# Moving in space by accelerating charged particles.

In the basis of idea there is a principle of reducing the fuel weight by increasing the velocity of its expiry.

Having substituted in the jet propulsion formula the mass changes formula depending on the velocity, we get such an expression.

VrMr =VfMf

With the velocities close to С

where; Mr – rocket mass, Vr – rocket velocity, Mf – fuel mass, Vf – fuel velocity, С – velocity of light.

It is supposed to accelerate electrons or nuclei of atoms, i.e., with a stable mass and charge.
For the implementation of the conceived, I consider to use a plate accelerator. According to the principle of charged particles acceleration between the capacitor plates.

The particle moves from cell to cell at the time of pole shift due to the AC power supply in shaped characteristic, Figures 1 and 2.

The force, that effects the particle F=eE, where Е=U/h, h- distance between plates.
F=eU/h.
The velocity in the outlet stage, and therefore in the inlet is equal to
V2=V1+at where а-particle acceleration, а=F/m=eU/hm .
We express h through t- time.
h=at2/2=eUt2/hm2
h2= eUt2/2m

We substitute in the expression for the velocity, where V 1 – initial speed for the first stage, or the incoming part of the next steps Vin.

In the course of velocity increasing the particle mass increases according to the expression

We find the arithmetic mean between the values in the inlet and outlet from the stage and put it into the formula.

Where Vin- velocity in the inlet of the stage, Vout- velocity in the outlet from the stage.

The task is solved by the selection of the velocity Vout in such a way that as a result of the solutions we obtain the same velocity value.
Distance covered h- distance between the holes in the plates and the time t, for which this is covered, is necessary for the selection of voltage generator characteristics, we determine by the expression

h=t(Vin + Vout)/2

Here is an example of calculating the accelerator for electrons.
We calculate the first stage.
We define the value of the outlet speed 3 10 7 m / sec.
The inlet rate of the first stage is set to 0.
We obtained the necessary voltage value equal to 2.5686 103 B.
We specify the length of the first stage 3 10-2 m. and obtain a frequency for generator equal to 250 MHz.
Only the particle velocity and length of the stages changes further.
2nd. Vout =0.5984 108 m/s.; h=8.9846 10-2 m.
3rd. Vout =0.8938 108 m/s.; h=14.92 10-2 m.
4th Vout =1.1844 108 m/s; h=20.78 10-2 m.
5th Vout =1.4686 108 m/s.; h=26.53 10-2 m.
6th Vout =1.7442 108 m/s.; h=32.12 10-2 m.
7th Vout =2.0088 108 m/s.; h=37.53 10-2 m.
8th Vout =2.2597 108 m/s.; h=42.68 10-2 m.
9th Vout =2.4929 108 m/s.; h=47.52 10-2 m.
10th Vout =2.7027 108 m/s.; h=51.95 10-2 m.
11th Vout =2.8763 108 m/s.; h=55.76 10-2 m.
12th Vout =2.9876 108 m/s; h=58.63 10-2 m.
13th Vout =2.9999 108 m/s.; h=59.87 10-2 m.
From these obtained results it is evident that we have “stop” to the growing mass of the particle. If we define the weight of the machine 10,000 kg., and the mass of the accelerated electrons is100 gr. (V = 2.999998) than from the formula (1) follows that the machine must reach a speed of 779395 m/s.
The greater the number of stages, the more the total force acting on the particles, and, accordingly, on the accelerator.
The force acting on the group of particles that are in the same group of stages equal to
F = neU /h, where n is the number of particles.

Undoubtedly, for the movement in space in this manner, a corresponding amount of energy is necessary. One of the ways of obtaining this energy I consider a controlled thermonuclear reaction.
My view on this issue is as follows.
If we consider two particles possessing charges q1 and q2 and flying parallel with the speed V1, than it is possible to calculate the velocity at which the particles have to fly, so the pole (compression force) generated by their movement surpassed repulsion force of particles (coulomb) and the particles got into the field of each other nuclear forces, which act at a distance of about 10-15 m. Since at this distance the maximum repulsive force will apply, the entire calculation should be made to overcome the maximum force before the “inlet” of nuclear.
The force acting on the two parallel current conductors.

Where, L- considered distance along the axis of charge propagation, Pi- Pi character.
To move from a current conductor to the particles, we define the current strength I as I = q / t, where t is the time in which the charge passes through this section. It can be expressed in terms of distance and velocity t = L / V, hence I = qV / L.
We place it in the formula.

Coulomb repulsion strength

We equate these two forces as a condition for convergence of the particles.

For the consideration conditions of the two particles, we neglect the values of the magnetic and dielectric constant of the substance M and E.

We also accept them as equals from the consideration condition of particles in a certain area in diameter R and length L.

Hence, from the resulting equation we express V

We take into account, that EoMo= 1/C2  where С – velocity of light, we get.

As a result of the calculation we found out that regardless of charge, the particles that have reached the velocity V (flying parallel) should fall under the influence of compressive forces in the field of action of atomic forces and leakage of thermonuclear reaction is possible. For this purpose an accelerator scheme proposed by me is the most suitable as particles move in groups therein.