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And btw it's a gun.

Small projectiles get launched to the moonhole. Passing close to the evnt horizon, they shoot ballast retrograde, accelerate and change their flight direction in desired manner.

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Shouldn't we be able to send in a spacecraft with a counterweight on a super-strong tether for a near pass of the lunar black hole, and separate the spacecraft at periapsis to get a nice boost from the tidal effect?

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3 hours ago, 5thHorseman said:

To get a specific Vinf for Earth, there is some specific Vinf at Moon. Does there exist an R at Moon (under its surface but not under the black hole's radius) where you can burn Z at it after burning Y in LEO, such that you can get the same Earth Vinf as a burn in LEO of X, such that Y+Z < X?

HMV,

 No, at least not in our solar system. There's only a couple targets that have a gate orbit anywhere near the moon, and you don't need to get close to the lunar surface to hit it. Farther targets are more efficiently reached by a LEO burn and simple slingshot.

3 hours ago, K^2 said:

We're talking about approaching "Moon" at 1km/s on trans-Lunar, picking up 100km/s in free-fall, kicking in engines to pick up 3km/s of extra delta-V at periapsis, loosing that 100km/s worth of KE climbing back out, and being left over with enough KE to be departing "Moon" at 25km/s, which is just about right for escaping Sol all together.

K^2,

 Again, that's not how it works. You don't actually pick up "extra" DV, even if you do your burn scraping an event horizon. The depth of the gravity well itself doesn't help or hurt you, it merely determines how far you can bend your trajectory at a high inbound velocity.
 This is the illusion of the Oberth effect; you get the idea that you get a DV payoff by burning at higher velocity. You don't. The payoff comes from making the transfer more eccentric. The energy of an orbit is equivalent to the area it contains. A long skinny orbit contains less area, but reaches the same Ap. Therefore requires less energy to achieve. *That's* how the Oberth effect actually works.

Best,
-Slashy

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1 minute ago, GoSlash27 said:

Again, that's not how it works. You don't actually pick up "extra" DV, even if you do your burn scraping an event horizon. The depth of the gravity well itself doesn't help or hurt you, it merely determines how far you can bend your trajectory at a high inbound velocity.

A 2kg rocket coasts at barely a few m/s towards a rogue neutron star with no significant neighborhood from a great distance. Both kinetic energy and potential energy are very nearly zero. As the rocket drops closer, it speeds up. Total energy stays roughly zero, with kinetic energy increasing and potential energy turning negative. It takes a while, but finally the rocket makes its closest approach. It is now traveling at 100km/s. The total kinetic energy is mv2/2 = 10GJ. And since total energy is still zero, its potential energy is -10GJ. At this point the rocket orients itself prograde and ignites its engines. After a brief burn, 1kg of fuel and oxidizer is expended, and the rocket is now traveling at 101km/s. So a 1km/s delta-V. Not a terribly good engine, I know, but we work with what we have. So the mass of the rocket is now 1kg. Its kinetic energy is now 5.1005 GJ. The potential energy is -5GJ, as the other -5GJ are in the exhaust. The total energy of the rocket is now 5.1005GJ - 5GJ = 100.5MJ, and the rocket continues on its escape trajectory.

Eventually, it gets far enough from the neutron star, where its potential energy is insignificant again. The total energy is still 100.5MJ, since no significant forces besides gravity acted on the rocket. Since potential energy is zero, this is all kinetic energy. 100.5MJ of kinetic energy on a 1kg rocket, which means the rocket is now hauling at 10.025km/s. This is despite the fact that it only had 1km/s of delta-V in it, and it was only going at a few m/s before it encountered the neutron star.

This is counter-intuitive, I understand that. Very few things in orbital mechanics make intuitive sense, which is why you have to take a problem and work through it. Obereth effect isn't an illusion. It's a very real way for rocket to gain "free" delta-V. If you have hard time picturing why this works, I challenge you to work through the problem without ignoring what happens to exhaust. What is its kinetic energy immediately after the burn? What is its total energy, then? How much of total energy of the system had to come from chemical energy of the fuel? What about conservation of momentum? Linear momentum is a pain in orbital mechanics, but you can just work with angular momentum instead.

What you will find is that we managed to extract significantly more energy from the fuel than there was chemical energy in it to start with. And the reason we were able to do it is because that fuel had a much higher potential energy before we entered the star system, then following the burn. A rocket on escape trajectory utilizing Obereth effect not only uses the chemical energy of the fuel, but its gravitational energy as well. It's a non-trivial quantity in ordinary orbital mechanics, but becomes just batty when neutron stars and black holes are involved, greatly exceeding the chemical energy available.

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I get where you're coming from. I used to think the same way. But (yet again) that's not how it actually works.
The goal isn't to inject energy as cheaply as possible, it's to get the vehicle to the destination as cheaply as possible. If you model the situation you're talking about, you'll see the pitfall in that line of reasoning.

Best,
-Slashy

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5 minutes ago, GoSlash27 said:

I get where you're coming from. I used to think the same way. But (yet again) that's not how it actually works.
The goal isn't to inject energy as cheaply as possible, it's to get the vehicle to the destination as cheaply as possible. If you model the situation you're talking about, you'll see the pitfall in that line of reasoning.

I have. That is the model. I computed final velocity given initial condition and correction burn. If you are telling me that the exit velocity will be less than 10km/s, you have to tell me where the energy change came from.

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7 hours ago, K^2 said:

I have. That is the model.

K^2,
 

I think this is where you went off the rails:

7 hours ago, K^2 said:

Eventually, it gets far enough from the neutron star, where its potential energy is insignificant again. The total energy is still 100.5MJ, since no significant forces besides gravity acted on the rocket.

This is not correct. The conservation of energy demands that the total energy remains the same in a closed system. You converted chemical energy into kinetic energy, adding 500 kJ. That's all you did. You should've algebraically added the kinetic and potential, which in this case is subtraction rather than addition. Whatever potential energy you have converted to kinetic energy on the way in will be converted back into potential energy on the way back out. Your total energy at the end of this maneuver isn't 100.5MJ, it's 0.5 MJ.

Best,
-Slashy

 

Edited by GoSlash27
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1 hour ago, GoSlash27 said:

The conservation of energy demands that the total energy remains the same in a closed system.

True. The only thing that increased is the orbital energy.

The increase looks "bigger" because it's the same stuff that we do at flybys without any extra energy. And because the cosine losses is smaller compared to burning radially, for instance (which is what would to happen earlier in the orbit if you don't want the overall trajectory to change, just how quickly you went through it).

Edited by YNM
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26 minutes ago, Xd the great said:

Well, you did not take into account the relativistic of the real world, because ksp is newton mechanics based.

I guess.

That's fine. If you get close enough to a moon mass black hole to see relativistic effects, how much they'll affect your eventual velocity will likely be the least of your worries.

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2 hours ago, GoSlash27 said:

 The conservation of energy demands that the total energy remains the same in a closed system. 

Exactly. And this closed system involves a hugely massive celestial body from where you "stoled" that energy. 

By the formula of the kinetic energy, you infer that the enormous mass of the body makes the delta-V it loose insignificant, barely computable. But it effects on your puny vessel are pretty visible. 

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This is one of the reasons why I love you guys.  I ask a question, in order to point out to non space people, that the Mun Moon turning into a black hole would not destroy the planet, nothing major would happen, etc etc, and it quickly de evolves into an orbital mechanics discussion.  

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2 hours ago, Lisias said:

Exactly. And this closed system involves a hugely massive celestial body from where you "stoled" that energy. 

By the formula of the kinetic energy, you infer that the enormous mass of the body makes the delta-V it loose insignificant, barely computable. But it effects on your puny vessel are pretty visible. 

Lisias, 

This is correct when the body is in motion. You're stealing kinetic energy from the moon. However, this doesn't apply to K^2's example. 

Best,

-Slashy

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2 hours ago, Gargamel said:

This is one of the reasons why I love you guys.  I ask a question, in order to point out to non space people, that the Mun Moon turning into a black hole would not destroy the planet, nothing major would happen, etc etc, and it quickly de evolves into an orbital mechanics discussion.  

And that’s why this is the best forum on the internet....we’re all nerds! :P

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34 minutes ago, GoSlash27 said:

This is correct when the body is in motion. You're stealing kinetic energy from the moon. However, this doesn't apply to K^2's example. 

But… But… But… Everything is in motion in the Universe. You can be (apparently) motionless in a given Reference Frame, but otherwise...

The very notion of Kinetic Energy is tied to a Reference Frame...

If I understood correctly, K^2 examples were given in two Frames - the Neutron Star one, and the one that rules the star's motion...

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1 hour ago, Lisias said:

But… But… But… Everything is in motion in the Universe. You can be (apparently) motionless in a given Reference Frame, but otherwise...

The very notion of Kinetic Energy is tied to a Reference Frame...

If I understood correctly, K^2 examples were given in two Frames - the Neutron Star one, and the one that rules the star's motion...

Lisias,

Not at all. K^2's example was given in the ship's frame. The example neutron star had no defined motion, thus no kinetic energy to steal.

I'm afraid you're confusing Oberth with gravity assist, which are 2 different subjects.

Best,

-Slashy

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46 minutes ago, GoSlash27 said:

The example neutron star had no defined motion, thus no kinetic energy to steal.

I'm afraid you're confusing Oberth with gravity assist, which are 2 different subjects.

Probably. But our main source of knowledge, Wikipedia :P , don't do much to clarify things:

Quote

The resulting maneuver is a more efficient way to gain kinetic energy than applying the same impulse outside of a gravitational well. The gain in efficiency is explained by the Oberth effect, wherein the use of an engine at higher speeds generates greater mechanical energy than use at lower speeds.

Source.

But, yeah, the source of the extra kinetic energy is "stolen" from the reaction mass (that became "slower" on the Reference Frame than outside the Oberth's window), not the body's - so, you are pretty right about this point:

Quote

It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's kinetic energy is balanced by a relative decrease in the kinetic energy the exhaust is left with (the kinetic energy of the exhaust may still increase but it does not increase as much).

Source.

 

Edited by Lisias
grammars. =/
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7 hours ago, GoSlash27 said:

This is not correct. The conservation of energy demands that the total energy remains the same in a closed system. You converted chemical energy into kinetic energy, adding 500 kJ. That's all you did. You should've algebraically added the kinetic and potential, which in this case is subtraction rather than addition. Whatever potential energy you have converted to kinetic energy on the way in will be converted back into potential energy on the way back out. Your total energy at the end of this maneuver isn't 100.5MJ, it's 0.5 MJ.

Ok, lets walk this back. Take these 0.5MJ and walk them back to the point just after the burn. The rocket's total energy is still 0.5MJ, and the potential energy here is -5GJ. So the total kinetic energy is 5.0005GJ. So the rocket is traveling at 100.005km/s. In other words, it has just expended 1kg of fuel and went from 100km/s to 100.005km/s, getting just 5m/s of delta-V. Does that sound right to you?

delta-V efficiency of the rocket doesn't change with where we fire it. If you are trying to understand where extra 100MJ of energy came from, look at what happened to the energy of the fuel/exhaust. Don't be lazy. Work through the whole thing.

You are making the same exact mistake that people make with rocket formula. Forget about gravity for a moment. If you start from rest and burn enough fuel to gain 1km/s, you increased rocket's energy by 500kJ per kg. If this same rocket is already traveling at 10km/s and it performs the same burn, its kinetic energy went from 50MJ/kg to 60.5MJ/kg, for a total gain of 10.5MJ. We burned the same exact fuel in exactly the same engine. Where did the extra 10MJ came from?

If you understand how energy conservation works for rocket in vacuum, you should be able to work through rocket in gravity well.

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3 hours ago, K^2 said:

Ok, lets walk this back. Take these 0.5MJ and walk them back to the point just after the burn. The rocket's total energy is still 0.5MJ, and the potential energy here is -5GJ. So the total kinetic energy is 5.0005GJ. So the rocket is traveling at 100.005km/s. In other words, it has just expended 1kg of fuel and went from 100km/s to 100.005km/s, getting just 5m/s of delta-V. Does that sound right to you?

K^2,

 No, it sounds completely wrong :)

 Your total energy at the bottom of the well prior to the burn is zero (or very close to nil). You had 5GJ of kinetic energy and -5GJ of potential energy, which cancel out. You then added .5MJ of kinetic energy by increasing your velocity 1,000 m/sec, which you keep as you leave. The kinetic energy is transferred back into potential energy to pay off the debt you incurred by falling into the hole, leaving your 1,000 m/sec DV.

Best,
-Slashy

 

3 hours ago, Lisias said:

But, yeah, the source of the extra kinetic energy is "stolen" from the reaction mass (that became "slower" on the Reference Frame than outside the Oberth's window), not the body's - so, you are pretty right about this point:

Lislias,
 Again you are confusing the mathematics of Oberth with the mechanics of gravity assists. Oberth doesn't "steal" anything from the parent body, it simply generates higher kinetic energy from travelling faster. Gravity assists *literally* steal kinetic energy from the body... but that body must be in motion in order to have any kinetic energy to steal. The neutron star in K^2's example is not in motion.

Best,
-Slashy

Edited by GoSlash27
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Uh… I'm pretty sure I had given the reason to the guy. Not sure about how to react….

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1 hour ago, GoSlash27 said:

 No, it sounds completely wrong :)

 Your total energy at the bottom of the well prior to the burn is zero (or very close to nil). You had 5GJ of kinetic energy and -5GJ of potential energy, which cancel out. You then added .5MJ of kinetic energy by increasing your velocity 1,000 m/sec, which you keep as you leave. The kinetic energy is transferred back into potential energy to pay off the debt you incurred by falling into the hole, leaving your 1,000 m/sec DV.

Kinetic energy is always given by mv2/2. If you say that speed has increased from 100km/s to 101km/s, you are saying that kinetic energy increased from 5GJ to 5.1005GJ.

If you are saying that kinetic energy increased by 0.5MJ to 5.0005GJ, then the speed increased from 100km/s to 100.005km/s, for a delta-V of 5m/s.

You have to pick one of these scenarios. You simply cannot have both. The first one is correct in this case, but it's useful to work out why. Again, look at the rocket in a vacuum (no gravity) example I've given in the previous post. Do you agree that kinetic energy gain is different there, depending on how fast the rocket was going before the engines kicked in?

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@K^2 @GoSlash27

I think the key to these questions lies in the fact that rocket engines preserve momentum, not energy.

But I've also read that "Δν" as is colloquially used is not always the same as "change in velocity".

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