Hohmann Δv calculation — both maneuvers: Visual walkthrough

We keep a running story: a little craft on an inner circle wants to reach an outer circle. Every step below adds one idea and one picture.

Step 1 — Two circles and one oval that kisses both

WHAT. We set the stage. There are three paths, not two:

the inner circle of radius $r_{1}$ (where we start),
the outer circle of radius $r_{2}$ (where we want to end, with $r_{2} > r_{1}$ ),
an ellipse (a stretched oval) that just touches the inner circle at its nearest point and the outer circle at its farthest point.

WHY. You cannot teleport from one circle to another. A free-falling object always follows a Kepler path (circle or ellipse). The cheapest bridge that touches both circles tangentially (moving the same direction as each circle where it touches) is this one ellipse. Tangential touching matters because then the engine only ever changes how fast, never which way — no thrust is wasted turning the velocity sideways.

PICTURE. The magenta inner circle, the violet outer circle, and the orange transfer ellipse. Notice the ellipse's closest point (periapsis) sits on the inner circle, its farthest point (apoapsis) sits on the outer circle. The planet is the orange dot at the shared focus.

Step 2 — Speed on a circle (gravity = the inward pull you need)

WHAT. Find how fast you move on a circle of radius $r$ . Call it $v_{circ}$ .

WHY. Both burns compare a circle-speed to an ellipse-speed at the same radius. So first we need the circle-speed itself. It comes from one idea: on a circle, gravity supplies exactly the inward force that bending your path in a circle requires.

Setting those two forces equal (see Circular orbital velocity):

$gravity pulling in \frac{GM m}{r ^{2}} = inward force a circle needs \frac{m v ^{2}}{r}$

$G$ = gravitation constant, $M$ = planet mass, $m$ = craft mass — together the pull.
$\frac{m v ^{2}}{r}$ = the centripetal force: how hard you must be yanked inward to keep curving at speed $v$ on radius $r$ .

Cancel $m$ , multiply both sides by $r$ :

$v_{circ}^{2} = \frac{GM}{r} = \frac{μ}{r} \Rightarrow v_{circ} = \frac{μ}{r}$

$μ = GM$ is the standard gravitational parameter — we bundle $G$ and $M$ because we never need them apart.

PICTURE. The inward gravity arrow and the "needed inward force" arrow are the same length on a circle — that balance is what a circular orbit is. The bigger the radius, the smaller $μ / r$ , so outer circles are slower.

Recall Why is the outer circle slower?

Because $v_{circ} = μ / r$ shrinks as $r$ grows ::: farther out, gravity is weaker, so you need less speed to keep from falling in — the outer merry-go-round spins slower.

Step 3 — Speed anywhere on any orbit (vis-viva)

WHAT. Get one master formula for speed $v$ at any distance $r$ on an orbit of a given size.

WHY. On the ellipse the speed changes — fast near the planet, slow far away. A single circle formula can't handle that. We need a rule that gives speed as a function of position. That rule is vis-viva, and it comes straight from energy bookkeeping (see Specific orbital energy and Vis-viva equation).

Energy per kilogram of the craft (its specific energy $ε$ ) is kinetic minus a gravity term:

$ε = motion energy \frac{v ^{2}}{2} - depth in the well \frac{μ}{r} = - \frac{μ}{2 a}$

$\frac{v ^{2}}{2}$ = kinetic energy per kg.
$- \frac{μ}{r}$ = gravitational potential per kg (negative: you're in a "well," deeper = more negative).
$a$ = semi-major axis, half the long width of the orbit — it alone fixes the total energy. This is a Kepler result: same $a$ ⇒ same energy, whatever the shape.

Solve for $v^{2}$ :

$v^{2} = μ (\frac{2}{r} - \frac{1}{a})$

Plug $a = r$ (a circle) and you get $μ (\frac{2}{r} - \frac{1}{r}) = \frac{μ}{r}$ — Step 2 pops right back out. ✔

PICTURE. A bar chart of energy at three spots on the ellipse. The total height ( $ε$ ) is the same everywhere (flat dashed line), but the split shifts: near the planet, lots of kinetic + very deep potential; far away, little kinetic + shallow potential.

Step 4 — The size of the transfer ellipse

WHAT. Find the semi-major axis $a_{t}$ of our specific bridge ellipse.

WHY. Vis-viva needs $a$ . Our ellipse's long axis runs straight from periapsis (distance $r_{1}$ from the planet) to apoapsis (distance $r_{2}$ ), both measured from the same focus — the planet's center.

$full long axis 2 a_{t} = near side r_{1} + far side r_{2} \Rightarrow a_{t} = \frac{r _{1} + r _{2}}{2}$

$2 a_{t}$ is the whole length of the ellipse the long way; splitting it in half gives $a_{t}$ , the average of the two radii.

PICTURE. The full long axis drawn as one straight line through the planet, labelled $r_{1}$ on the short side and $r_{2}$ on the long side, with $a_{t}$ marked as half the total.

Step 5 — The four speeds we will compare

WHAT. Line up the four speeds that matter, two per burn.

WHY. Each burn happens at one fixed radius, comparing the circle speed there with the ellipse speed there. That's four numbers total.

Where	Speed	Which formula	Role
Inner circle	$v_{c 1} = μ / r_{1}$	Step 2 at $r_{1}$	before burn 1
Ellipse periapsis	$v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}})$	Step 3 at $r = r_{1}$	after burn 1
Ellipse apoapsis	$v_{a} = μ (\frac{2}{r _{2}} - \frac{1}{a _{t}})$	Step 3 at $r = r_{2}$	before burn 2
Outer circle	$v_{c 2} = μ / r_{2}$	Step 2 at $r_{2}$	after burn 2

At the inner radius, the ellipse is faster than the inner circle ( $v_{p} > v_{c 1}$ ): it has extra energy to fling itself outward.
At the outer radius, the ellipse is slower than the outer circle ( $v_{a} < v_{c 2}$ ): it has crawled to apoapsis and lost speed climbing.

PICTURE. A speed-vs-radius plot with four dots. The magenta circle-speed curve and the orange ellipse-speed curve cross behaviours: at $r_{1}$ orange sits above magenta; at $r_{2}$ orange sits below magenta.

Step 6 — Burn 1: the periapsis kick

WHAT. The first velocity change, at $r_{1}$ .

WHY. You're on the inner circle at $v_{c 1}$ ; to enter the ellipse you must reach $v_{p}$ , which is larger. So you fire the engine prograde (forward, along your motion) — pure speed-up, no turn (velocities are parallel), so the vector change collapses to a subtraction of speeds.

$Δ v_{1} = v_{p} - v_{c 1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$

$μ / r_{1}$ = the inner circle speed, our baseline.
$\frac{2 r _{2}}{r _{1} + r _{2}}$ = the boost factor; the far radius $r_{2}$ sits on top because you're reaching out. Since $r_{2} > r_{1}$ this factor exceeds $1$ , so the bracket is positive.
The " $- 1$ " is the piece we already had (the circle speed we start with).

PICTURE. At $r_{1}$ : a short magenta arrow (current $v_{c 1}$ ) and a longer orange arrow (target $v_{p}$ ) both pointing the same way; the extra orange length is $Δ v_{1}$ , drawn in violet as the added thrust.

Step 7 — Burn 2: circularize at the top

WHAT. The second velocity change, at $r_{2}$ .

WHY. You coast up the ellipse to apoapsis and arrive at $v_{a}$ , which is slower than the outer circle's $v_{c 2}$ . If you did nothing you'd fall back down the ellipse. So you fire prograde again to top up from $v_{a}$ to $v_{c 2}$ — another pure speed-up.

$Δ v_{2} = v_{c 2} - v_{a} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})$

$μ / r_{2}$ = the outer circle speed, our target baseline.
$\frac{2 r _{1}}{r _{1} + r _{2}}$ = the apoapsis-speed factor; now the near radius $r_{1}$ sits on top (mirror of Step 6). Since $r_{1} < r_{2}$ this is below $1$ , so the bracket is positive — you add speed.
Order is $v_{c 2} - v_{a}$ (circle minus ellipse) because at apoapsis the ellipse is the slower one.

PICTURE. At $r_{2}$ : a short orange arrow ( $v_{a}$ ) and a longer magenta arrow ( $v_{c 2}$ ), same direction; the violet gap is $Δ v_{2}$ .

Step 8 — Edge and limiting cases (never leave a gap)

WHAT. Check the corners so no scenario surprises you.

WHY. A derivation you trust must survive its extremes.

$r_{2} = r_{1}$ (no transfer). Then $a_{t} = r_{1}$ , the "ellipse" is the circle, $v_{p} = v_{a} = v_{c 1} = v_{c 2}$ , so $Δ v_{1} = Δ v_{2} = 0$ . Zero cost to "move" to where you already are. ✔
$r_{2} \to \infty$ (escape-like reach). Burn 1's factor $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ , so $Δ v_{1} \to v_{c 1} (2 - 1)$ — exactly the extra kick from circular speed to escape speed. Meanwhile $v_{c 2} \to 0$ and $v_{a} \to 0$ , so $Δ v_{2} \to 0$ : there's no orbit to circularize onto at infinity. The whole cost lives in burn 1. ✔
Going inward ( $r_{2} < r_{1}$ ). The same algebra runs, but now both brackets flip sign — the burns become retrograde (you slow down twice). The magnitudes $Δ v$ are still what the formulas give (take absolute value); the direction reverses. So Hohmann works both ways.
Impulsive assumption. All of this treats each burn as instantaneous at a single point — the Impulsive maneuver approximation. Real engines burn over an arc, which slightly raises the true cost; the Oberth effect explains why burning deep and fast (like burn 1) is energetically efficient. For very large ratios $r_{2} / r_{1}$ , a Bi-elliptic transfer can even beat Hohmann.

PICTURE. Two mini-panels: left, the degenerate case where all three paths collapse onto one circle ( $Δ v = 0$ ); right, the $r_{2} \to \infty$ case where the ellipse opens into a near-parabola and burn 2 vanishes.

The one-picture summary

Everything compressed: the two circles, the transfer ellipse, both prograde arrows at their radii, and the two speed-differences labelled with their formulas. Trace it once and you've re-derived the whole thing.

Recall Feynman retelling of the whole walkthrough

You're circling a planet on a small, fast loop. Draw an oval whose near end kisses your loop and whose far end kisses a bigger loop you want. Push forward once where you are: this is cheap-to-describe but the biggest push, because down low everything moves fast and you have to add real speed to fling yourself out along the oval. Coast — you climb the oval, slowing the whole way, until you reach its far end, now moving too slowly to stay out there. Push forward again to top up your speed to match the big loop, and you settle into it. Two forward pushes, add their sizes, done. The reason the first push is bigger is that it does the heavy lifting of injecting the energy to reach far away, deep where gravity is strongest.

Active Recall

Why exactly one ellipse and not a straight hop?

Free-fall paths are Kepler conics; the ellipse tangent to both circles is the cheapest two-impulse bridge.

r_{1}

, is the ellipse faster or slower than the circle?

Faster —

v_{p} > v_{c 1}

, so burn 1 is a speed-up.

r_{2}

, is the ellipse faster or slower than the circle?

Slower —

v_{a} < v_{c 2}

, so burn 2 is also a speed-up.

What happens to

Δ v_{1}

r_{2} \to \infty

It approaches

v_{c 1} (2 - 1)

, the circular-to-escape kick.

What is

Δ v_{2}

when

r_{2} = r_{1}

Zero — no transfer needed.

Which burn dominates for LEO→GEO and why?

Burn 1, because it injects the energy to reach distant apoapsis while deep in the gravity well.