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I am trying to calculate an electromagnetic rail launch from the Moon but I need to know if humans can survive the launch. Please see the specs below.

150 miles of track

Proposed final launch speed of vehicle = 59, 808 miles/hour

I have tried calculating linear g-force on the Moon but I cannot be certain that the same principles even apply. Can anyone shed some light on the matter?

I am hoping that, if my calculations are correct, in the near future we could launch space craft from the Moon when Mars is closest to Earth traveling without resistance at a speed of 59, 808 miles per hour arriving in a little more than 24 days.

More information:

Speed of sound @ sea level (normal conditions) = 340.29 meters/second

Speed of sound in space = ~1300 meters/second (yes, there is sound in space. Although it is not perceptive, there are still percussion waves that travel faster than on Earth with no resistance.) = 73.82% increase of speed in space.

Current speed of electromagnetic railgun projectile @ sea level= 7000 meters/second

Taking into account the 73.82 % increase and applying it to the known projectile velocity...

7000/0.2618 (73.82% incr.) = 26,737.9679 m/s = 59,808.965 miles/hour

On July 27th, 2018, Mars will be ~35.8 million miles from Earth. 35,800,000/59,808.965 = 598 hours (approx. 24 days)

What would be the shortest length of track needed to make the launch survivable?

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  • $\begingroup$ What's the payload of that railgun projectile? Are you worrying about orbital parameters, other than straight-line distance from Earth to Mars? is 7 km/s even above the moon-earth system escape velocity, starting from the moon? Survivability would be governed mainly by the peak acceleration -- how long is the railgun? Why are you even citing the speed of sound? $\endgroup$ – Jerry Schirmer Jan 26 '15 at 20:09
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    $\begingroup$ How do you stop when you get to Mars? $\endgroup$ – JDługosz Jan 26 '15 at 20:15
  • $\begingroup$ You $v_{initial}$ and $v_{final}$ and the length of the track, seems like you can you use your SUVAT equations. $\endgroup$ – Kyle Kanos Jan 26 '15 at 20:18
  • $\begingroup$ From post below - I must say, I am not familiar with physics. Although I would like to be. I am simply asking for a blog that I will be posting. Personal interest. I have looked on Google how to calculate g-forces and wound up here. $\endgroup$ – Erik Burden Jan 26 '15 at 20:31
  • $\begingroup$ It is the idea of sending a small 5-10 person craft with payload to Mars. I would assume between 100-200 tonnes. $\endgroup$ – Erik Burden Jan 26 '15 at 20:32
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Newtonian mechanics works the same in all environments. The equation of motion v^2 = 2as (for an initially stationary object accelerating over distance s at acceleration a and ending at speed v).

Re-ordering the equation, a = v^2/2s, and substituting in v=27000 m/s (approx 60000 miles/hour) and s = 240000 m (approx 150 miles), results in a = 1520 m/s^2. This is approximately 150 times the acceleration due to gravity on the surface of the earth, so any human would be instantly squashed to a pulp.

The maximum sustained acceleration a normal human could cope with is smaller by a factor of 10, about 15 gravitys. So if you extended your ramp to 1500 miles long, you might have a 'viable' solution. Note that the moon is approximately 2200 miles in diameter, so by boring an appropriately oriented hole through the moon to act as your rail-gun cavity, you could reduce the acceleration to a nearly comfortable 10 gravitys.

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  • $\begingroup$ "Chuck Yeager, and became known as "the fastest man on earth" In one of his final rocket-propelled rides, Stapp was subjected to 46.2 times the force of gravity." $\endgroup$ – Erik Burden Jan 26 '15 at 20:54
  • $\begingroup$ @Eric Burden But Stapp was at that peak acceleration only momentarily (and at ~25 g for ~1 second). That was enough to cause long term damage to his eyes. Fighter pilots sustain turns of ~10 g while sitting without blacking out, so 10 g lying down should be OK for the 4 minutes 40 sec required to accelerate using the "through the moon" option. Stopping at the other end will be tricky... $\endgroup$ – penguino Jan 26 '15 at 22:31
  • $\begingroup$ ^ That's some good stuff right there. Thanks again for your input. Much appreciated. $\endgroup$ – Erik Burden Jan 27 '15 at 19:03
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Just from the first paragraph, what is the Acceleration given a Velocity of v at distance x? The basic physics formulas given in a pre-calculus physics class are enough to solve for $a$. There is not one listed for exactly that: use two or more with algebra.

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  • $\begingroup$ I must say, I am not familiar with physics. Although I would like to be. I am simply asking for a blog that I will be posting. Personal interest. I have looked on Google how to calculate g-forces and wound up here. $\endgroup$ – Erik Burden Jan 26 '15 at 20:28

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