... 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.