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> 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?

I'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/) 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.

One big factor that the Tsiolkovsky rocket equation fails to account for is the square–cube law. When you scale up a rocket, the mass of the structure of the rocket tends to grow more slowly than the mass of the fuel it contains, which more or less cancels out the log term in the rocket equation. Note that this does not necessarily hold true for extremely small or extremely large rockets.

But even if the mass of the rocket structure was zero, you still need fuel to move the fuel right?

………

> 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.

You 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).

This analysis falls short if you apply three body orbital mechanics. Going barely faster or slower than the Earth in close proximity to Earth will slowly cause Earth's gravity to pull the spacecraft back in. This can mean all sorts of funny shaped trajectories (horseshoe asteroids etc).

You need about 11-12 km/s to relative to the Earth to escape Earth's gravitational pull from LEO (only a bit more than it takes to reach the moon). Earth itself is orbiting the sun at about 30 km/s. To not get re-captured, the spacecraft needs to be far enough in distance or the difference in velocity needs to be large enough. How much is "enough" depends on the time scale, but for spacecraft we're talking several km/s.

That sounds like you're assuming you're standing still beforehand. Sure, the effect afterwards might look like you're going around the sun faster/slower than Earth but you started deep in Earth's gravitational well.

Here'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.

Please 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.

Huh, 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.

A 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.

Hell, my phone could fry bacon after 10 minutes in the sunlight and we're 18,000,000 km away from the source.

~150,000,000 km

Well, it would explain why their phone is toast after only 10 minutes in the sun...

That's what I get for Googling "8 light minutes in km".

One might employ a solar sail/parachute. When you get near Mercury, you’d have more than 4 times the force on the sail too.

How would the sun push a solar sail towards the sun?

You sail at night. It's not rocket science.

Obviously it is sail science and therefore boat stuff. This should be a joint NASA + NOAA mission.

"For outward bound trajectories, the sail force vector is oriented forward of the Sun line, which increases orbital energy and angular momentum, resulting in the craft moving farther from the Sun. For inward trajectories, the sail force vector is oriented behind the Sun line, which decreases orbital energy and angular momentum, resulting in the craft moving in toward the Sun. It is worth noting that only the Sun's gravity pulls the craft toward the Sun—there is no analog to a sailboat's tacking to windward. To change orbital inclination, the force vector is turned out of the plane of the velocity vector."

https://en.wikipedia.org/wiki/Solar_sail

Orbital 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.

In space you practically never ever want to accelerate radially away or towards whatever you're orbiting. Counterintuitively, on average that doesn't get your closer to or farther from the primary, it only serves to change the eccentricity of your orbit (raising the apoapsis and lowering the periapsis or vice versa).

What you want to do is accelerate either parallel or antiparallel to your orbital velocity (ie. tangent to the orbit). With a solar sail you do this simply by orienting the sail at a roughly 45° angle relative to the sun, allowing you to either accelerate or decelerate depending on which way you reflect the photons (the resulting force will be normal to the sail so in practice solar sailing is more complicated than that, but you get the point).

If only our solar sailing boats could have a keel.

https://en.m.wikipedia.org/wiki/Tacking_(sailing)

But 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.

On Earth you need a keel to oppose the force on the sail, which has a component in the direction of the wind. In orbit, the equivalent radial force is mostly perpendicular to the direction of motion, and so does very little work. Radial forces in orbit are much less efficient at changing orbital parameters than the equivalent axial (prograde/retrograde) acceleration. Moreover, if your orbit is roughly circular the applied force mostly averages out over each orbit.

That's a really good point. Here's what I was able to find about the matter from a brief search.

https://space.stackexchange.com/questions/370/can-you-tack-a...

Tacking not only uses a keel, it also uses the shape of a sail and Bernoulli's principle [0] to get the perpendicular and forward force.

I'm not versed in solar-particle-physics, but I doubt the Bernoulli's principle applies to solar particles.

[0] https://en.wikipedia.org/wiki/Bernoulli%27s_principle

Now 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.

I, too, use 'lithobr(e)aking' as often as possible. Thanks KSP, and your rapid unplanned disassemblies.

Rapid unplanned disassemblies are easy to avoid, though. You just have to add them to your plan :)

Landing 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.

You can always lithobrake on Mercury, depends on your goal :D

I 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?

It'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.

Right 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.

> we’re stuck with bringing our gas everywhere.

Bringing your own gas is fine. The real problem with space travel is you need to bring your own roads. Because what you're short of isn't energy, it's something to push against.

Good point!

Hey, this is just a share to see if anyone shares my interest. In my spare time, I have been kind of been making a videogame based on this question. (I’ve worked on sims professionally, but it is not most of my career.)

A “floating origin” and using alt dynamics is nothing new, but I’m a bit obsessed with weird sim physics in games. I had been playing around with a kinetic quadracopter dogfighting concept in godot engine for about a year.

Then the “tic-tac UFO” videos resurfaced, and I’ve been having a lot of fun imagining “how those physics would work” if reality happened to be as interesting as what Commander Fravor says he saw.

I doubt the UAP news is what anyone thinks it’s about, but I’ve made little simulations involving everything from spacetime warping to matrix-style “if you fly close to c.”

Sadly, it’s not a very rewarding game concept beyond “woah neat!” Gets me back to why I started programming, though.

The physics would be a mechanism, but to be a game, there needs to be some sort of conflict or problem to solve.

I think you should keep exploring it. Maybe this is how folks get an intuition for orbital mechanics.

Ah, got it - thanks.

https://en.wikipedia.org/wiki/Orbital_mechanics explains the mathematics behind the delta-v calculations quite well.

Orbital refueling is probably the most practical answer in the short run.

[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.

The chart in the link says: Delta-V in km/s.

Yup. I fixed the units.

How long until we can strap a massive laser to the moon and use it to push ships around the solar system?

Any 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); the Planetary Society has been looking into it (https://www.planetary.org/sci-tech/lightsail).

since it's on a rotating body it wouldn't always be pointing the direction we wanted it to

Could you put it on one of the poles?

It's also on a moving body, by reference to the sun

On 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.

Millenia, if ever?

That "27" in front of Venus has to be a typo, right?

Nope; Venus has a very thick atmosphere and thus a rocket ride off the surface would require a prodigious amount of energy.

Hm. The creator did label it "delta-v", but apparently they did mean "energy required by a simple rocket" instead.

No, 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.

> 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.

These 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.

You 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