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Old 11-05-2019, 01:51 PM
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Can somebody explain these space travel physics to me?


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.

Can somebody tell me where I'm wrong?
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Old 11-05-2019, 02:02 PM
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Minute physics explains it better than I could. Specifically, starting at the 2 minute mark.

Minute Physics (Youtube)

Last edited by Folly; 11-05-2019 at 02:04 PM. Reason: link would help
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Old 11-05-2019, 02:18 PM
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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.
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Old 11-05-2019, 02:34 PM
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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?

Last edited by scr4; 11-05-2019 at 02:34 PM.
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Old 11-05-2019, 02:39 PM
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Originally Posted by Little Nemo View Post
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.

Can somebody tell me where I'm wrong?
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.
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Old 11-05-2019, 03:47 PM
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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.
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Old 11-05-2019, 04:31 PM
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To clarify the issue, here's what Munroe wrote:

Quote:
A tip for dropping things into the Sun: launching directly to the Sun from Earth is really difficult - it actually takes more fuel than launching something out of the Solar System completely. A more efficient way to reach the Sun is to launch something to the far outer Solar System - possibly with the help of gravity assists from the planets. When it's far from the Sun, it will be moving very slowly, and it will only take a little extra fuel to slow it to a halt - after which it will fall directly towards the Sun. It takes much longer than a direct launch, but only requires a fraction of the fuel.
This seems to be the same thing that the video Folly linked to is saying.

Last edited by Little Nemo; 11-05-2019 at 04:33 PM.
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Old 11-05-2019, 05:55 PM
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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.
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Old 11-05-2019, 06:11 PM
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But it does. The fuel needed to get out there is less than half of what you'd need for the direct route.
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Old 11-05-2019, 06:11 PM
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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.
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Old 11-05-2019, 06:18 PM
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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.
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Old 11-05-2019, 08:31 PM
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Why does he say "launching directly to the Sun from Earth is really difficult"?

It's not more difficult, it just uses more fuel. Seems like it would be easier than calculating planetary gravity assists and what-not.
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Old 11-05-2019, 09:10 PM
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Why does he say "launching directly to the Sun from Earth is really difficult"?

It's not more difficult, it just uses more fuel. Seems like it would be easier than calculating planetary gravity assists and what-not.
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,
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Old 11-05-2019, 09:18 PM
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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.
So it's more difficult due to current limitations in propulsion technologies?
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Old 11-05-2019, 09:26 PM
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So it's more difficult due to current limitations in propulsion technologies?
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.
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Old 11-05-2019, 09:32 PM
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Well, I'm not the one who said "really difficult" Just trying to learn what 'really difficult' means in this context.
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Old 11-05-2019, 09:46 PM
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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.
"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.


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Well, I'm not the one who said "really difficult" Just trying to learn what 'really difficult' means in this context.
It's most likely "really difficult" meant "requires more delta-V than any spacecraft has had available so far" in this context.

Last edited by dtilque; 11-05-2019 at 09:46 PM.
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Old 11-05-2019, 10:37 PM
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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?
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Old 11-05-2019, 11:10 PM
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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.
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Old 11-06-2019, 12:35 AM
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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?
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.
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Old 11-06-2019, 12:52 AM
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Orbits don't spiral. once you've finished your burn, you have established your new orbit. If you are travelling away from the central mass (the sun) faster than escape velocity, it is a hyperbolic orbit and you never come back. If you are travelling less than escape velocity, it is an elliptical orbit. Your current position and velocity vector determine the path of the orbit. If the path of the ellipse takes you too close to the surface of the sun, you will not describe a complete ellipse...

Quote:
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.
But absent the slingshot effect - you spend 9km/s, then need to spend a bit more (guessing about 4km/s) to circularize the orbit there, then you need to shed 13km/s to fall into the sun. The only reason Jupiter is efficient is the slingshot effect, where you an Jupiter trade momentum vectors so that you are going in a more desirable direction, rather than spend all that Delta-V.

If as the OP suggests, you go to the edge of the solar system (but presumably, not beyond) then you will still have sufficient angular velocity and direction to swing around the sun like a comet instead of falling directly into it.
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Old 11-06-2019, 12:58 AM
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When you think about it, the only way to fall directly into the sun (on a purely radial path) is to have an angular velocity of zero relative to it. Everything else is an elliptical orbit.
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Old 11-06-2019, 02:28 AM
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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.
That seems to be pessimistic. A transfer orbit from the distance of Earth out to the distance of Jupiter only has an orbital speed at apogee of 7.7 km/s. That's what you would have to kill. So going out to the distance of Jupiter, then killing the orbital speed takes 16.7 km/s.

Going out to the distance of Saturn would take 10.3 + 4.0 = 14.3 km/s

Neptune: 11.7 + 1.4 = 13.1 km/s

Diminishing returns, but large savings over the 30 km/s of going direct into the Sun.

ETA: Sorry misread scr4's post, he said it's less than Jupiter's orbital speed. I got my orbital speed numbers from https://www.omnicalculator.com/physics/orbital-velocity

Last edited by Frankenstein Monster; 11-06-2019 at 02:30 AM.
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Old 11-06-2019, 04:08 AM
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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?
You do not seem to be missing that for the basic manoeuvres you want to point your rocket "prograde" or "retrograde", that is, in the direction of your orbit or opposite, in order to raise or lower your orbit. You would not be able to get to the sun by starting in orbit around it and firing your motor while pointing directly at the sun; think about it!

Next, objects really far from the sun orbit more slowly than objects close to the sun. So you might imagine that if you were hypothetically really out there, you would be barely moving, and it would not take much fuel to arrest any residual velocity to ensure it sucks you back in for a collision. So as an initial exercise you could try ignoring extra planets and approximately calculate how much fuel it would take to launch away from the Earth and out of the Solar system.

This is all pretty theoretical; the video tells us that the actual satellite is supposed to swing by Earth and Venus (not Jupiter) a few times in order to get to the Sun.
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Old 11-06-2019, 07:23 AM
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Quoth md2000:

But absent the slingshot effect - you spend 9km/s, then need to spend a bit more (guessing about 4km/s) to circularize the orbit there, then you need to shed 13km/s to fall into the sun.
Circularizing your orbit, once you're out at apohelion, would be counter-productive. You'd be boosting forward a bit, then boosting backwards that bit and more. You'd only do that if you wanted to hang out in the vicinity of Jupiter for a long time before continuing on to the Sun.
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Old 11-06-2019, 08:18 AM
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Note to self: ANOTHER topic to ask my rocket scientist son about when he visits over turkey day, because I just can't get my head around this.

I often wonder if I am lacking the education/vocabulary, whether something about me makes it impossible for me to grasp things that seem counter-intuitive (well, wouldn't the sun's gravity pull it in if you launched TOWARDS the sun?), or if somehow I'm willingly allowing myself to not understand it/not putting the mental effort in. I enjoy reading threads like this, but I consistently have a hard time understanding them. And I do not generally consider myself a stupid person.
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Old 11-06-2019, 09:14 AM
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Note to self: ANOTHER topic to ask my rocket scientist son about when he visits over turkey day, because I just can't get my head around this.

I often wonder if I am lacking the education/vocabulary, whether something about me makes it impossible for me to grasp things that seem counter-intuitive (well, wouldn't the sun's gravity pull it in if you launched TOWARDS the sun?), or if somehow I'm willingly allowing myself to not understand it/not putting the mental effort in. I enjoy reading threads like this, but I consistently have a hard time understanding them. And I do not generally consider myself a stupid person.
Start with this: Near Earth we have the space station. Why does the Earth's gravity not "pull it in"?

Your explanation of this question will help others structure their sun-launch explanation to you so that you can understand it.
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Old 11-06-2019, 09:42 AM
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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.
Would it be possible to use a solar sail to approximate this kind of drag? If the object had a sail which was at a 45 degree angle to the sun, would the energy from the sun pushing it out and back eventually slow it completely down?
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Old 11-06-2019, 09:59 AM
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So ... if there were really such a craft as the Starship Enterprise, or the Orville or whatever ... and that craft had "magic" propulsion and no real fuel concerns:

Said craft launches from Earth, with an aim to land neatly on the Sun's surface. The crew doesn't even give the slightest rip about conserving fuel, gravity assists, or any of that. So, the entire way to the Sun, is the crew constantly having to course-correct to actively prevent from going into orbit (even a really elongated one)? Does making the trip a presumably straight shot in fact require constant "steering"?
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Old 11-06-2019, 10:23 AM
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If the object had a sail which was at a 45 degree angle to the sun, would the energy from the sun pushing it out and back eventually slow it completely down?
"Completely" stop relative to the sun? Not sure. But you can use a solar sail to get in closer to the sun for sure. Might not even be necessary as long as you reduced your orbital velocity enough for gravity to do it's work.

It's possible to "tack" (or something like it) using the solar wind. You don't have to go straight out. The Japanese Ikaros probe tested solar sail technology to get to Venus.

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Does making the trip a presumably straight shot in fact require constant "steering"?
Basically, yes, you'll need constant thrust. Though the definition of 'straight' gets a bit wonky with all these celestial bodies in relative motion to each other and gravity 'curves' local spacetime.
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Old 11-06-2019, 10:42 AM
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Start with this: Near Earth we have the space station. Why does the Earth's gravity not "pull it in"?

Your explanation of this question will help others structure their sun-launch explanation to you so that you can understand it.
(As the creaky gears try to turn...) Because it is falling, right? Which pretty much strains my brain to the breaking point.

And which pretty much represents my miles-wide-but-inches-deep, superficial familiarity w/ so many scientific principles that really interest me. With other liberal arts bullshitters, I'm capable of spouting a factoid or name or 2, sufficient to convey the impression that I actually know something. When I read something, or someone intelligent explains it to me (another kid and her SO are molecular biologists! ), it makes sense to me in the moment, but just doesn't stick w/ me, and doesn't scale/transfer to other situations.

Same way I tried to read "A Brief History" several times, and can point out the exact page where it exceeded my ability (willingness?) to understand.

I think that now, in my late 50s, I'm reaping the results of having studiously avoided putting in the effort to study math and sciences in high school and college. My brain is warped toward bullshit rather than deep understanding of facts.

But I'll keep reading these threads, and reading books and watching programs that interest me but somehow don't seem to stick.
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Old 11-06-2019, 10:49 AM
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Originally Posted by Great Antibob View Post
Basically, yes, you'll need constant thrust. Though the definition of 'straight' gets a bit wonky with all these celestial bodies in relative motion to each other and gravity 'curves' local spacetime.
That might be one of the harder things to grok about orbital mechanics: A destination like the Sun is not actually stationary.

In the everyday earthbound human experience, the Sun is essentially reckoned as a never-changing point in space. Something stationary that you can aim at upon launching and just "stay the course" and make it there. An apparent "straight shot", as it were.

But in fact, the Sun is ALWAYS moving in relation to countless other objects in space -- The center of the Milky Way. I guess its own planets. Etc.

...

Thought experiment:

If we had a "magic tape measure" (or rope, string, whatever) that could instantaneously stretch from the center of the Earth to the center of the Sun ... and that tape measure could be stretched taut between the two objects ... the tape measure would not follow a straight line, but some kind of curve. Even when pulled taut. Right?
  #33  
Old 11-06-2019, 11:01 AM
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Would it be possible to use a solar sail to approximate this kind of drag? If the object had a sail which was at a 45 degree angle to the sun, would the energy from the sun pushing it out and back eventually slow it completely down?
If you had constant thrust, from a solar sail or an ion engine or similar, then you can indeed spiral down into the Sun. Essentially, every moment your thrust is moving you into a slightly lower orbit.
  #34  
Old 11-06-2019, 11:07 AM
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Start with this: Near Earth we have the space station. Why does the Earth's gravity not "pull it in"?
I thought they did. Like Skylab. I figured that you had to occasionally give them a burst of new thrust in order to keep them in orbit.
  #35  
Old 11-06-2019, 11:19 AM
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That might be one of the harder things to grok about orbital mechanics: A destination like the Sun is not actually stationary.
While that is true, it's a very minor consideration for things we've been discussing in this thread.

The most important effect is, when you start from Earth orbit, you are already moving at a significant speed, about 30 km/s. And there's no friction in space to slow you down. If you used a massive engine to come to a complete stop, you would fall straight down to the Sun. But that requires launching something at 30 km/s relative to the Earth. The fastest spacecraft ever to leave the Earth was the New Horizons, which left Earth at a speed of 16.3 km/s, so a little more than half the speed you need to fall down to the Sun. And this required launching this 1000-lb probe on top of an Atlas-V with 5 strap-on boosters. That's a rocket that can put 41,000 lb into low earth orbit.

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Thought experiment:

If we had a "magic tape measure" (or rope, string, whatever) that could instantaneously stretch from the center of the Earth to the center of the Sun ... and that tape measure could be stretched taut between the two objects ... the tape measure would not follow a straight line, but some kind of curve. Even when pulled taut. Right?
I think it would be straight. Every part of this "tape measure" is in a circular path around the Sun, but not moving fast enough to stay in that path without external force directly away from the Sun. That force is provided by the tension of the "tape measure". (Though the fact that the Earth's orbit is slightly elliptical may change this slightly.)
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Old 11-06-2019, 11:20 AM
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I thought they did. Like Skylab. I figured that you had to occasionally give them a burst of new thrust in order to keep them in orbit.
Only because the space stations are in low earth orbit, where the atmospheric drag is not negligible. For an object orbiting the Sun and not near any planet, the drag is not even measurable. (Unless it happens to collide with an asteroid.)

Last edited by scr4; 11-06-2019 at 11:22 AM.
  #37  
Old 11-06-2019, 11:23 AM
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I thought they did. Like Skylab. I figured that you had to occasionally give them a burst of new thrust in order to keep them in orbit.
Skylab was in an unstable low earth orbit. I can't explain what you're asking in general but he wants you to consider that anything launched from the earth has earth's orbital velocity and will tend to stay in earth's orbit around the sun. Aiming directly at the sun doesn't help to reduce that orbital velocity. To get to the sun you have to change the orbit to intersect the sun or get very close. It sounds like using the gravity of other planets to alter the orbit is the most efficient way to do that, and that means increasing the radius of the orbit initially. Or else I have no more understanding than you about this, or maybe less.
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Old 11-06-2019, 11:41 AM
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Originally Posted by bordelond View Post
If we had a "magic tape measure" (or rope, string, whatever) that could instantaneously stretch from the center of the Earth to the center of the Sun ... and that tape measure could be stretched taut between the two objects ... the tape measure would not follow a straight line, but some kind of curve. Even when pulled taut. Right?
Nah, you can still have 'straight' lines. This thought experiment gets a bit hairy, though, because you can't actually measure both ends simultaneously in the real world - there's enough distance that relativity is an actual concern.

But never mind that. The issue is that the Earth is in orbit around the sun. If you start from Earth orbit, you already have significant velocity relative to the sun. So, boosting 'straight' towards the sun is not efficient.

It's the old saying that you don't shoot where something is, you shoot where it's going to be.

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I thought they did. Like Skylab. I figured that you had to occasionally give them a burst of new thrust in order to keep them in orbit.
Yeah, that wasn't the greatest example. Even the ISS is constantly having to get little nudges to maintain its orbit. It's got a low orbit and is skimming the atmosphere constantly.

Satellites in geosynchronous orbit are probably a better example. There are several that can conceivably be up there for thousands of years in stable orbits if otherwise left alone.
  #39  
Old 11-06-2019, 02:35 PM
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(As the creaky gears try to turn...) Because it is falling, right? Which pretty much strains my brain to the breaking point.
Yes, they are falling. All objects in earth orbit are constantly accelerating towards earth due to the earth's gravity, which in low earth orbit (LEO) isn't much less than the 9.8m/s2 we experience on the surface. At 400km up, you're looking at 8.7m/s2. However, as an object in LEO falls towards the earth, it's also moving sideways. Quite rapidly. And as it moves sideways, the surface of the earth gets further away, because the earth is not flat. If an object in orbit were not subject to earth's gravity it would instead travel in a straight line. Imagine drawing a straight line next to a circle, and then measuring the "altitude" of points on the line as it moves away from the circle - that altitude increases. But of course objects in LEO are subject to gravity, so their trajectory is continually bent towards the surface of the earth - it's just never bent towards the surface faster than the surface bends away beneath.

So that's the basics of why things in orbit don't fall down, but it doesn't get into the orbital mechanics of raising and lowering orbits. Orbital mechanics aren't hard to understand, but they are very, very counter-intuitive if you haven't grasped how orbits work. To go "down" in an orbit, you actually need to slow down your sideways motion. If you're not moving sideways as fast, then the surface below you isn't bending away as fast, while you're still being accelerated towards it at the same rate. However, just slowing down a bit won't make your whole orbit lower. As you get lower, you'll pick up speed (like you're a marble rolling down a hill), and assuming you didn't slow down so much that your orbit now intersects the atmosphere, that speed will eventually become enough that you'll start going back up, and you'll end up at exactly the same place you were when you slowed down in the first place. If you want your whole orbit to be lower, you need to slow down your sideways motion a second time when you reach the lowest point. To go up you do the opposite - first you increase your sideways motion. This makes the surface curve away from you faster so that your altitude will increase. As your altitude increases, your speed will decrease until you start losing altitude again and you end up exactly where you were when you first sped up. To stay in a higher orbit, you have to increase your sideways speed a second time when you're at the high point of your orbit.

These pairs of accelerations are called Hohman transfers, and the orbit you're on in between them is called a transfer orbit. If you watch SpaceX launches, you've seen the first half of this when they launch satellites to "geostationary transfer orbits." First they have the initial low earth orbit insertion - the long initial burn of the second stage. Then they have the "coast phase" which is actually just waiting until the rocket is over the equator. They wait till they're over the equator because, if you haven't noticed by now, in orbit you always loop back around to where you were when you made your maneuver. So since these satellites are supposed to end up over the equator, you want to make your maneuver over the equator. Then the second burn of the second stage is picking up a bunch more speed so that the satellite will coast up to the altitude of geostationary orbits (36,000km or so). If nothing further happened, the satellite would then come falling back down to LEO altitude, and then coast back up to GSO altitude, and back down, and back up. What actually happens is that the satellite uses a maneuvering thruster to complete the second half of the Hohman transfer (also called "circularizing" as it turns the more elliptical transfer orbit into a nearly perfect circle). This of course happens long after the SpaceX broadcast ends, and may not be all done at once.

So all that said, why is it easier to get to the sun by initially moving out rather than in? Well, in some cases rather than using a Hohman transfer to raise or lower your orbit it's more efficient to use something called a bi-elliptical transfer instead. It's more efficient because when you're at the high point of a very elliptical orbit you're moving very slowly, and when you're moving very slowly you can change your orbit more with less delta-V. So to get to the sun, you're starting out by increasing your sideways speed relative to the sun so that you'll gain altitude (relative to the sun now, not earth) and as you gain altitude you'll slow down till you're moving much slower than the 30km/s that the earth is. You can then slow down that remaining sideways motion with less effort. And that's all before considering any planetary fly-bys to do gravitational slingshots.
  #40  
Old 11-06-2019, 03:16 PM
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Yes, they are falling. To go "down" in an orbit, you actually need to slow down your sideways motion. If you're not moving sideways as fast, then the surface below you isn't bending away as fast, while you're still being accelerated towards it at the same rate. However, just slowing down a bit won't make your whole orbit lower. As you get lower, you'll pick up speed (like you're a marble rolling down a hill), and assuming you didn't slow down so much that your orbit now intersects the atmosphere, that speed will eventually become enough that you'll start going back up, and you'll end up at exactly the same place you were when you slowed down in the first place. If you want your whole orbit to be lower, you need to slow down your sideways motion a second time when you reach the lowest point.
What happens if instead of a thruster burn (forwards) to slow down, you do a thruster burn 'up' (away from the Earth)? Thanks!
  #41  
Old 11-06-2019, 03:20 PM
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Yeah, that wasn't the greatest example. Even the ISS is constantly having to get little nudges to maintain its orbit. It's got a low orbit and is skimming the atmosphere constantly.
It's not a bad example. Because I would follow that response with another question:
What direction are those "little nudges" pushing the station? Why?
It's important to understand that these nudges are not pushing the station away from the Earth, the way someone might imagine nudging a falling object; like the way people nudge a balloon to keep it aloft, or nudge a concert beach ball to keep it from falling to the ground. The nudges are to increase the speed of the station. The drag from the atmosphere is slowing the station down, and if the station is not going fast enough, it will not stay in orbit.
This must be understood to move on to the sun launch scenario.

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Satellites in geosynchronous orbit are probably a better example. There are several that can conceivably be up there for thousands of years in stable orbits if otherwise left alone.
Maybe this is better. But I don't mind the conversation that follows from the ISS scenario.

Dinsdale, keeping on the subject of geosynchronous satellites, I have a question for you. If you have a satellite at the end of your rocket and you launch it straight up to an altitude of 36km above the Earth, what happens to the satellite? Does it stay up there in orbit? After all, you've reached the required altitude, right? What happens to it?

I'm seriously considering making a video for this thread and posting it online.
  #42  
Old 11-06-2019, 03:24 PM
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Originally Posted by Bear_Nenno View Post
I'm seriously considering making a video for this thread and posting it online.
I'd watch that video.

So, with all this great explanations everyone is giving, the scene in the Star Trek Generations movie where the guy launches the sun-killing rocket into the sun from the planet is not possible?
  #43  
Old 11-06-2019, 03:29 PM
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Dinsdale, keeping on the subject of geosynchronous satellites, I have a question for you. If you have a satellite at the end of your rocket and you launch it straight up to an altitude of 36km above the Earth, what happens to the satellite? Does it stay up there in orbit? After all, you've reached the required altitude, right? What happens to it?
Excuse me. I mean 36,000Km.

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Originally Posted by Folly View Post
What happens if instead of a thruster burn (forwards) to slow down, you do a thruster burn 'up' (away from the Earth)? Thanks!
Two possibilities. If you have a Scifi level ship with enough energy to reach escape velocity going straight up, then you will leave the orbit of the Earth.
Otherwise, you're going to "move your orbit" without changing it's shape. If you are in a circular orbit with a radius of 100,000km and you burn "up", you're still going to have a cicular orbit with the same radius. But now, one side of the circle will be further from the Earth, and the other side will be closer. the Earth is no longer in the center of your Orbit. In Gorsnak's examples, speeding up and slowing down changed the circle orbit to an oval orbit. One side stayed the same height, and the other side got higher (when speeding up). But burning straight up will not change your circle. So, it moves one side further away, while moving the other side of your orbit an equal distance closer. So, if burn "away" from the Earth hard enough (but not hard enough to reach escape velocity, of course), then you will crash into the Earth on the other side of your orbit. If you burn "up" at the 3 o'clock position, you will crash into the Earth at the 9 o'clock.

Last edited by Bear_Nenno; 11-06-2019 at 03:32 PM.
  #44  
Old 11-06-2019, 03:35 PM
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What happens if instead of a thruster burn (forwards) to slow down, you do a thruster burn 'up' (away from the Earth)? Thanks!
If you're currently in a circular orbit, you'll end up in an elliptical orbit where the high point of the orbit is higher than your current orbit, and the low point is lower. If you're already in an elliptical orbit, then you'll either end up more elliptical or less elliptical, depending on where you are in the orbit when you make the burn. In any case, if you burn at 90 degrees to your direction of travel you are neither adding nor subtracting energy from your orbit, so your average altitude (technically the semi-major access of your ellipse) won't change.
  #45  
Old 11-06-2019, 03:37 PM
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I'm seriously considering making a video for this thread and posting it online.
You could just link to a Scott Manley KSP tutorial video on Youtube.
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Old 11-06-2019, 03:38 PM
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Thanks for the effort, guys. Gotta tell you that right now my brain is fried at the end of a rough day at work, so I'm gonna have to get back to this later.

Gorsnak, your 1st 3 paragraphs make perfect sense, but my brain locked up on the final par. Same reason I can't even begin to answer your question B N. My ignorant assumption is that once something gets to the right altitude at the right speed, it keeps falling and missing, so it appears to stay there. But I know that is so simplistic as to distort. My kid used to work for ULA and now works for Ball, so believe me, he has tried to explain all this to me before - along w/ space elevators, Lagrange points, etc. But my brain tends to glaze over pretty quickly. Like I said, he's coming into town in a couple of weeks, so I'll try to bone up on some of this sort of thing to be better able to discuss it with him.

My kid who is a microbiologist used to work as a library shelver. One time she brought home a book which - in her words - did a pretty good job of explaining what she is interested in. And even tho it is written for the general public, it does not distort things overly. When I told her I made it thru 16 pages before I thought my head would explode. To which my loving spawn said, "Next time I'm working in juvie I'll see if they have anything for you w/ pop-ups!" It is frustrating to consider oneself reasonably competent and intelligent, yet feel SO STUPID when trying to figure out things like this. Don't even TRY to tell me about the pixies inside my computer
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  #47  
Old 11-06-2019, 03:40 PM
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Originally Posted by Bear_Nenno View Post
Otherwise, you're going to "move your orbit" without changing it's shape. If you are in a circular orbit with a radius of 100,000km and you burn "up", you're still going to have a cicular orbit with the same radius. But now, one side of the circle will be further from the Earth, and the other side will be closer. the Earth is no longer in the center of your Orbit. In Gorsnak's examples, speeding up and slowing down changed the circle orbit to an oval orbit. One side stayed the same height, and the other side got higher (when speeding up). But burning straight up will not change your circle. So, it moves one side further away, while moving the other side of your orbit an equal distance closer. So, if burn "away" from the Earth hard enough (but not hard enough to reach escape velocity, of course), then you will crash into the Earth on the other side of your orbit. If you burn "up" at the 3 o'clock position, you will crash into the Earth at the 9 o'clock.
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Originally Posted by Gorsnak View Post
If you're currently in a circular orbit, you'll end up in an elliptical orbit where the high point of the orbit is higher than your current orbit, and the low point is lower. If you're already in an elliptical orbit, then you'll either end up more elliptical or less elliptical, depending on where you are in the orbit when you make the burn. In any case, if you burn at 90 degrees to your direction of travel you are neither adding nor subtracting energy from your orbit, so your average altitude (technically the semi-major access of your ellipse) won't change.

These two answers seem to be different. Am I reading them wrong?
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Old 11-06-2019, 03:40 PM
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So, with all this great explanations everyone is giving, the scene in the Star Trek Generations movie where the guy launches the sun-killing rocket into the sun from the planet is not possible?
It's possible, given sci-fi level rocketry technology.

You can have it burn fuel throughout its flight. Or launch with a truly atrocious velocity. Or use that nifty warp technology they seem to have.
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Old 11-06-2019, 03:46 PM
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These two answers seem to be different. Am I reading them wrong?
They're pretty much the same, except Bear Nenno describes the ensuing orbit as "still circular but with one side of the orbit lower than the other" which isn't actually a thing. Orbits are only circular when they're exactly the same altitude all the way round. The ensuing orbit is higher on one side and lower on the other, but it's also squashed into an ellipse. This squashing might well be very minor if the difference between high and low points isn't very much, but it is always there.
  #50  
Old 11-06-2019, 03:50 PM
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Originally Posted by Gorsnak View Post
They're pretty much the same, except Bear Nenno describes the ensuing orbit as "still circular but with one side of the orbit lower than the other" which isn't actually a thing. Orbits are only circular when they're exactly the same altitude all the way round. The ensuing orbit is higher on one side and lower on the other, but it's also squashed into an ellipse. This squashing might well be very minor if the difference between high and low points isn't very much, but it is always there.
Ah, I see. And you could actually crash into the Earth if the "low side" were low enough?
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