What would be your vision of a better humanity?

Sorry to interject, but, how far is it in terms of light year to get to the next habitable planet ..... estimated to be?

Sorry to interject, but, how far is it in terms of light year to get to the next habitable planet ..... estimated to be?

Next habitable planet? I was unaware that any habitable planets outside earth had been discovered yet. But that is your point isn't it?

... An ion drive engine for a star drive, designed 1000 years from now can be assumed to be immensely effective. This means high exhaust velocity. A stream of charged ions leaving at in excess of 99% of light speed. For electrostatic forces to have an effect on slowing the vehicle, the ions would have to be very close to the vehicle (force decreases as the square of the distance). The time they would be that close would be measured in fractions of a femtosecond. i.e. effectively no braking effect whatever.
No. it makes zero difference how fast you climb a hill - you lose the same amount of energy in doing so. (You are somewhat confusing hill climbing speed with energy required.) That is true for hill made of dirt on earth, and hill made of gravity field, like trying to leave the sun, or a hill made of a retarding electric field.

What would happen in space is that at first turn on of the un-neutralized ion beam the retarding field's "hill" is small - only a few kilovolts of potential but it will rapidly become steeper - a trillion kilovolts for example. Then each ion will need to have more than a trillion kilovolts of kinetic energy to climb a trillion kilovolts electric field potential energy hill or it will fall back to the rocket ship producing zero net thrust.

Now to give that ion a trillion kilovolts of kinetic energy requires a very impressive accelerator, which will have huge mass and if is a linear acceleration to avoid kilo tons of mass (in the magnets that curve the ion around in a circle while it gains energy), it will be miles long.

These accelerators are very expensive also. After the circular Fermi one near Chicago (correct name escapes my memory just now - strange as I was once offered a summer job there.) some wag physicist said he was sure the next one, with higher energy capacity would be a linear one because of the expense. Only if it stretched across the country thru more than half the states would Congress appropriate the money. It turned out he was wrong. The LHC, part in France and part in Switzerland, is miles in diameter circular one but got international financing as not even the US could afford to build it alone.

... I agree that solar sails will be of limited value further from the sun. But again, remember that this is 1000 years in the future. We can expect the technology to be very advanced, and solar sail performance much better than anything we plan for today. However, I agree it will be of no value once the vehicle has gone more than a few AU from the sun. That is when the ion drive takes over.
Technology will not increase the solar flux pressure per unit area. You can only get 200% of that.

The thrust of ion dives, even if it were possible to throw ions out at 90% of C, is going to be more than a million times less than the solar weight of your huge 10 million tons ship at a few AU or less. If ions are ejected by an accelerator, the thrust would not even overcome the sun's gravitational attraction for the accelerator alone! No proponent of ion drives is crazy enough to suggest anything but a high voltage field be used to accelerate the ions. At a few AU from sun the net force on the space ship with accelerator produced ion thrust alone is back towards the sun (because of the huge mass of the accelerator). Technology is not going to reduce the mass of magnets significantly or the mass of 10+ mile long linear accelerator producing your relativistic ions.

... Actually, I suspect the 'departure point' will be somewhere in the Kuiper Belt, when the last of the extra reaction mass is taken aboard, which makes the solar sail at that point not useable. Instead, our space city will float alongside a Kuiper Belt object, such as a comet, to permit it to be mined.
by time you get the 10 million ton space craft out to the Kuiper belt you will have climbed more than 95% of the solar gravitational hill - How I don't know.

... Leaving the gravity wells of both Earth and the moon to take on board a payload is more than 1000 times different to a vehicle beginning in solar orbit, and never touching down inside a gravity well.
No. Earth plus moon makes a trivial gravitation hill to climb away from compared to the sun's gravitational hill if you are going out to the kuiper belt to collect more mass.

... Even the sun's gravity, to be overcome when leaving the solar system, is miniscule by comparison.
NONSENSE, even you should be able to show how extremely false this is:
The mass of the sun is 330,000 times greater than that of the Earth. Thus when the space craft is x away from the Earth, on the sun/earth line to make it simple, their gravity fields are equal when x, in AU, is solution to this equation:

M/r^2 = 330,000/(1+x)^2 = 1/(x)^2

Where this right side is the Earth's gravity. I won't solve this quadratic equation but just guess an x which makes the right side less than the sun's gravity.

For example if x = 0.01 AU then the sun's gravity is 330,000/1.021 = 323,498 vs. 1/(0.01)^2 = 10,000 for the earth's gravity or already at only 0.01 AU from earth the sun's gravity is more than 32 times stronger than the earth's gravity is! Because of this the sun is pulling on the moon stronger than the Earth is. I thus, say the moon is orbiting the sun, not the earth. The Earth just makes a very tiny "wobble" on the moon's basically elliptical orbit about the sun. If Earth and moon's orbits are plotted on 8.5 by 11 inch sheet of paper with typical pencil there is no difference in their over lapping plotted trajectories - that is how small the "wobble" is. I.e. the amplitude of the "wobble" is less than the width of the pencil line!

Quite the contrary to your statement, when climbing the gravity hill out to the Kuiper belt with lift off from either moon or Earth you can totally neglect the attraction of the earth and moon with less than 1 % error. You obviously have zero appreciation for how deep a gravity well the sun has dug in space.

... The time factor is large. If it takes 10 years to accelerate to 0.1C and 10 years to decelerate, then transit time Earth to Alpha Centauri is 55 years. Not too much of a problem on a self sufficient space city. ...
Again rather than make wild false statements, calculate a little. Neglect the gravitational attraction of the sun and use the ideal rocket equation to calculate how much mass you must throw out to get to 0.1C I won't do that but am quite confident it is much more than 100 times the mass of your space craft city. Don't forget that you need to accelerate that much mass plus the mass of your space craft to have it available when you near Alpha Centauri for the de-acceleration down from 0.1C Also consider the volume this mass taken to Alpha Centauri would occupy. I.e. 99% or more of your ship's volume going to Alpha Centauri is just for containing the de-acceleration mass.

If Alpha Centauri is more massive than the sun by X then that mass you must transport to Alpha Centauri for the de-acceleration is X times larger than the rocket exhaust mass you needed to escape from the sun. I will let you look up what X is, but I think it is much greater than 1.

These are some of the reasons why I said, several post back, that if man is every going to Alpha Centauri etc. it will be as information (not very massive) in a computer which is used by the on-board robot to reconstruct man in a series of test tubes and then incubators. No living woman will ever go to Alpha Centauri and certainly "useless unneeded men" will not.

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Sorry to interject, but, how far is it in terms of light year to get to the next habitable planet ..... estimated to be?
Assuming you would like it to have oxygen in the atmosphere (made and maintained* by the pre-existing green plants) and gravity not more than a factor of two greater than earth's gravity and at least some part of it with temperature where water can exist as a liquid, I think it likely you must leave our galaxy and explore some others. SETI has been looking for a few decades for any evidence that this is wrong, but many may not agree with me as there are a lot of stars in our galaxy.

* Oxygen has this unfortunate habit of chemically combining with things so even if you were to try to make some it would rapidly disappear into chemical compounds. BTW, too much O2 will make you blind and too little will make your die. You need "goldielocks" O2 levels. Also your local sun must be reasonable stable and not have any neighbors which are gamma ray bursters.

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Sorry to interject, but, how far is it in terms of light years to get to the next habitable planet ..... estimated to be?
Since we haven't found it yet, we don't have an estimate, just speculations. But we're pretty sure that it can't be any closer than 15LY, as Billy noted. As has also been noted, given our bodies' inability to withstand sustained acceleration much greater than one gee, not to mention the price of fuel these days, that minimal distance is far enough that a round-trip voyage will require more than one human lifetime.

So we have to develop a generation starship before we establish any colonies.

So let me see the hands of people who volunteer to ride in the first generation starship, one that has to not only reach its destination but send down landers and build a successful colony before it can turn around and come home, built by the same species who brought you Shrimp In Petroleum Sauce, the Giant Hokkaido Night Light, and the Incredible Shrinking Rain Forest.

To Billy

You are correct about the length of the linear accelerator used as the ion drive. I suggested ten kilometres. That is not so bad if it is part of a space city.

That means the ions are being accelerated over a ten kilometre length, and accelerating the space city during the whole ten kilometres that the ions are so driven. The electrostatic braking that you suggest can only happen whoile the ions are very, very close to the vehicle, since the force drops off so rapidly. A few millimetres.

So the ions accelerate the city for ten kilometres and brake it for 10 millimetres. Guess which dominates. So no. You do not need to neutralise those ions. Basic physics!

The mass of the magnets, especially with 1000 years improvement, is small compared to the mass of the reaction mass to be carried.

On gravity.
The basic equation that Newton came up with shows the force of gravity between two masses is inversely proportional to the distance squared. So again, the force falls off very quickly. If we start a vehicle from the Earth's surface, the source of gravity is very close, and hence strong. If we start accelerating away from the sun from a point well clear of the Earth, but still in the Earth's orbit, we are 93 million miles from the source of gravity. A tiny fraction of the Earth's gravity at ground level.

So the trip, from Earth surface to moon surface and back, involves breaking away from a massively greater gravitational force than simply cruising from solar orbit to clear of the sun's gravity. Miniscule by comparison. Again, basic physics.

On achieving 0.1C.
I am not a rocket scientist, but I am basing my descriptions on others who are. There was an article in Sciam about 10 years ago, written by two NASA rocket scientists, who predicted, within 500 to 1000 years, a star drive based on principles similar to what I am describing, and which could accelerate to between 0.1 and 0.2C.

To Fraggle

We will not need another Earth as a destination. As I described earlier, a society that lives in space cities will be quite capable of going to Alpha Centauri and continuing their space city way of life. They can use space debris as their main resource.

Today, many people are talking of a colony on Mars. There is no reason in theory why this should not be possible. Difficult, yes, and will need technology more advanced than we yet have. But possible in theory. If that can happen, then we can do the same on cold planetary bodies around one or more of the stars of Alpha Centauri.

Sorry to interject, but, how far is it in terms of light year to get to the next habitable planet ..... estimated to be?

Gliese 581G is 20 light-years away and is a potentially habitable planet. Gravity of around 1.5G's, and surface temps (assuming an atmosphere with similar greenhouse characteristics as Earth's) would be around -35F to +10F. It's tidally locked to its primary star which means one side always faces the sun. Thus you may be able to choose a temperature by just moving closer and closer to the sunward side of the planet.

Mars at its closest point is 25 million kilometres away and is a potentially inhabitable planet. We now know there is plenty of water. We would need to live in enclosed habitats to provide atmosphere and warmth, but it is definitely potentially habitable. Every bit as much as Gliese 581g, and an awful lot closer.

When humans leave Earth, we are seriously unlikely to find and colonise nice Earthlike planets, with high oxygen atmosphere, and pleasant temperatures. We will most likely find worlds more like Mars, and we will need to adapt those worlds to us.

... The electrostatic braking that you suggest can only happen while the ions are very, very close to the vehicle, since the force drops off so rapidly. A few millimetres.
So the ions accelerate the city for ten kilometres and brake it for 10 millimetres. Guess which dominates. So no. You do not need to neutralise those ions. Basic physics!
A totally false claim (and only an unsupported claim). For the ion to leave the ship the ENERGY of the ion must be greater than the electric potential hill it must climb. True that "hill" is steepest near the ship but it extends to infinity. The electric field does fall off approximately as 1/r^2 (would be exactly that if the ship were a point), but the potential hill only falls off as 1/r.

Yes it is basic physics, called "conservation of energy" not the nonsense you are suggesting about comparing the acceleration length to a few mm. If the kinetic energy of the ion is less than the potential energy of the hill, it will not climb out of that potential well, no more than a rock thrown up in Earth’s gravity hill will leave Earth if its KE is less than the depth of Earth’s gravitational PE well.- Rock will slow down, stop, and then accelerate back just as the ions will if they have less KE than the PE hill they must climb.

If you do throw an ion out with more KE than the PE of the hill, yes it will escape but also it will also increase the electrical charge on the space craft so that the height of the hill (the value of hill’s PE) is increased and then the next ions must be thrown out with more KE if they too are to escape. Yes it is the simple physics - the conservation of energy law.

After enough ions have been throw out to make the PE of the electrostatic hill equal to the peak KE the accelerator can produce, no more will leave. = zero thrust.
... The mass of the magnets, especially with 1000 years improvement, is small compared to the mass of the reaction mass to be carried.
That is no doubt true, but there will be very little improvement magnet strength to weight ratios even in a million years. (Less than 10% reduction in mass for a given B field from permanent magnets I would estimate. This is because you can only line up 100% of the atomic magnets and the best magnets have the atomic magnets already ~90% all aligned. Only a few atoms have any significant atomic magnets to align and they are heavy atoms.)*

I only said that the ion thrust would not support the mass of the accelerator alone when less than 2AU from the sun to try to get you to understand that the much more massive space ship would also have net force towards the sun; but yes I agree the space ship, especially including the mass it must carry to Alpha Centaury for the de-acceleration when it nears Alpha Centaurs, which is at least 100 times greater than the mass of the empty space craft, which in turn is much bigger than the mass of the accelerator.

... On gravity.
The basic equation that Newton came up with shows the force of gravity between two masses is inversely proportional to the distance squared. So again, the force falls off very quickly. If we start a vehicle from the Earth's surface, the source of gravity is very close, and hence strong. If we start accelerating away from the sun from a point well clear of the Earth, but still in the Earth's orbit, we are 93 million miles from the source of gravity. A tiny fraction of the Earth's gravity at ground level.
Yes near the Earth’s surface the gravity force is much stronger than the sun’s gravity. What this means is that to even leave the surface the thrust must be great (slightly greater than the vehicle’s weight). But the Earth’s gravity force is 75% LESS when only an earth radii (`4000 miles) away from earth. At that point the sun’s gravity is essentially unchanged. I.e. is reduced by only (93,000,000 / 93,004,000)^2.

While the strong Earth’s surface gravity does set the minimum thrust requirement for liftoff, the Earth’s gravity has very little effect on the energy required for your trip out to the Kuiper Belt to get more mass for expelling. I.e. the energy required would not even be reduced 1% if the Earth had no gravity and you were to leave for the Kuiper belt from it. To prove this requires integral calculus, so I will not do it but;
Perhaps you can understand that earth’s gravity is not important from an energy POV by considering just the trip out to Pluto, which is more than 5766 million miles from Earth (more as it is not in same plain but I only subtracted their distances from the sun to get 5766E6.). Using your 93E6 miles as 1 AU, then the trip to Pluto is 5766/93 = 62AU long. Recall I computed in post 103 that sun is more than 32 times stronger at 0.01AU from Earth and hence still 8 times stronger at 0.005 AU from Earth as there earth’s gravity is only 4 times stronger than at 0.01, so at about 0.002AU the sun’s gravity is equal to that of the earth. Hence in the 63AU long trip, the earth gravity is dominate for only 0.002AU of the trip.

Half way to Pluto, (32 AU from sun and 31AU from Earth) this is how their relative gravities compare: Sun has 330,000/ 32^2 and Earth has 1/ 31^2. Thus at trip’s half way point the sun is 330,000 x(31/32)^2 or more than 300,000 times more important than the earth when calculating the energy required to go the next mile. Now do you understand, without using calculus, why from and energy POV it is the earth that is insignificant, not the sun as you falsely assert?

...So the trip, from Earth surface to moon surface and back, involves breaking away from a massively greater gravitational force than simply cruising from solar orbit to clear of the sun's gravity. Miniscule by comparison. Again, basic physics.
No as just shown, by analysis, not your unfounded claim, you have it just backwards. You are confused in thinking that the large force very near Earth instead of energy required for the trip is what is important for escaping from the sun. It is energy you need to have to climb out of the sun’s deep potential energy well that makes leaving the solar system difficult. Again it is energy requirements you need to consider, not forces, not distance traveled during thrusting or accelerator lengths for accelerating ions, etc.

* True story is considerably more complex, as the quantum mechanical "exchange energy" in a few materials is what aligns the atomic magnets. I.e. all iron, for example is 100% magnetized on a very small scale, even if you don't notice any external magnetic field. That is all the atomic magnets of neighboring atoms in a "domain" all point the same way, but the various domains all point in different ways until the iron is "magnetized." When you magnetize iron etc. you are flipping the magnetic orientation of the domains. This happens very quickly, one domain at a time, so makes an observable dB/dt externally which a coil wound around the iron can via and audio amplifier produce a click sound. This audio noise is called Brakhausen noise (but I probably have it spelled incorrectly) is made while a piece of iron is being magnetized.

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An additional requirement not among those given in post 104 will rule out, I would guess, 90% of those, if any, planets that do meet all of the requirements given in post 104.

CO2 is a very common and typically large part of the atmosphere of planets. For example it is more than 95% of the atmosphere of Mars. Humans don't survive a day in even 5% CO2 as it makes the blood too acidic. Here is what wiki states:

(1) At about 5% it causes stimulation of the respiratory center, dizziness, confusion and difficulty in breathing accompanied by headache and shortness of breath.[53] Panic attacks may also occur at this concentration.[55][56]
(2)At about 8% it causes headache, sweating, dim vision, tremor and loss of consciousness after exposure for between five and ten minutes.[53]

Thus in many cases to make the planet a home for man, you would need to remove the existing atmosphere. We would like to remove some of the CO2 now in earth's atmosphere, but there is no affordable way to do that. How could a new colony do that?

Today, many people are talking of a colony on Mars. There is no reason in theory why this should not be possible. Difficult, yes, and will need technology more advanced than we yet have. But possible in theory. If that can happen, then we can do the same on cold planetary bodies around one or more of the stars of Alpha Centauri.
I thought the problem with Mars was no magnetic field and therefor no atmosphere?

My feeling is that by the time we build space cities, people probably won't be organic, at least most of the time :shrug:

We will most likely find worlds more like Mars, and we will need to adapt those worlds to us.
Or us to them...

Billy

You do not know what you are saying.

Let me nail down one point.

The force on an object at the Earth's surface from Earth's gravity is massively greater than the force of the sun's gravity on an object in solar orbit, away from Earth but at the same distance from the sun as Earth (one AU).

This can be calculated using Newton's equation.
F = GM1M2/d.d

G is the gravitational constant and is 6.7E-11 m3/kg/s2

Make it simple and calculate the relative force on a one kilogram mass on Earth versus solar orbit.

The force on one kilogram at the Earth's surface is one kilogram.

A bit more difficult for the solar orbit.

Mass of the sun is 2E30 kg
Distance to the sun is 1.5E11 metres

So F = 6.7E-11 times 1 times 2E30/150E9 times 1.5E11

Which comes to 0.006kgs.

So the gravitational force of the sun on a one kilogram mass in solar orbit at one AU, compared to that of the Earth at ground level is close to a thousand times less.

So do you still stick to your preposterous idea that escaping the sun's gravity would be more difficult than escaping the Earth's gravity? I might add that, if our space city were leaving from a point further out, such as the Kuiper Belt, the difference would be even more staggering.

A similar equation works for the braking force of an ion leaving an ion drive exhaust. Since the braking force is related to the inverse square of the distance, it pretty much instantly finds itself too far away to have any measurable braking force at all.

Gliese 581G is 20 light-years away and is a potentially habitable planet. Gravity of around 1.5G's, and surface temps (assuming an atmosphere with similar greenhouse characteristics as Earth's) would be around -35F to +10F. It's tidally locked to its primary star which means one side always faces the sun. Thus you may be able to choose a temperature by just moving closer and closer to the sunward side of the planet.
What is your reference for this? Parts are quite different from facts given here: http://www.sciforums.com/showpost.php?p=2762038&postcount=95

...
So do you still stick to your preposterous idea that escaping the sun's gravity would be more difficult than escaping the Earth's gravity? I might add that, if our space city were leaving from a point further out, such as the Kuiper Belt, the difference would be even more staggering.

A similar equation works for the braking force of an ion leaving an ion drive exhaust. Since the braking force is related to the inverse square of the distance, it pretty much instantly finds itself too far away to have any measurable braking force at all.
You still don't get it: Energy must be conserved, not force. Climbing away from just earth's gravitational field out to just Pluto takes energy but that energy is very small (much less than 1%, I think) of the energy required to climb away from just the sun's gravitational field out to Pluto. It is the sun, not the earth, that has made a very deep potential energy hole to climb out of to reach any point in space near or beyond Pluto.

I have shown this to you mathematically - what is it you don't understand about the mathematic demonstration? I can give another and different demostration based on fact the potential energy goes as 1/r and potentials are addative. I.e. calculate the energy required to travel from 4000 miles away from center of the earth to deep space and compare to energy required to travel from 1 AU away from sun to deep space.

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Here is the exact calculation I said I could do in last post, but I fear you will not understand it unless you know potential energy of inverse square field goes as 1/r and usually very large r is taken as the zero energy level.

In relative mass terms and miles, the sun’s potential energy well depth at 1AU from sun is M/r = 330,000/93,000,000 vs
Earth’s potential energy well depth at surface of the earth which is m/r = 1/4000 = 330,000/ 4x330,000,000

Multiply both by 1,000,000 /330,000 and then relative depth of wells is:
1/93 vs 1/(4x330) = 1/1320. Thus the energy to go to deep space from 1AU against the pull of the sun alone is larger than going from surface of the earth (4000 miles from it center) against the pull of the Earth alone to deep space by factor of 1320/93 = 14.2 times larger.

So my guess of 1% was too low. The correct fraction (using fact that potentials add) is 1/(1+14.2) = 6.5% or in words, 6.5% of the energy needed to go to Pluto or beyond from earth’s surface is required to climb against the Earth’s gravity field and 93.5 % of the energy is being used to climb against the sun’s gravity field.

SUMMARY: YES, I am sticking to my “preposterous idea that escaping the sun's gravity would be more difficult than escaping the Earth's gravity” because that is what the EXACT mathematical calculation proves.

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So do you still stick to your preposterous idea that escaping the sun's gravity would be more difficult than escaping the Earth's gravity? I might add that, if our space city were leaving from a point further out, such as the Kuiper Belt, the difference would be even more staggering.

Escape velocity from the surface of the Earth is 11.2km/s. Escape velocity from the Sun, assuming you start at the Earth's orbit, is 42km/s. Thus it is almost three times harder to escape the Sun's gravity than the Earth's.

Since the Earth is moving with respect to the Sun, you get a significant advantage there; it is moving at just under 30km/s. Compare that to the earth at the equator which is moving at .4 km/s. That helps (if you launch in the right direction of course!) but you still need to supply 12km/s vs 11.4km/s to escape from the sun vs the earth.

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Escape velocity from the surface of the Earth is 11.2km/s. Escape velocity from the Sun, assuming you start at the Earth's orbit, is 42km/s. Thus it is almost three times harder to escape the Sun's gravity than the Earth's.

Since the Earth is moving with respect to the Sun, you get a significant advantage there; it is moving at just under 30km/s. Compare that to the earth at the equator which is moving at .4 km/s. That helps, but you still need to supply 12km/s vs 11.4km/s to escape from the sun vs the earth.
I think you are almost completely correct as far as you go but we were speaking of the energy required so your velocity ratios must be squared.

Also the orbital speed of the earth is not much help in trying to climb out of the sun's gravity well as it is almost exactly in the useless* tangential direction about the sun. My answer about the relatively energy required is exactly correct if I made no math errors.

BTW, a fact not well known, is that it takes more energy to "fall" into the sun from an Earth launch than to escape from the sun to deep space! Or in other words, Earth is more than half way up the sun's potential energy hill.

* If you want to be clever, and time the launch from Earth so that as your ship goes forward along the Earth's trajectory it finds the moon near it (crossing Earth's trajectory) and you let the moon turn this useless speed into speed away from the sun. That is called a gravity assist boost. If memory serves me correctly, the space ship that went to Pluto first went towards the sun to get two gravity assist boost off of Venus and then a third gravity assist boost off of Jupiter or Saturn. It ain't easy to get way out to Pluto against the sun's gravity pull.

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Billy

In space travel, force is, in fact, more important than energy much of the time. To escape the Earth's gravity requires truly heroic engines to provide titanic thrust (force). To escape the sun's gravity, because of the enormously smaller force, requires an engine with much lower thrust.

For this reason, a space vehicle leaving the Earth has enormous engines and immense fuel consumption, all just to drive it a short distance. If a space vehicle is in solar orbit, but clear of the Earth, then it can move to a higher and higher orbit using relatively tiny amounts of thrust, from an ion drive engine. Low acceleration but a very high exhaust velocity, permitting massively greater efficiency in terms of mass accelerated per kilogram of reaction mass.

If you are talking of the difficulty of sending mass to, say, Pluto, the greatest problem is getting it all clear of the Earth. After that, it is just time and a tiny amount of thrust. To claim the sun's gravity is the problem is just plain wrong.

We have already seen this with the Voyager probes, which are leaving the bonds of the sun's gravity. The engines to send them clear of the sun was tiny and the fuel used minimal, compared to the amount used simply to get clear of Earth.

So your statement that escaping the sun's gravity is the bigger problem is just wrong. Greater energy, sure, but a way smaller problem. Energy is not the major problem in space travel. A much bigger problem is to use energy effectively to provide the degree of force or thrust needed.

Once clear of the Earth's gravity well, a space vehicle can use the far more efficient ion drive engines, rather than the incredibly inefficient chemical rockets (in spite of their greater thrust). The escape velocity from the sun is 42 km/sec, and that can be attained by slow and steady acceleration with a high velovity exhaust that uses far less fuel.

The other thing about the system I propose is that it can refuel from space debris, all the way out to the Kuiper Belt. Water ice can provide reaction mass, as well as deuterium for a fusion generator.

Billy ... In space travel, force is, in fact, more important than energy much of the time. To escape the Earth's gravity requires truly heroic engines to provide titanic thrust (force).
No. There are ways to reduce the force required. Force is NOT fundamental. - I.e. use many smaller weight rockets and go to the moon then launch your assembled big ship from there or even better just assemble it in space* so the ion engine is all the force you need for launching that big ship from orbit.

The energy requirement CAN NOT BE BY PASSED because it is fundamental. Energy is conserved so no matter how you climb out of the sun's potential energy well you MUST pay that price in energy.
... The other thing about the system I propose is that it can refuel from space debris, all the way out to the Kuiper Belt.
That is highly improbably and normally very "mass negative" I.e. to match the speed of your massive ship to that of the debris would require huge expenditure of energy. That is why I assumed in prior post that you would send small robotic "tug ships" from the big ship to go out to and match velocity (not just speed) with the debris and then re accelerate themselves and the debris they had captured back to the big ship and again match velocity with it. All this velocity matching and turn around to go back to mother ship will take a huge amount of energy - I think that the mass of the exhaust used for these maneuvers will be greater than the mass captured. I.e. "mass negative" - try to show conditions in which it is not.

* If your colony ship contains the 10Km long linear accelerator as you suggested, you have no choice but to assemble the big ship in space. 100,000 tiny fully robotic ships launched from earth to build it could be only a little bigger than a WWII V-2 rocket -very small forces required, but the energy requirement is fixed.

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