Hohmann Transfer Orbits and Delta-V
Hohmann Transfer Orbits
Getting from one orbit to another requires changing velocity — and in the vacuum of space, velocity changes require burning precious fuel. The most fuel-efficient way to transfer between two circular orbits is the Hohmann transfer, invented by German engineer Walter Hohmann in 1925. This elegant maneuver uses two brief engine burns to move a spacecraft from one circular orbit to another via an elliptical transfer orbit.
The Hohmann transfer works as follows. The spacecraft starts in a lower circular orbit of radius r1. It performs a prograde burn (in the direction of travel) to enter an elliptical transfer orbit whose periapsis is r1 and apoapsis is r2, the radius of the target orbit. The spacecraft then coasts along the ellipse for half an orbit until it reaches apoapsis at r2, where it performs a second prograde burn to circularize into the target orbit.
Each burn changes the velocity by a specific amount called delta-v. The total delta-v for the transfer is the sum of the two burns. The velocities on the transfer ellipse can be calculated using the vis-viva equation, which relates speed to position on any orbit: v squared equals GM times the quantity 2 over r minus 1 over a, where a is the semi-major axis of the orbit.
The vis-viva equation is perhaps the most useful single equation in orbital mechanics. It applies to circular orbits, elliptical orbits, parabolic escape trajectories, and hyperbolic flybys. For a circular orbit, r equals a and it reduces to v equals square root of GM over r. For escape, a is infinite and it gives the escape velocity formula.
$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$$
The most important equation in orbital mechanics. Works for any conic orbit.
Transfer orbit semi-major axis: $a_t = (r_1 + r_2)/2$
First burn (at $r_1$): $$\Delta v_1 = \sqrt{\mu\left(\frac{2}{r_1} - \frac{1}{a_t}\right)} - \sqrt{\frac{\mu}{r_1}}$$
Second burn (at $r_2$): $$\Delta v_2 = \sqrt{\frac{\mu}{r_2}} - \sqrt{\mu\left(\frac{2}{r_2} - \frac{1}{a_t}\right)}$$
Total: $\Delta v = |\Delta v_1| + |\Delta v_2|$
Calculate the delta-v to transfer from a 300 km LEO to geostationary orbit (35,786 km altitude).
Step 1: Find orbital radii.
$r_1 = 6371 + 300 = 6671$ km, $r_2 = 6371 + 35786 = 42{,}157$ km.
Step 2: Transfer orbit semi-major axis.
$$a_t = (6671 + 42157)/2 = 24{,}414 \text{ km}$$
Step 3: Velocities using vis-viva (all in meters):
$v_{\text{circ},1} = \sqrt{\mu/r_1} = \sqrt{3.986 \times 10^{14}/6.671 \times 10^6} = 7730$ m/s
$v_{t,1} = \sqrt{\mu(2/r_1 - 1/a_t)} = \sqrt{3.986 \times 10^{14}(2/6.671-1/24.414)\times 10^{-6}} = 10{,}252$ m/s
Step 4: First burn: $\Delta v_1 = 10252 - 7730 = 2522$ m/s
$v_{t,2} = \sqrt{\mu(2/r_2 - 1/a_t)} = 1622$ m/s
$v_{\text{circ},2} = \sqrt{\mu/r_2} = 3075$ m/s
Step 5: Second burn: $\Delta v_2 = 3075 - 1622 = 1453$ m/s
Step 6: Total $\Delta v = 2522 + 1453 = 3975$ m/s $\approx 4.0$ km/s
How long does a Hohmann transfer from LEO to GEO take?
Step 1: The transfer orbit is half an ellipse. Transfer time $= T_t/2$.
$$T_t = 2\pi\sqrt{a_t^3/\mu} = 2\pi\sqrt{(2.4414 \times 10^7)^3/3.986 \times 10^{14}}$$
$$T_t = 2\pi\sqrt{3.655 \times 10^{22}/3.986 \times 10^{14}} = 2\pi \times 9578 = 60{,}177 \text{ s}$$
Step 2: Transfer time $= T_t/2 = 30{,}089$ s $\approx 8.36$ hours.
Practice Problems
Show Answer Key
1. $r_1 = 6571$ km, $r_2 = 7371$ km. $a_t = 6971$ km. $\Delta v_1 = v_{t,1} - v_1 = 7826 - 7788 = 38$ m/s. $\Delta v_2 = v_2 - v_{t,2} = 7353 - 7316 = 37$ m/s. Total $\approx 75$ m/s.
2. $a_t = (6671+6779)/2 = 6725$ km. $T_t = 2\pi\sqrt{(6725000)^3/\mu} = 5492$ s. Transfer $= T_t/2 = 2746$ s $\approx 45.8$ min.
3. $v = \sqrt{\mu(2/r - 1/a)} = \sqrt{1.327 \times 10^{20}(2/1.496\times10^{11} - 1/50\times1.496\times10^{11})} = 41{,}892$ m/s $\approx 42$ km/s.
4. $v_{\text{circ}} = 7668$ m/s. $v_{\text{esc}} = \sqrt{2} \times 7668 = 10{,}843$ m/s. $\Delta v = 3175$ m/s.
5. This requires computing the heliocentric transfer, then the departure burn from LEO. Total departure $\Delta v \approx 3.6$ km/s from LEO.
6. Circular: $r = a$. $v^2 = \mu(2/r - 1/r) = \mu/r = \mu/a$. $v = \sqrt{\mu/a}$. ✓
7. $r_1 = 6871$ km, $r_a = 8371$ km. $a_t = 7621$ km. $v_1 = 7616$ m/s. $v_{t,1} = \sqrt{\mu(2/r_1 - 1/a_t)} = 7914$ m/s. $\Delta v_1 = 298$ m/s.
8. The tangential burns at periapsis and apoapsis are most efficient because they align with the velocity vector (Oberth effect). Any other two-burn transfer would require a component perpendicular to velocity, wasting fuel.
9. $\Delta v = 3000 \ln(10/1) = 3000 \times 2.303 = 6908$ m/s.
10. $a_t = (6671 + 42157)/2 = 24414$ km. $v = \sqrt{\mu(2/6671000 - 1/24414000)} = 10{,}252$ m/s.