Hohmann transfer — derivation, minimum energy transfer: Visual walkthrough

We are orbiting a planet. Call the planet's gravitational strength $μ = GM$ — this is just one number that says "how hard this planet pulls." $G$ is the universal gravity constant, $M$ is the planet's mass. Bigger $μ$ = stronger pull. That's all $μ$ means; nothing more is smuggled in.

Step 1 — Draw the two circles you're jumping between

WHAT. Two circular orbits around the same planet, sharing the same flat plane (like two rings drawn on one sheet of paper). The inner one has radius $r_{1}$ (distance from planet centre to the ring). The outer one has radius $r_{2}$ , with $r_{2} > r_{1}$ .

WHY start here. Before you can compute a single number, you must know what you're leaving and what you're arriving at. A radius is just "how far out." The whole problem is: get from the small ring to the big ring cheaply.

PICTURE. The planet sits at the centre (the dot). The inner ring is where you start; the outer ring is your destination. The two arrows show which way you're circling.

Step 2 — Find how fast you are ALREADY moving (inner circle)

WHAT. On any circular orbit, gravity's inward pull exactly supplies the "turning force" that keeps you looping instead of flying straight. Balancing them gives one clean speed: $v_{c 1} = \frac{μ}{r _{1}}$

Term by term:

$v_{c 1}$ — the " $c$ " means circular, the " $1$ " means inner ring. This is your speed right now, before you touch the engine.
$μ$ on top — stronger planet ⇒ you must move faster to keep from falling in.
$r_{1}$ on the bottom, under a square root — farther out ⇒ slower. Being deep in the well makes you race; being far out lets you dawdle.

WHY this tool (the square root). We want speed, but energy balance naturally gives us $v^{2}$ . To undo the square and read off $v$ , we take a square root — the operation that answers "what number, times itself, gives $v^{2}$ ?" That is the only reason appears.

PICTURE. The velocity arrow is tangent to the circle — it points along the ring, sideways, never inward or outward. That sideways direction matters enormously in a moment.

Step 3 — Draw the transfer ellipse that "kisses" both rings

WHAT. We now need a curve that bridges the two rings. That curve is an ellipse. An ellipse is more than a "squashed circle" — it has a precise defining rule, and that rule is what makes it the natural shape of orbits.

We choose the ellipse whose nearest point (periapsis) grazes the inner ring at $r_{1}$ and whose farthest point (apoapsis) grazes the outer ring at $r_{2}$ .

The semi-major axis $a_{t}$ is half of the ellipse's longest diameter. That longest diameter stretches from the inner grazing point all the way across to the outer one: $2 a_{t} = r_{1} + r_{2} ⟹ a_{t} = \frac{r _{1} + r _{2}}{2}$

Term by term:

$r_{1}$ — the closest distance (periapsis).
$r_{2}$ — the farthest distance (apoapsis).
Their sum is the full long diameter; halve it to get $a_{t}$ . So $a_{t}$ is literally the average of the two radii.

WHY this ellipse and not some other. Among all ellipses that could bridge the two rings, this bi-tangent one touches each ring exactly once and shares each ring's velocity direction there (tangent to tangent). That shared direction is what makes the burns cheap — we'll cash that in at Step 5.

PICTURE. The planet sits at one focus (the filled dot); the second focus is the hollow dot inside the ellipse. The string-sum property is shown by two lines from a sample point to the two foci. Periapsis is the low point on the inner ring; apoapsis is the high point on the outer ring; the dashed line is the long diameter $2 a_{t}$ .

Step 4 — Find the speed needed at the START of the ellipse

WHAT. To ride this ellipse you must be moving at a specific speed as you leave the inner ring. The vis-viva equation gives it: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$

Intuition WHY vis-viva looks like this (energy bookkeeping)

A moving spacecraft owns two kinds of energy per unit mass: motion energy $\frac{1}{2} v^{2}$ (bigger when you're fast) and height energy $- μ / r$ (a debt that shrinks toward zero as you climb far away, and is deeply negative when you're close in). Their sum never changes on a coasting orbit — that's energy conservation: $\frac{1}{2} v^{2} - \frac{μ}{r} = ε (constant) .$ Now, a separate geometric fact ties that constant to the orbit's size: for any ellipse the total energy depends only on the semi-major axis, $ε = - μ /2 a$ (a big orbit is a shallow, low-energy orbit; a tight orbit is deeply bound). Set the two expressions for $ε$ equal and solve for $v^{2}$ : $\frac{1}{2} v^{2} - \frac{μ}{r} = - \frac{μ}{2 a} ⟹ v^{2} = μ (\frac{2}{r} - \frac{1}{a}) .$ So vis-viva is just "kinetic + potential = constant, and that constant is fixed by $a$ ." (Full algebra in Vis-viva equation.)

Evaluate it at radius $r = r_{1}$ using the ellipse's $a = a_{t}$ : $v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = μ (\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}})$

Term by term:

$v_{p}$ — speed at periapsis of the ellipse.
$\frac{2}{r _{1}}$ — the "how close am I" term (from the height-energy $- μ / r$ evaluated at $r_{1}$ ); being deep gives lots of speed.
$\frac{1}{a _{t}}$ — the "how big is my whole orbit" term (from $ε = - μ /2 a_{t}$ ); a bigger orbit stores more energy and slightly reduces this subtraction. Since $a_{t} = (r_{1} + r_{2}) /2$ , $\frac{1}{a _{t}} = \frac{2}{r _{1} + r _{2}}$ .

WHY $v_{p} > v_{c 1}$ . Compare Step 4 with Step 2. The circle used $a = r_{1}$ , so its term was $\frac{2}{r _{1}} - \frac{1}{r _{1}} = \frac{1}{r _{1}}$ . The ellipse uses $a_{t} > r_{1}$ , so $\frac{1}{a _{t}} < \frac{1}{r _{1}}$ — we subtract less, leaving a bigger number under the root. Physically: to swing all the way out to $r_{2}$ you need extra energy, and at periapsis that shows up as extra speed.

PICTURE. Two velocity arrows from the same periapsis point: the short one is the circular speed $v_{c 1}$ you have, the long one is $v_{p}$ you need. Both point the same way (tangent).

Step 5 — The FIRST burn (why it's just a subtraction)

WHAT. Fire the engine prograde (in the direction you're already going) to make up the speed gap: $Δ v_{1} = v_{p} - v_{c 1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$

Term by term:

$Δ v_{1}$ — "delta-vee," the change in speed the engine must deliver. This is the fuel cost currency of spaceflight (see Delta-v budget).
$v_{p} - v_{c 1}$ — needed speed minus current speed.
The bracketed form: factor out $μ / r_{1} = v_{c 1}$ ; the leftover $(\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ is always positive because $r_{2} > r_{1}$ , so the burn always speeds you up.

WHY it's a plain subtraction, not vector math. Both arrows in Step 4 pointed the same direction. When two velocities are collinear, the change in the vector is just the change in length: $Δ v = ∣ v_{p} ∣ - ∣ v_{c 1} ∣$ . This is the entire reason we chose tangent burns — it turns hard vector subtraction into easy arithmetic.

PICTURE. The green $Δ v_{1}$ arrow is exactly the gap between the two arrows from Step 4 — laid end to tip along the same line.

Step 6 — Coast the half-ellipse; arrive SLOW at the top

WHAT. Engine off. You climb from $r_{1}$ to $r_{2}$ along half the ellipse. Energy is conserved, so as $r$ grows your speed drops. At apoapsis vis-viva at $r = r_{2}$ (still on the same ellipse, so still $a = a_{t}$ ) gives: $v_{a} = μ (\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}})$

Term by term:

$v_{a}$ — speed at apoapsis (top of the climb).
$\frac{2}{r _{2}}$ — the "how close am I" term, now evaluated far out at $r_{2}$ ; it's small, so this alone gives little speed. You've "run out of steam."
$\frac{2}{r _{1} + r _{2}}$ — this is exactly $\frac{1}{a _{t}}$ , the same orbit-size term from Step 4 (because you never left the transfer ellipse — its semi-major axis $a_{t}$ is unchanged the whole coast). It is the fixed energy debt set by the ellipse's size; subtracting it is what encodes "this is an ellipse, not a circle at $r_{2}$ ."

WHY you slow down. Climbing out of a gravity well trades kinetic energy (motion) for potential energy (height), exactly like a ball thrown up slows as it rises. Same total energy, redistributed. (This is Orbital energy & semi-major axis in action.)

PICTURE. The coasting arc, with the velocity arrow shrinking as the craft climbs from periapsis to apoapsis. No engine anywhere on this arc.

Step 7 — The SECOND burn (speed UP, don't brake!)

WHAT. At apoapsis you're on the outer ring's radius but far too slow to hold a circle there. The outer circular speed is $v_{c 2} = \frac{μ}{r _{2}},$ and since $v_{a} < v_{c 2}$ , you must fire prograde again to catch up: $Δ v_{2} = v_{c 2} - v_{a} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})$

Term by term:

$v_{c 2}$ — circular speed on the outer ring (Step 2's formula with $r_{2}$ ).
$v_{a}$ — your slow arrival speed from Step 6.
The bracket $(1 - \frac{2 r _{1}}{r _{1} + r _{2}})$ is positive because $r_{1} < r_{2}$ , so again the burn is prograde (speed-up).

WHY not brake. The car-intuition ("going up ⇒ slow down") is a trap. You're not stopping at $r_{2}$ ; you're trying to stay in circular motion there, which requires a definite speed $v_{c 2}$ that is higher than your arrival speed $v_{a}$ . Give energy, don't take it.

PICTURE. The short arrival arrow $v_{a}$ , the longer circular arrow $v_{c 2}$ , and the orange $Δ v_{2}$ gap between them — all tangent to the outer ring.

Step 8 — Edge cases: check the formula never breaks

WHAT. Push the inputs to their limits and confirm the picture still makes sense.

Case A — $r_{2} = r_{1}$ (rings coincide). Then $a_{t} = r_{1}$ , and both brackets collapse: $\frac{2 r _{1}}{2 r _{1}} - 1 = 0$ . So $Δ v_{1} = Δ v_{2} = 0$ . Picture: the ellipse degenerates back into the circle you're already on — no burn needed, exactly right.

Case B — $r_{2} ≫ r_{1}$ (jump to enormous orbit). The periapsis factor $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , so $Δ v_{1} \to v_{c 1} (2 - 1) \approx 0.414 v_{c 1}$ — the first burn saturates. Meanwhile $v_{c 2} \to 0$ , so $Δ v_{2} \to 0$ : arriving at a huge, faint orbit costs almost nothing at the top. Picture: an ellipse so stretched it's nearly a straight shot outward.

Case C — going INWARD ( $r_{2} < r_{1}$ ). Re-label so the bigger radius is the start. Now both burns are retrograde (you fire backward to subtract speed): you brake at the high circle to drop onto the ellipse, coast down gaining speed, then brake again at the bottom. The formulas are the same; only the direction of the burns flips from prograde to retrograde. Picture: the same diagram run in reverse, arrows pointing backward.

Why this is the minimum-energy two-burn transfer

The one-picture summary

Everything above, compressed into one frame — showing both directions. Read it left-to-right for a raise (both burns prograde, green), or right-to-left for a lower (both burns retrograde/braking, slate): same ellipse, same magnitudes, opposite burn directions.

$Δ v_{total} = burn 1: leave inner ring \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) + burn 2: hold outer ring \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})$

For a raise ( $r_{2} > r_{1}$ ) both terms are positive and both burns are prograde. For a lower ( $r_{2} < r_{1}$ ) the same two magnitudes are burned retrograde (backward), braking you down from the big ring onto the ellipse and again onto the small ring.

Recall Feynman retelling of the whole walkthrough

You're circling a planet on a small ring, moving sideways at some fixed speed (Step 2). You want the big ring outside you. You can't drive straight out — gravity curves everything. So you draw a squashed loop, an ellipse (a curve where the two distances to two fixed pins always add to the same total), that just touches your small ring at its low point and the big ring at its high point (Step 3). To ride it you need to be moving a bit faster than you are (Step 4), so you give one prograde shove — and because you're pointing the same way you're already going, the cost is just the speed difference, a simple subtraction (Step 5). Then you cut the engine and coast. Climbing out, you slow down, trading motion for height, and arrive at the big ring going too slow to stay there (Step 6). One more prograde shove speeds you up to the big ring's circular speed and you're done (Step 7). Two pushes, both forward, cheapest ride. If the rings were the same size you'd need no push at all; if you wanted to drop inward instead, the same two pushes become gentle brakes fired backward — retrograde (Step 8). That's the entire Hohmann transfer.

Recall Active recall — cover the answers

Why is $Δ v$ a plain subtraction and not vector math? ::: Because the burns are tangent, so the two velocities are collinear; the change in a vector's length equals the change in its magnitude. What is the semi-major axis of the transfer ellipse? ::: $a_{t} = (r_{1} + r_{2}) /2$ , the average of the two radii. Why do you speed UP at apoapsis instead of braking? ::: You arrive slower than the outer circular speed ( $v_{a} < v_{c 2}$ ), so you must add energy to hold the larger circle. What is an ellipse's defining property? ::: The sum of distances from any point on it to the two foci is constant; the planet sits at one focus. What happens to the burns when $r_{2} = r_{1}$ ? ::: Both go to zero — the ellipse collapses into the circle you're already on. For a huge outward jump, what does $Δ v_{1}$ approach? ::: $v_{c 1} (2 - 1) \approx 0.414 v_{c 1}$ — it saturates. In what sense is Hohmann optimal? ::: It is the minimum-total- $Δ v$ transfer among all two-burn transfers between coplanar circular orbits (up to $r_{2} / r_{1} \approx 11.94$ ).

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