I was reading Randall Munroe’s new book How To and he talked about launching something from the Earth into the sun. And he said that the best way (assuming you’re not in a hurry) would be to launch it away from the sun. The sun’s gravity would pull on your item and eventually slow its speed down to zero and then begin pulling it back towards the sun where it would eventually end up. He said that launching the item directly towards the sun would be quicker but would require more energy.
He did not elaborate on this point. And I am confused. I’m assuming he’s right - he used to work for NASA and I’m sure he knows a lot more about space travel than I do. But I don’t see why it would take more energy.
I’m assuming (perhaps wrongly) that it takes the same amount of energy to launch an item off the Earth regardless of which direction it’s being sent to. And it seems to me that if you launched towards the sun initially, your launch speed would be working with the sun’s gravity rather than against it.
You don’t have to completely negate your angular momentum to get to the sun – you could make a Hohmann orbit in which your inner radius is about at the surface of the sun, which is effectively getting you to the sun. But it still requires more delta v than going out of the solar system and falling back. But it’d be a lot quicker than going all the way out and coming back. Actually, if you’re satisfied with just getting to the orbit of Mercury (which is plenty hot enough for most purposes), it’s both less “expensive” in terms of energy and much faster to just Hohmann transfer inwards.
I don’t have the book with me right now, did he really say to launch it away from the Sun? It thought it would be better to launch it backwards (opposite of orbital motion) to kill the orbital motion as much as possible, lowering the perihelion all the way down to the surface of the Sun.
Although I think the most energy-efficient way to get something to the Sun may involve at least one Jupiter flyby (gravity assist) and probably a few more planetary flybys. Is that what he was referring to?
I’m not sure what “away” means in this context- the directly opposite side of the earth that the sun is facing? If so that’s just going to launch it into an elliptical orbit, unless it manages to get a gravitational slingshot off of some other planetary body. Same thing with launching directly towards the sun. The only real way you’re going to be able to do it is by launching it at the exact opposite direction the earth is moving along its orbital path, at the same speed (66,000 mph), and let gravity pull it in.
It’s kinda like driving 60 mph down the street and trying to throw a ball at a target along side the road right as you pass by it, by aiming directly at it. Once the ball leaves your hands, its still flying 60 mph down the street, tangential to your target. No matter how fast you throw, you’re going to miss the target. The only way to hit the target is to either:
A: throw it long before you pass the target, while the car’s path is pretty close (in angle) relative to the path to the target.
B: throw it 60mph out the back of the car (counteracting the car’s velocity) plus a little bit of nudge velocity in the direction of the target.
Unfortunately, option A is not available for launching things from the earth to the sun, because you are always “passing” your target. It’s a circular orbit. You’re going to have to lose that 66,000 mph tangential to the sun, regardless of what you do. You could get there faster by adding some speed vector towards the sun, but that would be in addition to the 66,000 mph.
If you’re just giving your object a single burst of delta-V, like launching it out of a cannon, then the best you can do, energy-wise, is to just launch straight “back” along Earth’s orbit, to get its orbital velocity to zero. But that takes a lot of energy, more than twice as much energy as getting it to escape speed, the speed you’d need to (eventually) get an infinite distance from the Sun. This is because orbital speed is already about 71% of the way to escape speed, so you only need to add an extra 29% to escape.
But that amount of energy you need to kill your “sideways” motion depends on your distance from the Sun. Get far enough away, and your “sideways” motion will be very small, and thus easy to kill. So you can launch something at just shy of escape speed, wait until it’s at apohelion (the point furthest from the Sun), when its speed is tiny, and then make one more little tiny burn with your rocket to kill that little bit of remaining sideways velocity, and after that it’ll (eventually) fall back in on its own. This is definitely cheaper than the “throw it backwards” option, though I’m not sure precisely how far out you’d want to go for best results.
Of course, all of this is assuming that the Sun, Earth, and rocket are the only objects in the universe. In the real world, as soon as you can reach any other celestial body (say, the Moon), you can start using gravitational slingshots in complicated maneuvers to get pretty much anywhere you want.
OK, that makes sense with gravity assist from Jupiter (as I wrote earlier). But going to the outer solar system and then using your own fuel to stop orbital motion and falling down to the Sun? I’m pretty sure that won’t work. Sure you’re going slower, but don’t think that would make up for the fuel needed to get out there in the first place.
Jupiter or no Jupiter, once you are far away from the Sun (after entering a hyperbolic orbit starting from low-Earth orbit) and accelerate in a retrograde direction, the Sun’s gravity will suck you in without having to burn any more fuel. So if launching from Earth + escaping from Earth + escaping from the Solar system only takes 18 km/s worth of fuel, or whatever it is, it will not take much more than that to fall back into the Sun. Of course you are not going to waste fuel by shooting away from the Sun at 100 km/s and then slowing right back down.
You can save a lot of delta-V by doing multiple gravitational slingshots at inner system planets. For example the Messenger space probe of Mercury did a flyby of Earth once and Venus twice to get to Mercury. It then did three flybys of Mercury to reduce its speed relative to that planet so it could go into orbit about it. Do something similar, but use flybys of Mercury to get even closer to the Sun may be the best way to go. However, note that Messenger took almost 7 years to get into Mercury orbit, so, while it will likely be faster than the outer solar system trajectory, it still won’t be very fast.
In space, Delta-V is everything. The more of it you need to change, the more difficult it becomes. So yes, a place that requires you to change the most Delta-V can be fairly said to be more difficult to get to.
Note that’s not the same as saying it’s the most complex way to get there. Gravitational slingshots are not difficult at all - we can compute them and use them routinely, and we do. We can do it with small, cheaper satellites. The flight is easier, but it takes longer and is more complex.
But Delta-V is hard to achieve, because you need mass for thrust, And to get mass into the spaceship you need more lifting capability, which means a bigger rocket, and to fly a bigger rocket you need even more fuel to not just lift the fuel for the final spaceship, but the extra fuel needed for the bigger, heavier rocket.
This is why our deep space probes take so long to get where they are going - it’s easier and cheaper to just use gravitational slingshots than it is to carry the extra fuel required for a direct Hohmann transfer to say, Saturn. Or the sun.
The rocket equation is a bitch.
If SpaceX gets Starship flying, it will have the Delta-V after in–orbit refuelling to get anywhere in the solar system directly, if the payload is light enough. We could cut years off the time it takes to get to the outer planets. Or, it could launch large space probes that carry more fuel or even advanced nuclear propulsion units. Until then, slingshots are still the easiest way to get out to the outer planets or down to the Sun or Mercury,
There’s nothing difficult about a gravitational slingshot per se. It’s a solved problem.
But I’m not sure ‘more difficult’ or ‘less difficult’ really applies here. It’s more about tradeoffs. Using gravitational slingshots simply trades time for fuel. It can make a mission more difficult if it’s time sensitive or because it’s harder to design a vehicle that can last longer, but we are pretty good at long duration space missions.
But consider a manned mission. Now using slingshots would be more difficult, because you would have to keep the crew alive during the years it would add to the mission.
So whether one method is more difficult than the other comes down to whatever other requirements and constraints the mission has.
“Light enough” is the critical parameter there. I understand that the SpaceX Starship will require 4 launches of just fuel to get a typical load of colonists to Mars.
It’s most likely “really difficult” meant “requires more delta-V than any spacecraft has had available so far” in this context.
I guess I’m just missing some piece of the big picture.
I understand that it’s not as simple as pointing a rocket at the sun and pushing the launch button. The Earth is traveling very fast in orbit around the Sun and any object launched from Earth will have that speed in addition to whatever speed it’s launched with.
But I’m not seeing how using fuel to travel away from the Sun makes more sense than using that same amount of fuel to travel towards the Sun.
If I’m following the video correctly, it’s saying that if we launched a rocket from Earth towards the sun, it would just end up orbiting around the Sun due to the velocity it picked up from Earth.
I understand how that works for Earth; it’s traveling perpendicular to the Sun’s gravitational pull. But the Earth has no force pushing it towards the Sun other than the Sun’s gravity. So it maintains a stable orbit because its perpendicular movement balances out the Sun’s gravitational pull.
But that wouldn’t be the case with the rocket launched towards the Sun. It would be experiencing the Sun’s gravitational pull and would have the Earth’s perpendicular speed. But it would also have its launching speed. So why would it have a stable orbit like the Earth instead of an orbit that spirals inward towards the Sun?
OK, I stand corrected. Going from Earth orbit to Jupiter orbit takes about 9 km/s of delta-V. In this elliptical transfer orbit, you’d be going slower than Jupiter’s orbital speed which is about 13 km/s. So it takes less than 21 km/s to go to Jupiter orbit then kill the orbital speed. Whereas killing the orbital speed while in earth’s orbit takes 30 km/s of delta-V.
I guess this makes sense because once you carry the fuel to Jupiter orbit, the same amount of fuel expended has a much greater angular momentum around the Sun than if you used it in Earth orbit.
Orbits don’t spiral. They ellipse. If you accelerate towards the sun, you don’t actually change the total energy of your orbit, you change the eccentricity of your orbit - that is, instead of being roughly circular you’ll now have a more oblong orbit with a lower perihelion and actually a higher aphelion than what you started with. Yes, that’s right - if you thrust towards the sun some significant but realistic amount, you’ll end up with a stable orbit that is closer to the sun about a quarter of an orbit from your maneuver point, back up to the same height on the opposite side of the orbit, and actually further from the sun 3/4 of the way around.
The only way orbits can “decay” is if the orbiting object is experiencing some sort of drag. This is a thing for satellites in low Earth orbit, as there is some tenuous atmospheric drag at those altitudes, but it is not a thing in interplanetary space.
Orbital mechanics are weird and not very intuitive. If you really want to understand this sort of thing, the very best way (that is, the way that is most fun and filled with explosions) is to play Kerbal Space Program.