Hohmann transfer — derivation, minimum energy transfer: Exercises

The core toolkit — restated in words below (the first three lines are the true "big three"; the remaining lines are just named shorthands built from those three):

Formula The toolkit (all from

Vis-viva equation and Orbital energy & semi-major axis) The big three:

Circular speed at radius $r$ : $v_{circ} = μ / r$ . "How fast to hold a perfect circle of size $r$ ." We write $v_{c 1} = μ / r_{1}$ for the inner circle and $v_{c 2} = μ / r_{2}$ for the outer.
Vis-viva (speed anywhere on any orbit): $v = μ (\frac{2}{r} - \frac{1}{a})$ , where $r$ is your current distance and $a$ is the semi-major axis (half the longest diameter of the ellipse).
Transfer ellipse size: $a_{t} = \frac{r _{1} + r _{2}}{2}$ — its low point kisses the smaller radius, its high point kisses the larger radius.

Named shorthands (derived from the big three):

Periapsis speed $v_{p}$ = the transfer-ellipse speed at its closest point (evaluate vis-viva at the smaller radius, $a = a_{t}$ ). Apoapsis speed $v_{a}$ = the transfer-ellipse speed at its farthest point (vis-viva at the larger radius, $a = a_{t}$ ). $v_{p}$ is the fast end, $v_{a}$ is the slow end.
The two burns (raising): $Δ v_{1} = v_{p} - v_{c 1}$ (at $r_{1}$ , speed up onto the ellipse), $Δ v_{2} = v_{c 2} - v_{a}$ (at $r_{2}$ , speed up to circularize). For a lowering transfer the signs flip — see L2.4.
Half-orbit coast time: $t = π a_{t}^{3} / μ$ (from Kepler's Third Law).

See the map of a raise in Figure 1 below — every problem is a variation on this one picture.

Figure 1 — Hohmann geometry (raise).

Figure — Hohmann transfer — derivation, minimum energy transfer

Level 1 — Recognition

L1.1

A satellite is in a circular orbit at radius $r = 8000$ km around Earth. What is its speed?

Recall Solution

This is the circular speed formula — the question "how fast to hold a circle" has exactly one answer. $v_{circ} = \frac{μ}{r} = \frac{398600}{8000} = 49.825 = 7.059 km/s$

L1.2

A Hohmann transfer runs from $r_{1} = 7000$ km to $r_{2} = 105000$ km. What is the semi-major axis $a_{t}$ of the transfer ellipse?

Recall Solution

The ellipse's longest diameter reaches from the inner circle to the outer circle, so its length is $r_{1} + r_{2}$ , and $a_{t}$ is half of that. $a_{t} = \frac{r _{1} + r _{2}}{2} = \frac{7000 + 105000}{2} = 56000 km$

L1.3

For the transfer in L1.2, which end of the ellipse is periapsis (the closest point), and which is apoapsis (the farthest point)?

Recall Solution

Periapsis = closest to Earth = the smaller radius $= r_{1} = 7000 km$ . Apoapsis = farthest from Earth = the larger radius $= r_{2} = 105000 km$ . The satellite is fastest at periapsis and slowest at apoapsis (it trades speed for height as it climbs).

Level 2 — Application

L2.1

Raise a spacecraft from $r_{1} = 6800$ km to $r_{2} = 20000$ km around Earth. Find $Δ v_{1}$ , the first burn.

Recall Solution

Step A — current speed (on the inner circle): $v_{c 1} = \frac{398600}{6800} = 58.618 = 7.656 km/s$ Step B — transfer ellipse size: $a_{t} = \frac{6800 + 20000}{2} = 13400$ km. Step C — speed needed at periapsis (vis-viva at $r = r_{1}$ , $a = a_{t}$ ): $v_{p} = 398600 (\frac{2}{6800} - \frac{1}{13400}) = 398600 (2.9412 \times 1 0^{- 4} - 7.4627 \times 1 0^{- 5})$ $= 398600 \times 2.1949 \times 1 0^{- 4} = 87.489 = 9.354 km/s$ Step D — the burn (both velocities point the same way, so subtract magnitudes): $Δ v_{1} = v_{p} - v_{c 1} = 9.354 - 7.656 = 1.698 km/s$

L2.2

Same transfer as L2.1. Find $Δ v_{2}$ , the second burn, and the total $Δ v$ .

Recall Solution

Step A — speed at apoapsis (vis-viva at $r = r_{2} = 20000$ , still $a = a_{t} = 13400$ ): $v_{a} = 398600 (\frac{2}{20000} - \frac{1}{13400}) = 398600 (1.0 \times 1 0^{- 4} - 7.4627 \times 1 0^{- 5})$ $= 398600 \times 2.5373 \times 1 0^{- 5} = 10.114 = 3.180 km/s$ Step B — target circular speed at $r_{2}$ : $v_{c 2} = \frac{398600}{20000} = 19.93 = 4.464 km/s$ Step C — the burn (you arrived too slow, so speed up): $Δ v_{2} = v_{c 2} - v_{a} = 4.464 - 3.180 = 1.284 km/s$ Total: $Δ v = 1.698 + 1.284 = 2.982 km/s$ .

L2.3

For the same transfer, how long is the coast between the two burns?

Recall Solution

You fly half the ellipse, so the time is half the orbital period. Half-period formula: $t = π \frac{a _{t}^{3}}{μ} = π \frac{1340 0 ^{3}}{398600} = π \frac{2.406 \times 1 0 ^{12}}{398600}$ $= π 6.037 \times 1 0^{6} = π \times 2457.0 = 7719 s \approx 2.14 h$

L2.4

Edge case — lowering the orbit. A spacecraft must go from an outer circle $r_{1} = 20000$ km down to an inner circle $r_{2} = 6800$ km (the reverse of L2.1). Find both burns (with their signs/direction) and the total $Δ v$ .

Recall Solution

Sign convention first. $Δ v$ as a fuel cost is always the positive magnitude $∣ v_{after} - v_{before} ∣$ ; the direction (prograde = speed up, retrograde = brake) is what flips when lowering. The transfer ellipse is the same ellipse as in L2.1 — only the travel direction reverses: now periapsis is the target $r_{2}$ and apoapsis is the start $r_{1}$ . $a_{t} = \frac{20000 + 6800}{2} = 13400$ km (unchanged — averaging is symmetric).

Burn 1 — at the outer circle $r_{1} = 20000$ (this is apoapsis of the ellipse). You are currently on the outer circle at $v_{c 1} = 398600/20000 = 4.464$ km/s. To drop onto the ellipse you must slow to the apoapsis speed $v_{a} = 3.180$ km/s (same $v_{a}$ as L2.2). Since you go slower, this is a retrograde (braking) burn: $Δ v_{1} = v_{c 1} - v_{a} = 4.464 - 3.180 = 1.284 km/s (brake)$ Burn 2 — at the inner circle $r_{2} = 6800$ (periapsis of the ellipse). You arrive moving at the periapsis speed $v_{p} = 9.354$ km/s (same $v_{p}$ as L2.1), which is faster than the inner circular speed $v_{c 2} = 398600/6800 = 7.656$ km/s. To circularize you must brake again: $Δ v_{2} = v_{p} - v_{c 2} = 9.354 - 7.656 = 1.698 km/s (brake)$ Total: $Δ v = 1.284 + 1.698 = 2.982 km/s$ .

Key symmetry: identical total to the raise in L2.1–L2.2 (both burns are the same magnitudes, just swapped and now retrograde). Lowering and raising between the same two circles cost exactly the same fuel — you simply brake instead of boost.

Level 3 — Analysis

L3.1

A transfer goes from $r_{1} = 7000$ km to $r_{2} = 42164$ km (LEO→GEO). Someone claims "since we only raise by a factor of $6$ , the total $Δ v$ should be about $6 \times$ some base value." Compute the actual total $Δ v$ and comment on why the raise ratio does not scale the cost linearly.

Recall Solution

$a_{t} = (7000 + 42164) /2 = 24582$ km. $v_{c 1} = 398600/7000 = 7.546 km/s$ $v_{p} = 398600 (2/7000 - 1/24582) = 398600 (2.85714 \times 1 0^{- 4} - 4.06802 \times 1 0^{- 5}) = 97.616 = 9.880 km/s$ $Δ v_{1} = 9.880 - 7.546 = 2.334 km/s$ $v_{a} = 398600 (2/42164 - 1/24582) = 398600 (4.74338 \times 1 0^{- 5} - 4.06802 \times 1 0^{- 5}) = 2.6920 = 1.641 km/s$ $v_{c 2} = 398600/42164 = 3.075 km/s$ $Δ v_{2} = 3.075 - 1.641 = 1.434 km/s$ $Δ v_{total} = 2.334 + 1.434 = 3.769 km/s$ Comment: each burn's size depends on differences of square roots of the radii, not on the raise ratio itself. Doubling $r_{2}$ does not double $Δ v$ ; in fact as $r_{2} \to \infty$ the total $Δ v$ tends to a finite limit (see L4.1). Cost is sub-linear in the ratio — orbits are cheap to make big, expensive to make deep.

L3.2

Two spacecraft each raise their orbit by the same height $Δ r = 3000$ km. One starts low ( $r_{1} = 7000 \to r_{2} = 10000$ km), the other starts high ( $r_{1} = 40000 \to r_{2} = 43000$ km). Which transfer costs less total $Δ v$ , and why?

Recall Solution

Low starter ( $7000 \to 10000$ , $a_{t} = 8500$ ): $v_{c 1} = 398600/7000 = 7.546, v_{p} = 398600 (2/7000 - 1/8500) = 8.183$ $Δ v_{1} = 8.183 - 7.546 = 0.637$ $v_{a} = 398600 (2/10000 - 1/8500) = 5.729, v_{c 2} = 398600/10000 = 6.313$ $Δ v_{2} = 6.313 - 5.729 = 0.584, total = 1.221 km/s$ High starter ( $40000 \to 43000$ , $a_{t} = 41500$ ): $v_{c 1} = 398600/40000 = 3.157, v_{p} = 398600 (2/40000 - 1/41500) = 3.212$ $Δ v_{1} = 3.212 - 3.157 = 0.0546$ $v_{a} = 398600 (2/43000 - 1/41500) = 2.988, v_{c 2} = 398600/43000 = 3.045$ $Δ v_{2} = 3.045 - 2.988 = 0.0568, total = 0.111 km/s$ The high starter is over 10× cheaper for the same height gain. Why: deep in the gravity well speeds are large and gravity is steep, so changing your orbit's energy requires big velocity changes. Far out, everything is slow and gravity is gentle, so the same height costs almost nothing. This is the geometric face of the Oberth effect running in reverse — being deep amplifies energy per $Δ v$ but it also raises the baseline speeds you must overcome to reshape the orbit.

Level 4 — Synthesis

L4.1

Show that as the outer radius grows without bound ( $r_{2} \to \infty$ ) for a fixed inner radius $r_{1}$ , the first burn $Δ v_{1}$ approaches the escape-boost: $Δ v_{1} \to (2 - 1) v_{c 1}$ . Then evaluate it for $r_{1} = 7000$ km.

Recall Solution

First, rederive the first-burn formula from the toolkit. $Δ v_{1} = v_{p} - v_{c 1}$ with $v_{c 1} = μ / r_{1}$ and $v_{p} = μ (2/ r_{1} - 1/ a_{t})$ , $a_{t} = (r_{1} + r_{2}) /2$ . Substitute $1/ a_{t} = 2/ (r_{1} + r_{2})$ : $v_{p} = μ (\frac{2}{r _{1}} - \frac{2}{r _{1} + r _{2}}) = \frac{μ}{r _{1}} \frac{2 r _{2}}{r _{1} + r _{2}}$ (the last step factors $μ / r_{1}$ out of the bracket). Hence the working form: $Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) = v_{c 1} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)$ Take the limit. Divide top and bottom inside the root by $r_{2}$ : $\frac{2 r _{2}}{r _{1} + r _{2}} = \frac{2}{r _{1} / r _{2} + 1} r_{2} \to \infty \frac{2}{0 + 1} = 2$ So $Δ v_{1} \to v_{c 1} (2 - 1)$ This is exactly the burn that turns a circle into a parabola (escape orbit): $2 v_{c 1}$ is escape speed. Interpretation: the periapsis burn of an infinitely-large Hohmann is the escape burn — sensible, since an infinite ellipse is a parabola. Numbers: $v_{c 1} = 398600/7000 = 7.546$ km/s, so $Δ v_{1} \to (2 - 1) (7.546) = 0.41421 \times 7.546 = 3.126 km/s$

L4.2

For a planet-to-planet Hohmann around the Sun, take Earth's orbit $r_{1} = 1.496 \times 1 0^{8}$ km and Mars's $r_{2} = 2.279 \times 1 0^{8}$ km, with $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ . Find both heliocentric burns and the transfer time in days. (This is the Delta-v budget a probe needs on top of leaving Earth.)

Recall Solution

$a_{t} = (1.496 + 2.279) \times 1 0^{8} /2 = 1.8875 \times 1 0^{8}$ km. $v_{c 1} = \frac{1.327 \times 1 0 ^{11}}{1.496 \times 1 0 ^{8}} = 887.03 = 29.783 km/s$ $v_{p} = 1.327 \times 1 0^{11} (\frac{2}{1.496 \times 1 0 ^{8}} - \frac{1}{1.8875 \times 1 0 ^{8}}) = 1.327 \times 1 0^{11} \times (1.33690 \times 1 0^{- 8} - 5.2980 \times 1 0^{- 9})$ $= 1.327 \times 1 0^{11} \times 8.0710 \times 1 0^{- 9} = 1071.0 = 32.726 km/s$ $Δ v_{1} = 32.726 - 29.783 = 2.943 km/s$ $v_{c 2} = \frac{1.327 \times 1 0 ^{11}}{2.279 \times 1 0 ^{8}} = 582.27 = 24.130 km/s$ $v_{a} = 1.327 \times 1 0^{11} (\frac{2}{2.279 \times 1 0 ^{8}} - \frac{1}{1.8875 \times 1 0 ^{8}}) = 1.327 \times 1 0^{11} \times (8.7758 \times 1 0^{- 9} - 5.2980 \times 1 0^{- 9})$ $= 1.327 \times 1 0^{11} \times 3.4778 \times 1 0^{- 9} = 461.50 = 21.482 km/s$ $Δ v_{2} = 24.130 - 21.482 = 2.648 km/s$ Transfer time: $t = π \frac{a _{t}^{3}}{μ _{⊙}} = π \frac{( 1.8875 \times 1 0 ^{8} ) ^{3}}{1.327 \times 1 0 ^{11}} = π \frac{6.7255 \times 1 0 ^{24}}{1.327 \times 1 0 ^{11}} = π 5.0682 \times 1 0^{13}$ $= π \times 7.1191 \times 1 0^{6} = 2.2366 \times 1 0^{7} s$ Convert to days by dividing by the number of seconds in a day, $86400 s/day$ (because $60 s \times 60 min \times 24 h = 86400$ ): $t = \frac{2.2366 \times 1 0 ^{7} s}{86400 s/day} = 258.9 days$

Level 5 — Mastery

L5.1

Prove the classic threshold: a bi-elliptic transfer starts to beat a Hohmann transfer when the ratio $R = r_{2} / r_{1}$ exceeds ≈ 11.94 (for a bi-elliptic whose intermediate apoapsis $r_{b} \to \infty$ ). Do this by comparing total $Δ v$ 's as functions of $R$ and finding where they cross.

Recall Solution

First define the bi-elliptic and its symbol. A Bi-elliptic transfer uses three burns: burn out onto a big ellipse whose apoapsis reaches an ==intermediate radius $r_{b}$ == chosen larger than $r_{2}$ ; coast to $r_{b}$ and burn again to raise periapsis up to $r_{2}$ ; coast back down to $r_{2}$ and burn a third time to circularize. Here $r_{b}$ is that intermediate apoapsis radius — the "detour distance." We analyze the limiting case $r_{b} \to \infty$ (fling out to infinity), which gives the best the bi-elliptic can ever do.

Work in units where $r_{1} = 1$ and $v_{c 1} = μ / r_{1} = 1$ . Let $R = r_{2} / r_{1}$ .

Hohmann total (sum of the two burns, with $v_{c 1} = 1$ , $v_{c 2} = 1/ R$ , $a_{t} = (1 + R) /2$ ): $Δ v_{H} = Δ v_{1} (\frac{2 R}{1 + R} - 1) + Δ v_{2} \frac{1}{R} (1 - \frac{2}{1 + R})$

Bi-elliptic with infinite intermediate apoapsis ( $r_{b} \to \infty$ ). Burn 1 raises the circle at $r_{1} = 1$ onto an escape-like ellipse: from L4.1 that costs $2 - 1$ . As the apoapsis recedes to infinity the middle burn cost tends to $0$ . The final burn drops you from an infinite-apoapsis ellipse down to a circle at $r_{2}$ : apply the same escape argument at radius $R$ , where circular speed is $1/ R$ , giving cost $\frac{1}{R} (2 - 1)$ . Total: $Δ v_{B E}^{\infty} = (2 - 1) + \frac{1}{R} (2 - 1) = (2 - 1) (1 + \frac{1}{R})$

Find the crossover $Δ v_{H} (R) = Δ v_{B E}^{\infty} (R)$ . This is transcendental, so solve numerically. Tabulating the two expressions:

$R = 10$ : $Δ v_{H} = 0.3944$ , $Δ v_{B E}^{\infty} = 0.4550$ → Hohmann wins.
$R = 11.94$ : $Δ v_{H} = 0.3785$ , $Δ v_{B E}^{\infty} = 0.3785$ → tie (this is the root).
$R = 15$ : $Δ v_{H} = 0.3606$ , $Δ v_{B E}^{\infty} = 0.3211$ → bi-elliptic (infinite) wins.

The root sits at $R^{⋆} \approx 11.94$ Reading it: for $R < 11.94$ no bi-elliptic can beat a Hohmann; for $R > 11.94$ an infinite (and hence a sufficiently large finite) $r_{b}$ does. Between $R \approx 11.94$ and $R \approx 15.58$ only a finite $r_{b}$ wins; above $\approx 15.58$ even the infinite case wins outright. See Bi-elliptic transfer. The figure shows both curves crossing.

Figure 2 — Hohmann vs bi-elliptic total $Δ v$ vs ratio $R$ .

L5.2

Design check. You must raise a satellite from $r_{1} = 6600$ km to $r_{2} = 90000$ km ( $R \approx 13.6$ ). Compute the Hohmann total $Δ v$ , then the infinite bi-elliptic bound, and state which strategy you would investigate further.

Recall Solution

Use $v_{c 1} = 398600/6600 = 7.771$ km/s and $R = 90000/6600 = 13.636$ . Hohmann: $a_{t} = (6600 + 90000) /2 = 48300$ km. $v_{p} = 398600 (2/6600 - 1/48300) = 398600 (3.0303 \times 1 0^{- 4} - 2.0704 \times 1 0^{- 5}) = 112.98 = 10.629 km/s$ $Δ v_{1} = 10.629 - 7.771 = 2.858$ $v_{a} = 398600 (2/90000 - 1/48300) = 398600 (2.2222 \times 1 0^{- 5} - 2.0704 \times 1 0^{- 5}) = 0.6053 = 0.7780 km/s$ $v_{c 2} = 398600/90000 = 2.104 km/s, Δ v_{2} = 2.104 - 0.778 = 1.326$ $Δ v_{H} = 2.858 + 1.326 = 4.184 km/s$ Infinite bi-elliptic bound: $Δ v_{B E}^{\infty} = (2 - 1) v_{c 1} (1 + \frac{1}{R}) = 0.41421 \times 7.771 \times (1 + 1/ 13.636)$ $= 3.219 \times (1 + 0.27080) = 3.219 \times 1.27080 = 4.091 km/s$ Verdict: Since $R = 13.6 > 11.94$ , the bound already sits below Hohmann ( $4.09 < 4.18$ ). A finite- $r_{b}$ bi-elliptic would do even better — so this raise is a genuine candidate for Bi-elliptic transfer. But weigh it against the transfer time, which for bi-elliptic can be many times longer.

Recall

Recall One-line self-test

Circular speed at $r$ ? ::: $μ / r$ .
Speed at periapsis of transfer? ::: vis-viva at $r_{1}$ with $a = a_{t}$ .
Why is the far-out same-height raise cheap? ::: speeds fall as $1/ r$ , so the same $Δ r$ spans a tiny speed change.
Does lowering cost the same as raising? ::: yes, same magnitudes — you just brake instead of boost.
Bi-elliptic threshold? ::: $r_{2} / r_{1} \approx 11.94$ .
Limit of $Δ v_{1}$ as $r_{2} \to \infty$ ? ::: $(2 - 1) v_{c 1}$ , the escape boost.

Related: 3.2.19 Hohmann transfer — derivation, minimum energy transfer (Hinglish) · Vis-viva equation · Kepler's Third Law · Oberth effect · Two-body problem.