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It's about 1.7km/s delta-v to land on the moon from LLO. If you are assuming an orbital station in the lunar gateway, that's another ~0.7km/s. From LEO that's a
by CuriousCosmic 4y ago
It's about 1.7km/s delta-v to land on the moon from LLO. If you are assuming an orbital station in the lunar gateway, that's another ~0.7km/s. From LEO that's another ~3.6km/s (3.2 to TLI and 0.4 to gateway). That gives you around 6km/s delta-v flat. Of course you can optimize this a bit by using a more direct route but it's probably cheaper if you leverage existing infra like the gateway.
Compare this to anything on mars. It'll be 3.6-4.0km/s to mars injection. Then 2.0km/s to low orbit (can be optimized with clever aerobreaking down to 1.0km/s). To the surface will be a minimum of another 0.4km/s (no engine, parachute only) and closer to 1km/s if you want a soft landing. So you can expect 5km/s at the bare minimum and ~7km/s along established "routes" leveraging to-be-existing infra.
So then the question becomes whether lower delta-v returns, regular launch opportunities, and shorter comms delay are more or less important than 1km/s delta-v savings.
I'd argue the delta-v savings aren't worth it on their own. Mars will eventually be worth exploring and it's a valuable opportunity but by every measure other than raw "throw it at the orbital body with no support or return plan" delta-v savings, the moon is a better site for initial extraterrestrial habitation studies.
- hoseja 4y agoThis of course doesn't mention the ~10km/s you need to actually get off Earth.
- paledot 4y ago"If you can get your ship into orbit, you're halfway to anywhere." —Robert Heinlein
- sandworm101 4y ago>> It'll be 3.6-4.0km/s to mars injection. Then 2.0km/s to low orbit (can be optimized with clever aerobreaking down to 1.0km/s). Except that all that delta at mars is done with aerobraking/aerocapture. Inbound probes can crash strait into the mars atmosphere without slowing down for orbital insertion. That's why it takes far less fuel to get to surface of mars than the moon, at least with objects car-sized or smaller. Only with larger/heavier objects is mar's atmosphere so thin that fuel must be used for capture prior to descent. In the following deltaV map, much of the trip to mars is a free ride so long as you follow the red arrows. There are no free rides on the way to the moon. https://space.stackexchange.com/questions/2046/delta-v-chart-mathematics https://space.stackexchange.com/questions/2046/delta-v-chart...
- m4rtink 4y agoFor the record we have not done aero capture at Mars just yet - but aerobraking has been used a lot, after first capturing into a high eccentric orbit with an engine burn. In any case, bodies with atmosphere do enable you to effectively shave up to half of the delta-v for a transfer when you know what you are doing. :)
- mlindner 4y ago> It's about 1.7km/s delta-v to land on the moon from LLO. Looking at it from LLO isn't the number you should be using. It's the number from Lunar intercept orbits to landing.
- CuriousCosmic 4y agoI was just illustrating the path along the lunar gateway (LEO -> TLI -> Gateway -> LLO -> surface). Even if you do a direct (LEO -> TLI -> LLO -> surface) route, you still have to go through LLO as you'll need to be able to pick your landing spot and potentially even delay a landing (which is not possible if LLO is not budgeted in). Actually now that I think about it, if you want to save delta-v, you can get an extra 25% off the TLI leg of the direct route with a low energy transfer using weak stability boundary trajectories. These savings however come at the cost of a significantly longer flight time and far slimmer margins for failure.
- simonebrunozzi 4y agoI'm not an expert, but I would explore the possibility of a space elevator going from an spaceship orbiting the Moon to the Moon surface, and back.