Of course, it all depends on the spacecraft size.

The advantage of magnetic acceleration is that the mass of the spacecraft remains more or less constant through the acceleration, unlike the chemical rockets where the initial rocket size is huge compared to the final payload put into orbit.

Voyager 1 was about 722 Kilos (for the deep space probe).

The average communication satellite today is 2,000 to 5,000 kilos, with estimates that they will be on the order of 10,000 kilos in the near future.

Let me try to work out some units (and try out this TEX stuff too)

[tex]1g[/tex] = [tex]\frac{9.8m}{s^2}[/tex]

[tex]1N[/tex] = [tex]\frac{1 kgm}{s^2}[/tex]

[tex]1J[/tex] = [tex]1Nm[/tex] = [tex]\frac{1 kgm^2}{s^2}[/tex]

[tex]1W[/tex] = [tex]\frac{1J}{s}[/tex] = [tex]\frac{1Nm}{s}[/tex] = [tex]\frac{1 kgm^2}{s^3}[/tex] = [tex]1 kg \frac{m^2}{s^3}[/tex]

Ok, so I had estimated a maximum of 1000g of acceleration, over 26 seconds, and 3510 km, and a final velocity of final velocity is 260,000m/s.

Hmmm, how does that all fit in?

Over the first second, one uses the power:

[tex]1 kg \frac{10,000 m}{s^2}[/tex]x[tex]\frac{10,000}{2}\frac{m}{s}[/tex]

And, I think one gets [tex]kg[/tex] x [tex]50,000,000W[/tex] = [tex]kg[/tex] x [tex]50MW[/tex]

I am having a little troubles with my units conversions, and why the power requirements appear to be velocity dependent (and thus would increase over time).

But, anyway, it would take about 50MW per kilo to send the payload off at a 1000g acceleration.

If my final velocity is 260,000m/s, then does that mean that my power requirements ramps up from 50MW per kilo to 2.6GW per kilo?

I guess I'm having troubles understanding why I need to add more energy to maintain the constant acceleration the faster I go [

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Hmmm, for Kinetic Energy:

[tex]E_k[/tex] = [tex]\frac{1}{2}mv^2[/tex]

So, my final Kinetic Energy is:

[tex]E_k[/tex] = [tex]kg[/tex] x [tex]\frac{1}{2}[/tex][tex](260,000^2) \frac{m^2}{s^2}[/tex] = [tex]kg[/tex] x [tex](33,800,000,000) \frac{m^2}{s^2}[/tex]

Ok... getting close.

I should be able to divide that by my 26 seconds of acceleration, and get my power requirements as:

Average Power = [tex]kg[/tex] x [tex]1,300,000,000W[/tex] = [tex]kg[/tex] x [tex]1.3GW[/tex]

Hmmm, so maybe I was right with the increasing power requirements.

That is going to take a lot of solar panels, although it is only a 26 second launch period. I can imagine a 26 second blackout for my entire lunar colony for every satellite launch!!

I could buffer the power drain somewhat. If a Lead Acid battery can put out 12V x 1000A, or 12KW for a few seconds. I would only need about 217,000 car batteries per kilo to power the satellite launch [

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