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https://www.reddit.com/r/space/comments/1ktjfi/deltav_map_of_the_solar_system/ https://www.reddit.com/r/space/comments/1ktjfi/deltav_map_of... is another way of
by sxp 5y ago
https://www.reddit.com/r/space/comments/1ktjfi/deltav_map_of_the_solar_system/ https://www.reddit.com/r/space/comments/1ktjfi/deltav_map_of... is another way of looking at the problem. Once you're in Earth orbit, you need 2.4+.68+.14+.68+1.73 = 5.63 km/s to get to the moon. But you need 2.4+.68+.09+.28+2.06+6.31+1.22+3.06= 16.1 km/s to get to Mercury.
So if you don't use gravitational slingshots (which can be modeled as ramming your spaceship into a planet's gravitational field and bouncing off in the other direction), then you need ~3x the delta-v to get to Mercury compared to the moon. And getting to the Moon required one of the most powerful rockets in human history.
In theory, you could just use a rocket that's 3x as powerful as the Saturn V to get enough fuel into orbit to get to Mercury via a direct route. But this runs into engineering issues with creating a rocket that big. Alternatively, you could develop a way to refuel rockets in orbit and then launch 3x Saturn Vs with one space probe and 2 fuel tanks. This is the plan for SpaceX where they will launch a Starship with humans/robots and a Starship with fuel to get enough fuel into orbit for a trip to various parts of the Solar System.
[edited to fix units]
- firebaze 5y agoPlease fix your units :) You don't need 16.1 m/sec to get to mercury, there's a 'k' missing. To put it another way: that's off by about a factor of 1000.
- keanebean86 5y agoHow long until we can strap a massive laser to the moon and use it to push ships around the solar system?
- shadowgovt 5y agoAny old time. Powering and maintaining it would be a chore, and since it's on a rotating body it wouldn't always be pointing the direction we wanted it to, but solve those problems and we're great. Also somewhat fun is the idea of "solar sailing." Both the particulate solar wind and the raw solar radiation have momentum, and by redirecting either one you can get a push. Since increasing / decreasing the "height" of your orbit is actually a function of causing acceleration in the direction of orbit (or opposite that direction), we should be able to change the shape of a vessel's orbit by deflecting outward-streaming sunlight and particles so they're facing along the orbital path (i.e. a 90-degree turn, with a 45-degree-angle mirror). You'd need a lot of surface area for a reasonable amount of delta-V in a human timeframe, but nothing we know of makes this approach impossible. JAXA demonstrated it can work (https://en.wikipedia.org/wiki/IKAROS); https://en.wikipedia.org/wiki/IKAROS); the Planetary Society has been looking into it (https://www.planetary.org/sci-tech/lightsail https://www.planetary.org/sci-tech/lightsail).
- coldtea 5y agoMillenia, if ever?
- sfifs 5y agoOn the side facing the earth probably never as it will become trivial to threaten earth with a giant laser on the moon. Same objection with kinetic launchers (aka flinger, mass drivers) on the moon. On the side facing away, maybe but you lose half the utility.
- amanzi 5y agoI don't fully understand that chart, but it seems like it's showing that it's harder to get to smaller planets than larger ones - i.e. takes more effort (fuel) to get into a smaller planet's gravitational pull, but you get a lot less benefit since the gravitational pull is less. Is that a reasonable assessment?
- firebaze 5y agohttps://en.wikipedia.org/wiki/Orbital_mechanics https://en.wikipedia.org/wiki/Orbital_mechanics explains the mathematics behind the delta-v calculations quite well.
- bananabreakfast 5y agoIt's actually the other way around. If takes more effort to lower yourself into a large gravity well than a small one. Traveling anywhere outside of Earth's gravity means you are always going to fall into a well, so the smaller the better. The problem with Mercury is to get there you have to fall into the Sun's gravity well. The sun has the biggest gravity well around, and Mercury is deep down inside of it. If Mercury was bigger it would actually be even harder to get there, since you'd then also have to fall down it's large well too.
- b33j0r 5y agoRight on. Someone has mentioned aerobraking somewhere above, but for those with fewer hours spent in Kerbal, this may be of interest ;) A delta-v map will usually include atmosphere as a separate route from the raw energy requirements of entering and exiting gravity wells. Some of the counterintuitive aspects of these flight plans involve “very nice, ok! But how do you plan to slow down when you get there?” On arrival to the moon or mercury (or Minimus) you have to be able to stop under your own power, which you have necessarily brought in your tanks. And therefore mass. Too bad the resonant microwave propulsion thing was more likely a poorly designed experiment (acceleration without propellant). But… it violated our understanding of physics, so we’re stuck with bringing our gas everywhere.
- 5y ago
- robotresearcher 5y ago[edit: typo was corrected, nevermind] 5.63 m/2 vs 16.1 m/s ? Neither 5.63 meters divided by 2, nor 16.1 meters per second can be right, from context.
- state_less 5y agoOne might employ a solar sail/parachute. When you get near Mercury, you’d have more than 4 times the force on the sail too.
- samplatt 5y agoHow would the sun push a solar sail towards the sun?
- giantrobot 5y agoYou sail at night. It's not rocket science.
- fuzzylightbulb 5y agoObviously it is sail science and therefore boat stuff. This should be a joint NASA + NOAA mission.
- deleted 5y ago[deleted]
- tomatotomato37 5y agoOrbital manuevers are kinda counter-intuitive in that you when you want to get closer/farther from whatever you're orbiting you actually want to change the speed at which you're orbiting. So in the case of solar sails, you would tack in a retrograde direction.
- Teever 5y agohttps://en.m.wikipedia.org/wiki/Tacking_(sailing) https://en.m.wikipedia.org/wiki/Tacking_(sailing)
- JohnJamesRambo 5y agoBut to tack you need a centerboard or keel in the water to push against. I can't think of a way to do this in space.
- ithkuil 5y ago> then you need ~3x the delta-v to get to Mercury compared to the moon. [..] > In theory, you could just use a rocket that's 3x as powerful as the Saturn V to get enough fuel into orbit to get to Mercury via a direct route I'm a bit confused. I have a vague memory that due to the Tsiolkovsky rocket equation the amount of fuel needed to reach a delta-v grow exponentially. How is it possible that you need a 3x bigger rocket to reach 3x the delta-v?
- sxp 5y agoI'm fuzzing the math a bit, but assume you have a Saturn V style rocket that can get 100 tons into orbit. And that consists of 50 tons of spaceship and 50 tons of fuel. Then you get something like V_e * ln((50+50)/50) = .69 * V_e of delta-v. But if you had two more rockets that launched an additional 2 * 100 tons of fuel into orbit, you get V_e * ln((250+50)/50) = 1.79 * V_e. This gives you 2.5x the delta-v. Decrease the propellant mass fraction and this value goes up. The numbers are completely made up and based on my time playing KSP (https://xkcd.com/1356/ https://xkcd.com/1356/) rather than real rockets so change the numbers as you see fit. But my core point is that orbital refueling greatly extends the usable delta-v of a rocket once it's prepared in orbit since it's easier to build 3x of a normal rocket than a rocket with 3x the payload.
- croutonwagon 5y ago………
- thaumasiotes 5y ago> When you escape earth orbit you have to exceed its orbital velocity around the sun. This doesn't sound right. Orbital and escape velocities for the earth should depend only on the mass of the earth.
- bagels 5y agoYou can escape earth's orbit going barely faster, or barely slower than earth relative to the sun, depending on where you're going. This stuff can be confusing, because you're mixing two different reference frames (earth and the sun).
- bee_rider 5y agoHuh, LEO really is just about halfway to almost anywhere if you are willing to aerobrake, other than Mercury. I wonder why the chart doesn't consider the possibility to aerobreak into the sun.
- bagels 5y agoA lot more delta-v to get a transfer to the sun, then you have to deal with the thermal problems which are pretty difficult to overcome. If you're close enough to the sun to be in its atmosphere, it's going to be pretty hot.
- medstrom 5y agoHell, my phone could fry bacon after 10 minutes in the sunlight and we're 18,000,000 km away from the source.
- Natsu 5y agoNow where's the version of the map that accounts for lithobreaking[1] as well as aerobraking? :) [1] I feel that lithobreaking is slightly more correct than lithobraking.
- vikingerik 5y agoHere's another angle: Look at the orbital velocities. Mercury's is 48 km/s, Earth's is 30, and an object at infinity would be zero. Kinetic energy is proportional to velocity squared. Square those orbital velocities, and you see it takes more kinetic energy to change from Earth's orbit to Mercury's than to go from Earth's orbit to escaping the solar system.
- dheera 5y ago> 16.1 km/s to get to Mercury That's if you want to slingshot yourself to Mercury with no fuel though, right? If you have fuel then you can keep going to Mercury at a steady 0.1 km/s by constantly firing rockets to maintain that speed, and get there eventually. You don't need escape velocity to leave the Earth either; you only need escape velocity if you want to maintain orbit without constantly firing rockets.
- bagels 5y agoThese are delta-v numbers. How much do you need to change your velocity by to get there, basically. I think your comment misunderstands inertia and orbital dynamics, but it's hard to follow what you propose doing differently than your basic earth escape + transfer.
- neurostimulant 5y agoYou can burn your rocket engine in short powerful burst or in long gentle burst, but delta-v calculation still applies. https://en.wikipedia.org/wiki/Delta-v https://en.wikipedia.org/wiki/Delta-v
- randallsquared 5y agoThat "27" in front of Venus has to be a typo, right?
- raldi 5y agoNope; Venus has a very thick atmosphere and thus a rocket ride off the surface would require a prodigious amount of energy.
- randallsquared 5y agoHm. The creator did label it "delta-v", but apparently they did mean "energy required by a simple rocket" instead.
- blorgleblop 5y agoNo, they mean Δv. When launching from atmosphere the losses of doing so are included in the calculation. The losses from gravity drag are also included but that’s a different topic. The numbers on the map are a low guideline and are nearly ideal vehicles. Depending on the drag coefficient of the vehicle and such, actual results will be worse. Often significantly. Saturn V punched its way to orbit and wasn’t exactly svelte. I understand the confusion, though, because Δv in a space dynamics context is different than in a physics one. It is not a direct measure of added velocity as you’d expect from kinematics, but instead required impulse per unit of mass in order to achieve the desired outcome, which is ever so slightly different and considers additional impactful variables. Remember, the question being answered is really about fuel, so how much impulse you lose to any number of factors goes into your Δv budget as if you needed the extra anyway. I’m assuming LEO is 9.4 on the map (won’t load for me for some reason), and if it is, that’s the best number for space people. Physics would instead tell you that it can be done in 8. Different questions.
- xvilka 5y agoLanding is even worse - Mercury has no atmosphere so braking would also require big amount of fuel or more gravitational braking, which takes a lot of calculation and time.
- iso1210 5y agoYou can always lithobrake on Mercury, depends on your goal :D
- inglor_cz 5y agoOrbital refueling is probably the most practical answer in the short run.