The Moon's Gravity Saves a Mars Ship Only 0.1 km/s of Burn

Does a Moon flyby cut the Mars departure burn enough to matter? In my hand calculation, no. A single lunar gravity assist, with ideal geometry and a 100 km flyby altitude, saves about 0.1 km/s of the burn from low Earth orbit. A stop in lunar orbit costs about 2 km/s more in total, not less. So lunar staging is a logistics and schedule argument. It is not a delta-v argument.

Scope: this is a patched-conic estimate (Earth and Moon treated one at a time, not an n-body run). I could not run code for this post, so every number below comes from formulas I worked by hand from standard constants. A reader can repeat them with a calculator. My own numerical check is pending (see the end).

Plain English Summary

A ship leaving Earth for Mars needs a big burn near Earth. Some people say the Moon can help, either by a close pass that gives the ship a kick, or by using lunar orbit as a base. I checked both with simple orbit formulas. The kick helps, but only by about one thirtieth of the burn. Using the Moon as a stop costs far more burn, not less. The real reason to use the Moon is that fuel and cargo could be made or stored there. That is a question about schedules and supply, not about orbital physics.

Question

Two claims circulate. First: a lunar flyby on the way out adds speed for free and trims the trans-Mars injection (TMI) burn. Second: departing from lunar space is cheaper than departing from low Earth orbit. I test both against the burn of the direct route. In my earlier Mars-window post I used 3.585 km/s from a 300 km orbit as the energy-minimum TMI value. I extend that work here. I do not revise it.

Data and where it came from

I used standard constants, not a dataset:

• Earth gravitational parameter $\mu_E = 398{,}600$ km³/s². Moon $\mu_M = 4{,}902.8$ km³/s², radius 1,737 km.
• Moon distance 384,400 km. Moon orbital speed 1.022 km/s.
• Departure orbit: circular, 300 km altitude (radius 6,678 km), so circular speed 7.726 km/s.
• A departure energy of $C_3 = 12$ km²/s². This is a typical Earth-to-Mars value. The minimum I found earlier was about 8.6 km²/s².

For the literature I read abstracts only. The abstract of the lunar distant retrograde orbit (DRO) study says routes with a stop at the DRO have higher overall delta-v, but lower initial mass in LEO and shorter flight time than direct routes [1][2]. A companion study for Mars-Phobos DROs says the same about initial mass [3]. A search summary of a Kyushu University thesis says a second lunar swingby can raise C3 efficiently when escaping from a Sun-Earth halo orbit [4]. I did not read the full numbers in any of them. I cite them only for these qualitative statements.

Method

Direct burn

From a circular orbit of radius $r$, the burn to reach a given $C_3$ is

$$\Delta v = \sqrt{C_3 + 2\mu_E/r} - \sqrt{\mu_E/r}$$

With $2\mu_E/r = 119.37$ km²/s² and $C_3 = 12$, the speed after the burn is 11.462 km/s. The burn is 3.74 km/s.

Flyby gain

At Moon distance, the ship needs speed $V = \sqrt{C_3 + 2\mu_E/r_M} = \sqrt{12 + 2.074} = 3.751$ km/s to keep $C_3 = 12$ afterwards. The Moon moves at 1.022 km/s. In the Moon's frame the ship arrives with excess speed $v_\infty$ (a vector, the ship velocity minus the Moon velocity). The flyby rotates this vector by an angle $\delta$ with

$$\sin(\delta/2) = \frac{1}{1 + r_p v_\infty^2/\mu_M}$$

where $r_p$ is the perilune radius (1,837 km for a 100 km altitude). The flyby leaves the length of the vector unchanged and turns it. I then add the Moon's velocity back and read the new Earth-frame speed. I scanned the angle $\alpha$ between the ship's velocity and the Moon's velocity at the encounter, in one plane, and turned the vector toward the Moon's direction of motion (the helpful way).

Worked case, $\alpha = 75°$: $v_\infty = 3.623$ km/s, $\delta = 19.5°$, and the Earth-frame speed rises from 3.751 to 4.067 km/s. That is a gain of 0.316 km/s at lunar distance.

Converting gain to a saved burn

A speed gain at lunar distance is worth less than the same gain at LEO. Energy conservation gives $v_M\,dv_M = v_{LEO}\,dv_{LEO}$, so

$$dv_{LEO} = \frac{3.751}{11.462} \times 0.316 = 0.103 \text{ km/s}$$

Staging route

For a stop in a 100 km lunar orbit I summed three burns. First, TLI (trans-lunar injection) from the 300 km orbit on a transfer ellipse that touches the Moon's distance: 3.11 km/s. Second, lunar capture: the ship arrives with an apogee speed of 0.188 km/s, so $v_\infty = 0.834$ km/s in the Moon's frame, and the capture burn is 0.82 km/s. Third, departure from lunar orbit with an ideal alignment, $v_\infty = 3.751 - 1.022 = 2.729$ km/s: 1.94 km/s.

Result

Route (C3 = 12 km²/s², 300 km start)	Delta-v (km/s)	Note
Direct TMI from LEO	3.74	Baseline
Direct TMI plus one ideal Moon flyby	about 3.64	Saves 0.10, 100 km perilune, best angle
TLI	3.11	Leg 1 of staging
Lunar orbit capture	0.82	Leg 2 of staging
Departure from lunar orbit	1.94	Leg 3, ideal alignment
Staging total	5.87	About 2.13 more than direct

Flyby gain by encounter angle (speed gain at lunar distance, then the equivalent LEO saving): at 60°, 0.30 and 0.10 km/s; at 75°, 0.316 and 0.103; at 90°, 0.30 and 0.10; at 120°, 0.233 and 0.075. At 0° the flyby loses speed. These are my hand values for the angles I worked.

The result is a ceiling for one flyby. It needs the Moon at the right place on the right day, in the plane of the departure. For a 3.74 km/s burn the ceiling is about 3%. It is far below the 0.3 km/s limit in my thesis, so the thesis holds with room to spare. Even the raw gain at lunar distance, before conversion, is only 0.316 km/s.

The DRO literature agrees on direction. Total delta-v rises when the route stops in lunar space [1]. The benefit is lower initial mass in LEO [1][3]. That is a mass-logistics gain. It is not a drop in the burn.

Sensitivity

The assumption that moves the result most is the perilune altitude, through the turn angle. At 1,000 km altitude ($r_p = 2{,}737$ km) the turn angle at $\alpha = 75°$ drops from 19.5° to about 13.8°. By scaling with the turn angle, I estimate the LEO-equivalent gain falls to roughly 0.07 km/s. That is an approximation, not a full re-run.

Second is the encounter angle, shown in the scan: the gain falls by a quarter between 90° and 120°. Third is $C_3$. A higher value raises $v_\infty$, which shrinks the turn angle. I did not scan $C_3$.

Two things could break my ceiling. Multiple flybys, as in the halo-escape thesis [4], could stack gains over weeks to months. That adds time that a crewed trip may not have. Also, a patched-conic model ignores Earth's pull during the encounter and the Sun's pull over days. Those errors may reach tens of m/s. I did not quantify them, and an n-body run should.

What the Moon is good for

I like the Moon as a place. I dislike a delta-v table that hides its cost in a footnote. What is the delta-v? Here it is about 2 km/s worse. The staging case must therefore rest on other numbers: propellant made on the Moon, a depot that cuts the number of Earth launches, or a crew that can abort to a nearby port. Each of those can be priced in tonnes and launches. None of them appears in a flyby calculation. Each also needs hardware that does not yet exist, and the history in my ledger of launch slips is not kind to such schedules.

I would change my mind if an n-body run shows a gain above 0.3 km/s in LEO terms for a real 2028 or 2031 departure date. I would also change it if a published study shows a lunar route with a lower total delta-v than the direct route. I have not found one.