. Let's confirm with numbers — and watch for a twist.

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. Why this step? Baseline two-burn cost to beat.

Step 2 — Bi-elliptic (three burns via $r_{b} = 560000$ km). A bi-elliptic uses two big ellipses: first out to a far point $r_{b}$ , then back in to $r_{2}$ .

Burn 1 (at $r_{1}$ , onto the outbound ellipse with $a_{A} = (r_{1} + r_{b}) /2 = 283500$ km): $v = 398600 (2/7000 - 1/283500) = 10.578$ km/s; $Δ v_{A} = 10.578 - 7.546 = 3.032$ km/s. Why this sub-step? One deep, fast burn deep in the well (Oberth effect) flings you all the way out to $r_{b}$ — this is the expensive one.
Burn 2 (at $r_{b}$ , raising the other end from $r_{1}$ up to $r_{2}$ ; second ellipse $a_{B} = (r_{b} + r_{2}) /2 = 336000$ km): speed on the outbound ellipse at $r_{b}$ is $398600 (2/560000 - 1/283500) = 0.4676$ km/s; needed on the inbound ellipse is $398600 (2/560000 - 1/336000) = 0.9926$ km/s; $Δ v_{B} = 0.9926 - 0.4676 = 0.5250$ km/s. Why this sub-step? Way out at $r_{b}$ you are barely moving, so a small nudge cheaply lifts the near-side of the orbit from $r_{1}$ to $r_{2}$ — this is where bi-elliptic saves fuel.
Burn 3 (at $r_{2}$ , circularize — a brake): speed on the inbound ellipse at $r_{2} = 112000$ is $398600 (2/112000 - 1/336000) = 2.398$ km/s; $v_{c 2} = 1.887$ km/s; $Δ v_{C} = 2.398 - 1.887 = 0.511$ km/s (retrograde). Why this sub-step? Falling back in from $r_{b}$ you arrive too fast for the $r_{2}$ circle, so you brake — the only retrograde burn of the three.

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

Step 3 — read the result carefully. Hohmann $= 4.048$ km/s, this bi-elliptic $= 4.068$ km/s. Despite $R = 16 > 11.94$ , this particular bi-elliptic is slightly worse. Why the twist? The " $R > 11.94$ " rule only guarantees a sufficiently distant $r_{b}$ exists that beats Hohmann — it does not say every choice of $r_{b}$ wins. Our $r_{b} = 560000$ km ( $= 5 r_{2}$ ) is not far enough. Push $r_{b} \to \infty$ and the bi-elliptic total drops below Hohmann.

Verify: With $r_{b} = 560000$ km the totals are $4.048$ vs $4.068$ km/s — Hohmann still wins by $20$ m/s ✓ (see the machine check). A second check with $r_{b} = 1 0^{9}$ km gives a bi-elliptic total below Hohmann ✓, confirming the threshold rule is about the best bi-elliptic, at the price of a vastly longer flight time. Exam trap dodged: " $R > 11.94$ " ⇒ a better bi-elliptic exists, not that any given one is cheaper.

Cell H — Transfer-time question

Worked example Ex 8 · How long does the LEO→GEO trip take?

Use Ex 1's transfer ellipse: $a_{t} = 24582$ km, $μ_{\oplus} = 398600 km^{3} / s^{2}$ .

Forecast: You fly half the ellipse, so the time should be $\frac{1}{2}$ the full orbital period. Guess: a few hours.

Step 1 — period from Kepler's Third Law. $T = 2 π a_{t}^{3} / μ = 2 π 2458 2^{3} /398600$ . Why this step? Kepler's Third Law ties an orbit's period to its semi-major axis alone — no other orbit shape info needed. This is the right tool because we only know $a_{t}$ .

Step 2 — take half of it. A Hohmann arc runs from periapsis to apoapsis — exactly half the ellipse — so $t_{transfer} = \frac{T}{2} = π \frac{a _{t}^{3}}{μ} = π \frac{2458 2 ^{3}}{398600} = π 3.727 \times 1 0^{7} = π (6105) = 19180 s .$ Why this step? You don't fly the whole ellipse — you leave it at apoapsis with the second burn, so only half the period elapses.

Step 3 — convert to hours. $19180 s \div 3600 = 5.33$ h. Why this step? Human-readable units for a mission planner.

Verify: $19180$ s $= 5.33$ h ✓ — a reasonable half-day-ish transfer to GEO, matching published GTO coast times. Units: $km^{3} / (km^{3} / s^{2}) = s^{2} = s$ ✓.

Recall Cover the answers — did you hit every cell?

Raise vs lower: signs? ::: Raise = both burns positive (prograde); lower = both negative (retrograde); magnitudes are identical.
What happens at $r_{2} = r_{1}$ ? ::: $a_{t} = r_{1}$ , both $Δ v = 0$ ; formulas stay finite.
Ceiling on $Δ v_{1}$ as $r_{2} \to \infty$ ? ::: $v_{c 1} (2 - 1)$ — the escape (parabolic) kick.
Threshold where bi-elliptic beats Hohmann? ::: $r_{2} / r_{1} ≳ 11.94$ — but only a sufficiently distant $r_{b}$ realises the saving.
Transfer time formula? ::: $t = π a_{t}^{3} / μ$ (half the period).