The distance between moon and earth is just 384,400 km. So, if we can travel (from moon) to another planet, we also can travel (from earth) to the same planet. But I've read that some scientists want a moon base although we can have an earth base. Why? What are the advantages of a moon base.
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4$\begingroup$ One of the things i know is because gravity is lower on the moon it would take less fuel to get out of orbit, so it would be cheaper to travel to other planets $\endgroup$– DriesCommented Dec 9, 2018 at 10:29
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4$\begingroup$ If we put observatories on the far side of the moon, we reduce (by a large factor), the amount of noise (in the form of unwanted electromagnetic radiation) caused by human activity, that interferes with very delicate observation. $\endgroup$– StudyStudyCommented Dec 9, 2018 at 10:56
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1$\begingroup$ The moon is closer to home. It may help facilitate our developments into becoming independent on other worlds due to the moon base having direct—more or less—access to Earth's resources. $\endgroup$– BenjaminFCommented Dec 9, 2018 at 11:24
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1$\begingroup$ @Dries yes, but how would the fuel reach the moon? Only for light sails this would hold. ( or if oil or uranium is discovered on the moon) $\endgroup$– anna vCommented Dec 9, 2018 at 15:43
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$\begingroup$ Without answering the question (which is not a good fit for Physics) I'll note that there are engineering reasons and social/policy reasons and engineering reasons that come to be in response to social/policy limits. $\endgroup$– dmckee --- ex-moderator kittenCommented Dec 9, 2018 at 20:21
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4 Answers
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The reason many scientists want to create a Moon base isn't because of distances in space, it's because of gravity wells. The amount of energy to escape the gravity well of a body in space depends on the mass of said body. For example, the amount of energy required to take off from Earth and go to low Earth orbit is MUCH higher than the amount of energy it takes to take off from the moon and reach low Earth orbit. The Moon has a much lower mass and therefore a much lower gravity well.
The practical XKCD 681 gives a very intuitive example of the amount of energy it takes to get out of a planet (or moon's!) gravity well.
The amount of fuel a rocket needs to "escape" the planet is shown by the depth of the well in this chart. On the bottom right you can see that the gravity well of earth is much deeper than that of the Moon.
Another easy example is in fuel costs. This image shows fuel costs of the Apollo Moon missions:
Don't be fooled by the broken bar for launch. If you were to actually draw this chart to scale, it would be impossible to read as launch fuel costs were around 96% of fuel costs for the entire mission. This means that the Apollo mission used 96% of it's fuel taking off and getting into low Earth orbit. With the remaining 4% of fuel the astronauts went to the Moon, entered orbit, landed, took off, and flew back to earth.
In short, having a Moon base which produces fuel from resources available on the Moon (water ice probably) would mean that you could save enormous amounts of fuel on all missions in space. Instead of having to launch all the fuel you will need on a Mission from Earth (where the launch is very expensive), you could simply refuel at the Moon.
In fact, if asteroid mining ever becomes a reality, it will likely be cheaper to import metal from asteroids for use in earth orbit than to send the metal up from Earth because although asteroids are much further away, the energy requirement to move there and back is far lower than a launch from the surface of Earth to space, or as the old maxim goes: "Once you've reached orbit, you're halfway to anywhere"
answered Dec 10, 2018 at 15:56
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$\begingroup$ However, getting TO the moon to refuel, is still very expensive - more expensive than going to Mars. And while it's cheaper in terms of delta-v to bring propellant from the Moon surface to LEO, than from Earth surface to LEO, on Earth surface we have huge amounts of infrastructure, it's also very hard putting infrastructure on the Moon. $\endgroup$ Commented Dec 11, 2018 at 13:40
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$\begingroup$ @BlakeWalsh how is getting to the moon more expensive than getting to Mars? Are you talking about fuel costs? $\endgroup$– benCommented Feb 8, 2019 at 5:05
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$\begingroup$ @ben It is fuel, but it's not about the cost of the fuel itself, but the size of rocket needed to have tanks big enough to hold that much fuel, In space, going distance doesn't really use any fuel, you just coast, which uses fuel is speeding up and slowing down, which is referred to as "delta-V" -- change in velocity. Extra delta-V needs extra fuel, plus you have to go back and recalculate everything you did so far, to allow for carrying that fuel, so it mounts up really fast, Look up "rocket equation" $\endgroup$ Commented Feb 8, 2019 at 7:32
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2$\begingroup$ @ben Because you can use the atmosphere of mars to slow down. To land on the moon requires roughly 5670 m/s of delta-v, from LEO. To encounter the atmosphere of Mars requires a minimum of 4270 m/s. You then have to consider heat shield and braking strategies (i.e. parachutes or whatever) but in general that kind of stuff is much lighter than carrying propellant. If you send a ship to the moon which has the shielding and such it needs to land on Mars you definitely can't win because you'd use less fuel going directly to Mars. $\endgroup$ Commented Feb 8, 2019 at 14:19
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2$\begingroup$ @ben yes. The more the ship at Mars slows down using rocket engines rather than plowing into the atmosphere the less favorable it is. SpaceX Starship would probably use 600 m/s to land. Note that a lot of the speed a ship has when arriving at a world, is due to that world's gravity, escape velocity from the moon is about 2.4km/s, so a ship approaching the Moon's surface will be going at least that fast and have to burn off at least that much velocity. Tiny objects like near-earth-asteroids and even the tiny moons of Mars require less propellant to land on than the moon. $\endgroup$ Commented Feb 9, 2019 at 11:37
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If we can get some part of fuel production going on the moon this could be a huge help. For example if we can mine water that can be split into hydrogen and oxygen with solar energy.
Given local fuel production we have a fuel supply in a much shallower gravity well.
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The lower gravity and lack of atmosphere appear well suited to launching rockets. The assumption is that available resources that would not have to be brought from Earth will be obtainable on The Moon at lower cost. However the infrastructure required for making effective use of Lunar resources at the level of sophistication needed for a major space launch facility - and the distance to major centres of high tech industry to support it - means doing so is likely to remain prohibitively expensive for the foreseeable future.
A lot of hypothetical preliminary steps need to be achieved successfully for The Moon to be demonstrably better or cheaper than other options for a space launch facilities - ie better than ongoing direct launches from Earth or from orbital facilities.
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One reason could be the weather is better on the Moon.
Rocket launches from Earth are complicated by weather, which could lead to launch postponements. Meanwhile celestial mechanics marches on, oblivious to the puny clouds on a portion of one small planet, so launch windows are limited. Going to the Moon is less weather-sensitive because the Moon is bound to Earth. Then, there are fewer weather issues for the subsequent interplanetary launch from the Moon.
answered Feb 8, 2019 at 3:34
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$\begingroup$ There is such a thing as space weather though that term may apply only to Earth's upper atmosphere. Nonetheless launching from the Moon in sunlight does expose you to stuff thrown out by the Sun during a Coronal mass ejection during a critical phase when a computer glitch could pose a serious problem that you could't just shut-down and reboot from. $\endgroup$– uhohCommented Feb 10, 2019 at 10:43
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$\begingroup$ Can the term “space weather” refer to the environment in deep space, or is it only in reference to Earth's (or another planet's) upper atmosphere? $\endgroup$– uhohCommented Feb 10, 2019 at 10:53