Worked Examples: Earth–Mars Hohmann Transfers
These examples show how you actually calculate a Hohmann transfer between Earth and Mars, both outbound and return. You’ll see transfer time, orbital speeds, delta-v, and the all-important launch phase angle that lets you “aim where the planet will be” instead of where it is now.
Concept illustration of an early human mission to Mars.
Key Constants and Orbital Radii
We work in SI units, but think in astronomical units for intuition. One astronomical unit (AU) is the average Earth–Sun distance.
rM ≈ 1.52 AU ≈ 2.28 × 10¹¹ m (Mars orbital radius)
μ☉ ≈ 1.327 × 10²⁰ m³/s² (Sun’s gravitational parameter)
A Hohmann transfer between two circular orbits is an ellipse whose semi-major axis is just the average of the radii:
The transfer time is half the period of that ellipse:
Whether you go Earth → Mars or Mars → Earth on a Hohmann path, the time of flight is about eight and a half months.
Here you start in low Earth orbit, escape Earth’s gravity, and insert into a solar transfer ellipse that intersects Mars’s orbit exactly when Mars arrives there.
First compute circular orbital speeds of Earth and Mars around the Sun, then the spacecraft’s speed on the transfer ellipse at those same radii (vis-viva equation).
vM = √(μ☉ / rM) ≈ 24.1 km/s
vtr,E = √( μ☉(2/rE − 1/atr) ) ≈ 32.7 km/s
vtr,M = √( μ☉(2/rM − 1/atr) ) ≈ 21.5 km/s
In heliocentric terms, you speed up at Earth to climb outward, then speed up again at Mars to match its faster local orbital speed.
This is the “pure Sun-frame” cost to move from an Earth-like orbit to a Mars-like orbit.
Real missions begin in low Earth orbit (LEO), then perform a trans-Mars injection (TMI) burn to reach the required hyperbolic escape trajectory.
vesc,LEO = √2 · vcirc,LEO ≈ 10.9 km/s
v∞ ≈ 2.9 km/s
vp = √(vesc² + v∞²) ≈ 11.3 km/s
ΔvTMI = vp − vcirc,LEO ≈ 3.6 km/s
- Launch to LEO.
- Perform a ~3.6 km/s TMI burn to escape Earth with v∞ ≈ 2.9 km/s.
- Coast ~259 days on the transfer ellipse.
- Burn at Mars (~2.6 km/s) or use aerobraking to enter Mars orbit.
Launch Phase Angle (Earth → Mars)
During the 259-day cruise, Mars moves about 135° around the Sun. The transfer geometry demands a 180° separation between your departure and arrival points along the orbit, so Mars must start about 45° ahead of Earth at launch.
In other words, you depart when Mars is already leading Earth in its orbit. You aim for where Mars will be 259 days in the future, not where it is today.
Now you begin in low Mars orbit, drop inward toward the Sun, and time the transfer so Earth arrives at the intersection point just as you do.
The transfer ellipse is the same, but now you are leaving from Mars’s orbit and falling inward to Earth’s orbit.
vE ≈ 29.8 km/s (Earth circular)
vtr,M ≈ 21.5 km/s (transfer at Mars)
vtr,E ≈ 32.7 km/s (transfer at Earth)
At Mars you must slow down (retrograde burn) to drop inward. At Earth you must slow down again to match Earth’s slower circular speed at that radius.
Same total cost as the outbound trip, but paid at different points.
Assume a low Mars orbit (LMO) about 300 km above the surface. You need enough hyperbolic excess speed relative to Mars to match the inward transfer ellipse.
vesc,LMO ≈ 4.8 km/s
v∞ ≈ |vM − vtr,M| ≈ 2.6 km/s
vp = √(vesc² + v∞²) ≈ 5.46 km/s
ΔvTEI = vp − vcirc,LMO ≈ 2.1 km/s
- Start in low Mars orbit.
- Perform a ~2.1 km/s trans-Earth injection (TEI) burn to depart Mars with v∞ ≈ 2.6 km/s.
- Coast inward ~259 days on the transfer ellipse.
- Burn at Earth (~3.6 km/s) or use aerobraking/aerocapture to enter LEO.
Departure Phase Angle (Mars → Earth)
During the 259-day cruise, Earth moves about 256° around the Sun, while Mars moves about 135°. For the transfer to meet Earth at the perihelion end of the ellipse, the geometry requires that at departure, Mars is roughly 75–76° ahead of Earth in their orbits.
You leave Mars when Earth is lagging far behind, knowing that the faster inner planet will sweep forward and arrive at the intersection point just as you fall inward.