Hohmann Δv calculation — both maneuvers: Worked examples

Before anything, the symbols we reuse constantly:

Definition The recurring cast

$μ$ (mu) = standard gravitational parameter $= GM$ of the central body. Units $km^{3} / s^{2}$ . See Standard gravitational parameter.
$r_{1}$ = radius of the starting circular orbit (measured from the planet's center, not its surface).
$r_{2}$ = radius of the target circular orbit.
$a_{t} = \frac{r _{1} + r _{2}}{2}$ = semi-major axis of the transfer ellipse (half the periapsis-to-apoapsis span).
$v_{c 1} = μ / r_{1}, v_{c 2} = μ / r_{2}$ = the circular speeds on orbit 1 and orbit 2.
$v_{p}$ = periapsis speed = the transfer ellipse's speed at its near point (radius $r_{1}$ ). "Periapsis" = the point of an orbit closest to the planet. It is the fastest point of the ellipse.
$v_{a}$ = apoapsis speed = the transfer ellipse's speed at its far point (radius $r_{2}$ ). "Apoapsis" = the point of an orbit farthest from the planet. It is the slowest point of the ellipse.
$Δ v_{1}, Δ v_{2}$ = the speed change of burn 1 (at $r_{1}$ ) and burn 2 (at $r_{2}$ ).
$Δ v_{tot} \equiv Δ v_{1} + Δ v_{2}$ = the total speed budget. We define it as the plain sum because the two burns are separate, independent impulses (see Impulsive maneuver approximation); there is no vector cancellation to worry about, so the scalar magnitudes simply add.

All four speeds come from just two tools, which we lean on the whole page:

The scenario matrix

Every Hohmann problem is one (or a blend) of these cells. The examples below hit all of them. The diagram shows the shared geometry — refer back to it as each cell picks out its own start circle (blue), target circle (green), transfer ellipse (dashed orange) and the two red burn points.

Cell	What makes it special	Example
A · Standard raise	$r_{2} > r_{1}$ , both burns prograde, ordinary numbers	Ex 1
B · Small raise	$r_{2}$ only slightly $> r_{1}$ ; both $Δ v$ tiny and nearly equal	Ex 2
C · Large raise / "which burn bigger" flip	very large $r_{2}$ ; check whether burn 1 or burn 2 dominates	Ex 3
D · Lowering (going inward)	$r_{2} < r_{1}$ ; burns become retrograde (speed decreases)	Ex 4
E · Degenerate $r_{1} = r_{2}$	no transfer needed; must give $Δ v = 0$	Ex 5
F · Limit $r_{2} \to \infty$	escape-like; $Δ v_{1} \to (2 - 1) v_{c 1}$ , $Δ v_{2} \to 0$	Ex 6
G · Real-world word problem	radii given as altitudes, must add planet radius first	Ex 7
H · Exam twist	transfer time (Ex 8a), or given $Δ v$ and solve backward (Ex 8b)	Ex 8

Read this figure first. The blue circle is orbit 1, the green circle is orbit 2, the dashed orange oval is the transfer ellipse that touches blue at its near point (periapsis, right, where $v_{p}$ lives) and green at its far point (apoapsis, left, where $v_{a}$ lives). The two red dots are exactly where the engine fires — nowhere else. The blue and green arrows mark $r_{1}$ and $r_{2}$ measured from the planet at the focus. Every cell below is just this picture with different radii.

Figure — Hohmann Δv calculation — both maneuvers

Cell A — the standard raise

Look at the figure below — this exact case. The red dot on the right (periapsis, at $r_{1}$ ) is where $v_{c 1}$ jumps to $v_{p}$ ; the red dot on the left (apoapsis, at $r_{2}$ ) is where $v_{a}$ jumps to $v_{c 2}$ . The labelled speeds show why burn 1 is the tall one.

Step 1 — semi-major axis of the transfer. $a_{t} = \frac{6678 + 20000}{2} = 13339 km$ Why this step? Vis-viva needs $a$ ; the ellipse spans $r_{1}$ (periapsis) to $r_{2}$ (apoapsis), so $a_{t}$ is their average.

Step 2 — circular speed at $r_{1}$ . $v_{c 1} = \frac{398600}{6678} = 7.726 km/s$ Why? That is the speed we start with, riding circle 1.

Step 3 — transfer speed at periapsis ( $r = r_{1}$ , $a = a_{t}$ ). $v_{p} = 398600 (\frac{2}{6678} - \frac{1}{13339}) = 9.462 km/s$ Why? This is the speed we need after burn 1 to be on the ellipse. Because $a_{t} > r_{1}$ , the ellipse is faster than the circle here.

Step 4 — burn 1. $Δ v_{1} = v_{p} - v_{c 1} = 9.462 - 7.726 = 1.736 km/s$ Why subtract this way? Both velocities point the same direction (prograde), and $v_{p} > v_{c 1}$ , so $Δ v = ∣ v_{p} - v_{c 1} ∣ = v_{p} - v_{c 1}$ — larger minus smaller.

Step 5 — circular speed at $r_{2}$ . $v_{c 2} = \frac{398600}{20000} = 4.464 km/s$ Why? This is the speed we must end up at to ride circle 2, so we need it to compare against the ellipse's apoapsis speed in burn 2.

Step 6 — transfer speed at apoapsis ( $r = r_{2}$ , $a = a_{t}$ ). $v_{a} = 398600 (\frac{2}{20000} - \frac{1}{13339}) = 3.159 km/s$ Why? The ellipse is slower at its far point, so this is below $v_{c 2}$ — we will need to speed up.

Step 7 — burn 2 and total. $Δ v_{2} = v_{c 2} - v_{a} = 4.464 - 3.159 = 1.305 km/s, Δ v_{tot} = 1.736 + 1.305 = 3.041 km/s$ Why $v_{c 2} - v_{a}$ and why add? At apoapsis $v_{a} < v_{c 2}$ , so $Δ v_{2} = ∣ v_{c 2} - v_{a} ∣ = v_{c 2} - v_{a}$ (larger minus smaller, a prograde speed-up). We add the two because they are separate impulses at separate points, per the definition of $Δ v_{tot}$ .

Verify: Both burns positive ✔ (prograde as expected for a raise). $Δ v_{1} > Δ v_{2}$ — the forecast payoff: the deeper burn does more work (this is the seed of the Oberth effect). Units: $km^{3} s^{- 2} / km = km/s$ ✔.

Cell B — the small raise

Step 1 — $a_{t}$ . $a_{t} = (7000 + 7100) /2 = 7050$ km. Why? same averaging rule — the transfer ellipse spans periapsis $r_{1}$ to apoapsis $r_{2}$ .

Step 2 — the four speeds. $v_{c 1} = 398600/7000 = 7.546 km/s — circular speed we start on (v_{circ} at r_{1})$ $v_{p} = 398600 (2/7000 - 1/7050) = 7.573 km/s — vis-viva at r = r_{1} on the ellipse (a = a_{t})$ $v_{c 2} = 398600/7100 = 7.493 km/s — circular speed we must reach (v_{circ} at r_{2})$ $v_{a} = 398600 (2/7100 - 1/7050) = 7.466 km/s — vis-viva at r = r_{2} on the ellipse$ Why these formulae here? Each burn compares a circle speed to the ellipse speed at the same radius: circular speed uses $v_{circ} (r)$ , the ellipse speed uses vis-viva with $a = a_{t}$ . That is why $v_{c}$ uses $μ / r$ and $v_{p}, v_{a}$ use $μ (2/ r - 1/ a_{t})$ .

Step 3 — the two burns. $Δ v_{1} = v_{p} - v_{c 1} = 7.573 - 7.546 = 0.0267 km/s, Δ v_{2} = v_{c 2} - v_{a} = 7.493 - 7.466 = 0.0264 km/s$ Why so close? When $r_{2} \approx r_{1}$ , the ellipse is barely eccentric, so periapsis and apoapsis excesses are almost mirror images.

Verify: $Δ v_{1} \approx Δ v_{2}$ (differ by <2%), confirming the 50/50 forecast. Sanity: total $\approx 53 m/s$ , the right ballpark for a 100 km station-keeping raise ✔.

Cell C — the large raise & the "which burn is bigger" flip

Step 1 — $a_{t}$ . $a_{t} = (6678 + 400000) /2 = 203339$ km. Why average? The transfer ellipse still touches periapsis $r_{1}$ and apoapsis $r_{2}$ , so its semi-major axis is their midpoint — the rule never changes, only the numbers get large.

Step 2 — burn 1 (near-escape kick). $v_{c 1} = 7.726 km/s, v_{p} = 398600 (2/6678 - 1/203339) = 10.847 km/s$ $Δ v_{1} = v_{p} - v_{c 1} = 10.847 - 7.726 = 3.121 km/s$ Why so big, and why this subtraction? With $r_{2}$ huge, the periapsis speed approaches escape speed $2 v_{c 1} = 10.926$ km/s. Since $v_{p} > v_{c 1}$ , $Δ v_{1} = ∣ v_{p} - v_{c 1} ∣ = v_{p} - v_{c 1}$ .

Step 3 — burn 2 (deep-space circularize). $v_{c 2} = 398600/400000 = 0.998 km/s, v_{a} = 398600 (2/400000 - 1/203339) = 0.181 km/s$ $Δ v_{2} = v_{c 2} - v_{a} = 0.998 - 0.181 = 0.817 km/s$ Why $v_{c 2} - v_{a}$ ? At the far apoapsis the ellipse crawls ( $v_{a} ≪ v_{c 2}$ ), so we must speed up: $Δ v_{2} = ∣ v_{c 2} - v_{a} ∣ = v_{c 2} - v_{a}$ , larger minus smaller.

Verify: $Δ v_{1} ≫ Δ v_{2}$ — burn 1 still wins, and even more lopsidedly than Ex 1. Lesson: for a raise, the periapsis burn always dominates as $r_{2}$ grows. (This is exactly why Bi-elliptic transfer can beat Hohmann for very large ratios — it splits that expensive kick.) Units ✔.

Cell D — lowering the orbit (going inward)

Step 1 — $a_{t}$ . $a_{t} = (42164 + 6678) /2 = 24421$ km. Why? the transfer ellipse is the same shape as LEO→GEO — Hohmann is symmetric; averaging the two radii still gives the semi-major axis.

Step 2 — speeds at the (now outer) start radius $r_{1} = 42164$ . $v_{c 1} = 398600/42164 = 3.075 km/s — circular speed we start on$ $v_{a} = 398600 (2/42164 - 1/24421) = 1.607 km/s — ellipse speed here$ Why is this the ellipse's apoapsis speed, and why these formulae? Starting high, we drop onto the ellipse at its far point, so the transfer speed here is the slow apoapsis speed — use vis-viva with $r = r_{1}$ , $a = a_{t}$ . The circular speed still uses $v_{circ} (r_{1})$ . Note $v_{a} < v_{c 1}$ .

Step 3 — burn 1 magnitude. $Δ v_{1} = v_{c 1} - v_{a} = 3.075 - 1.607 = 1.468 km/s$ Why $v_{c 1} - v_{a}$ ? Here the before speed is $v_{c 1}$ and the after speed is $v_{a}$ , with $v_{c 1} > v_{a}$ . By the definition $Δ v = ∣ v_{after} - v_{before} ∣ = ∣ v_{a} - v_{c 1} ∣ = v_{c 1} - v_{a}$ — again larger minus smaller, so positive. Physically we slow down (retrograde) to drop inward.

Step 4 — speeds at the low target radius $r_{2} = 6678$ . $v_{p} = 398600 (2/6678 - 1/24421) = 10.15 km/s — ellipse periapsis speed$ $v_{c 2} = 398600/6678 = 7.726 km/s — circular speed we must reach$ Why? Arriving low, we reach the ellipse's near point (periapsis), fastest of all, via vis-viva at $r = r_{2}$ ; the target circle speed is $v_{circ} (r_{2})$ . Note $v_{p} > v_{c 2}$ .

Step 5 — burn 2 magnitude. $Δ v_{2} = v_{p} - v_{c 2} = 10.15 - 7.726 = 2.421 km/s$ Why $v_{p} - v_{c 2}$ ? Now the before speed is $v_{p}$ (on the ellipse) and after is $v_{c 2}$ (the circle), with $v_{p} > v_{c 2}$ . So $Δ v = ∣ v_{after} - v_{before} ∣ = ∣ v_{c 2} - v_{p} ∣ = v_{p} - v_{c 2}$ — larger minus smaller, positive. Physically another retrograde brake.

Verify: Total $= 1.468 + 2.421 = 3.889$ km/s — identical to the LEO→GEO total in the parent note (3.89 km/s) ✔. Hohmann is reversible: lowering costs the same as raising, just with retrograde burns. Both $Δ v$ 's stayed positive because we always wrote larger-minus-smaller, exactly what $∣ v_{after} - v_{before} ∣$ demands. The ordering of burn sizes flips too (big burn now second).

Cell E — degenerate case $r_{1} = r_{2}$

First we must earn the closed-form formula this cell tests. Take burn 1 and substitute the vis-viva expression for $v_{p}$ with $a_{t} = (r_{1} + r_{2}) /2$ :

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

Why does that simplify? Factor $\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}} = \frac{2}{r _{1}} \cdot \frac{r _{2}}{r _{1} + r _{2}}$ , then pull $μ / r_{1}$ out front. Subtracting $v_{c 1} = μ / r_{1}$ gives the factored form of burn 1.

Now derive $Δ v_{2}$ the same way, step by step. Start from the apoapsis speed $v_{a}$ (vis-viva at $r = r_{2}$ ): $v_{a} = μ (\frac{2}{r _{2}} - \frac{1}{a _{t}}) = μ (\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}})$ Why? same $a_{t} = (r_{1} + r_{2}) /2$ substitution. Factor the inside: $\frac{2}{r _{2}} - \frac{2}{r _{1} + r _{2}} = \frac{2}{r _{2}} \cdot \frac{r _{1}}{r _{1} + r _{2}}$ , so $v_{a} = \frac{μ}{r _{2}} \frac{2 r _{1}}{r _{1} + r _{2}}$ Then $Δ v_{2} = v_{c 2} - v_{a} = \frac{μ}{r _{2}} - \frac{μ}{r _{2}} \frac{2 r _{1}}{r _{1} + r _{2}}$ , and pulling $μ / r_{2}$ out front gives the boxed form below.

Step 1 — $a_{t}$ . $a_{t} = (10000 + 10000) /2 = 10000 = r_{1} = r_{2}$ . Why check? The "ellipse" degenerates to the circle itself.

Step 2 — plug into the factored formula just derived. $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) = \frac{μ}{r _{1}} (\frac{2 r _{1}}{2 r _{1}} - 1) = \frac{μ}{r _{1}} (1 - 1) = 0$ $Δ v_{2} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{2 r _{1}}) = 0$

Verify: $Δ v_{tot} = 0$ ✔. No division-by-zero anywhere ( $a_{t}$ is finite and nonzero). The formula is well-behaved at the degenerate boundary — a good correctness check.

Cell F — the limit $r_{2} \to \infty$

Step 1 — burn 1 limit. In $Δ v_{1} = μ / r_{1} (2 r_{2} / (r_{1} + r_{2}) - 1)$ , as $r_{2} \to \infty$ the ratio $\frac{2 r _{2}}{r _{1} + r _{2}} \to 2$ . So $Δ v_{1} \to \frac{μ}{r _{1}} (2 - 1) = (2 - 1) v_{c 1}$ Why? $2 v_{c 1}$ is escape speed; the extra push beyond circular is $(2 - 1) v_{c 1}$ .

Numerically: $(2 - 1) \times 7.726 = 3.200$ km/s.

Step 2 — burn 2 limit. In $Δ v_{2} = μ / r_{2} (1 - 2 r_{1} / (r_{1} + r_{2}))$ , the leading $μ / r_{2} \to 0$ while the bracket $\to 1$ . So $Δ v_{2} \to 0$ Why? At infinite radius the circular speed is zero, and the ellipse's apoapsis speed is also zero, so there is nothing left to burn.

Verify: $Δ v_{1} \to 3.200$ km/s, $Δ v_{2} \to 0$ ✔. Matches the parent's forecast box exactly. This is why an interplanetary escape burn is essentially just "burn 1 to infinity."

Cell G — real-world word problem (altitudes, not radii)

Step 1 — convert altitudes to orbital radii. $r_{1} = 6371 + 500 = 6871 km, r_{2} = 6371 + 800 = 7171 km$ Why this step? Gravity depends on distance from the center of mass, so every $r$ in vis-viva is center-relative. Forgetting $+ R_{\oplus}$ is the #1 real-world error.

Step 2 — $a_{t}$ and speeds. $a_{t} = (6871 + 7171) /2 = 7021 km$ $v_{c 1} = 398600/6871 = 7.617 km/s, v_{p} = 398600 (2/6871 - 1/7021) = 7.696 km/s$ $v_{c 2} = 398600/7171 = 7.456 km/s, v_{a} = 398600 (2/7171 - 1/7021) = 7.375 km/s$ Why these formulae? Circular speeds use $v_{circ} (r)$ ; ellipse speeds use vis-viva with $a = a_{t}$ — exactly as in every raise cell.

Step 3 — burns. $Δ v_{1} = v_{p} - v_{c 1} = 7.696 - 7.617 = 0.0791 km/s, Δ v_{2} = v_{c 2} - v_{a} = 7.456 - 7.375 = 0.0805 km/s$ $Δ v_{tot} = 0.1596 km/s \approx 160 m/s$

Verify: Small raise ⇒ tiny, nearly-equal burns (consistent with Cell B behavior) ✔. If you had wrongly used altitudes (500 and 800), $v_{c 1} = 398600/500 = 28.2$ km/s — absurd, faster than any real orbit — an instant red flag that you forgot $R_{\oplus}$ .

Cell H — the exam twists

Ex 8a — transfer time

Step 1 — transfer semi-major axis. $a_{t} = (6678 + 42164) /2 = 24421$ km. Why? period needs $a$ , and the transfer is half of this ellipse.

Step 2 — full period of the transfer ellipse (Kepler III). $T = 2 π \frac{a _{t}^{3}}{μ} = 2 π \frac{2442 1 ^{3}}{398600} s$ Why this tool? Specific orbital energy fixes speed but not timing; Kepler's third law $T = 2 π a^{3} / μ$ converts orbit size into elapsed time.

Step 3 — evaluate the full period. $T = 2 π \frac{1.4566 \times 1 0 ^{13}}{398600} = 2 π 3.654 \times 1 0^{7} = 3.798 \times 1 0^{4} s$

Step 4 — Hohmann uses only the half-ellipse (periapsis → apoapsis). $t_{transfer} = \frac{T}{2} = π \frac{a _{t}^{3}}{μ} = 1.899 \times 1 0^{4} s \approx 5.28 hours$ Why half, and hence the $π a_{t}^{3} / μ$ form? You ride from periapsis to apoapsis — exactly one half-loop of the ellipse — so the coast time is $T /2$ , which turns $2 π$ into $π$ : the standard transfer-time formula $t_{transfer} = π a_{t}^{3} / μ$ .

Verify: $\approx 5.3$ h is the textbook LEO→GEO transfer time ✔. Units: $km^{3} / (km^{3} s^{- 2}) = s$ ✔.

Ex 8b — solve backward from a given Δv

Step 1 — write $Δ v_{1}$ in factored form and isolate the square root. $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) \Rightarrow \frac{2 r _{2}}{r _{1} + r _{2}} = 1 + \frac{Δ v _{1}}{μ / r _{1}}$ Why? $r_{2}$ only appears inside that one square root, so we peel everything else off first.

Step 2 — plug numbers. $v_{c 1} = 398600/6678 = 7.726$ km/s, so $\frac{2 r _{2}}{6678 + r _{2}} = 1 + \frac{2.42}{7.726} = 1.3133 \Rightarrow \frac{2 r _{2}}{6678 + r _{2}} = 1.7248$ Why square? To remove the root and get a linear equation in $r_{2}$ .

Step 3 — solve the linear equation. $2 r_{2} = 1.7248 (6678 + r_{2}) \Rightarrow (2 - 1.7248) r_{2} = 1.7248 \times 6678 \Rightarrow r_{2} = \frac{11518}{0.2752} = 41857 km$ Why linear? After clearing the denominator both sides are first-degree in $r_{2}$ .

Verify: $r_{2} \approx 41900$ km — essentially GEO (42 164 km), matching the forecast ✔. Forward-check: $Δ v_{1} = 7.726 (2 \cdot 41857/ (6678 + 41857) - 1) = 2.42$ km/s ✔ (round-trip consistent).

Recall Which cell is which? (self-test)

Given only altitudes, first step? ::: Add planet radius to get center-relative $r_{1}, r_{2}$ (Cell G). Lowering an orbit — are the burns prograde or retrograde? ::: Retrograde (you slow down), but $Δ v$ magnitudes are still positive (Cell D). $Δ v_{1}$ as $r_{2} \to \infty$ ? ::: $(2 - 1) v_{c 1}$ , the near-escape kick (Cell F). $Δ v_{tot}$ when $r_{1} = r_{2}$ ? ::: Exactly zero (Cell E). For a raise, which burn is bigger and does that change with $r_{2}$ ? ::: Burn 1 (periapsis), and it dominates more as $r_{2}$ grows (Cells A, C). Transfer time formula? ::: $t_{transfer} = π a_{t}^{3} / μ$ (half the ellipse period) (Cell H). To find $r_{2}$ from a given $Δ v_{1}$ ? ::: Isolate the square root, square, solve the resulting linear equation (Cell H, Ex 8b).