# Traveling to Mars in 24 Days

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.

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?

• 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? – Jerry Schirmer Jan 26 '15 at 20:09
• How do you stop when you get to Mars? – JDługosz Jan 26 '15 at 20:15
• You $v_{initial}$ and $v_{final}$ and the length of the track, seems like you can you use your SUVAT equations. – Kyle Kanos Jan 26 '15 at 20:18
• 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. – Erik Burden Jan 26 '15 at 20:31
• It is the idea of sending a small 5-10 person craft with payload to Mars. I would assume between 100-200 tonnes. – Erik Burden Jan 26 '15 at 20:32

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.

• "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." – Erik Burden Jan 26 '15 at 20:54
• @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... – penguino Jan 26 '15 at 22:31
• ^ That's some good stuff right there. Thanks again for your input. Much appreciated. – Erik Burden Jan 27 '15 at 19:03

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.

• 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. – Erik Burden Jan 26 '15 at 20:28

I know it's been years, but I can't help myself. @penguino's answer is good, but ignores some important aspects of the scenario.

Unless one wishes to arrive on the surface of Mars at 60,000mph (ouch!), you need to account for deceleration. You won't have a railgun to do it for you, so you'll need rocket engines and fuel tanks onboard your projectile, with the ability to orient the craft in space so that you can use the rockets to decelerate.

At launch time, rendering the occupants unconscious by making them sustain 15G for a few minutes is borderline acceptable, as they'll have a couple weeks to recover during the flight and very little else to worry about during that time. However, when it comes time to land, I think it's ideal if a pilot can remain conscious and able to control the ship's systems during descent, so maybe we want to limit ourselves to something more like 1-44G. Doing the math, decelerating to zero at 4G would take eleven minutes; 22 minutes at 2G, 45 minutes at 1G. So, none of these have a significant impact on travel time. I'd be tempted to pick 1G just to minimize the chance of issues and maximize opportunities to make corrections. You'll get there the same day regardless.

Now we get to the fun part. Using a rocket to create a ΔV of 27,000m/s, imagining that the projectile itself (with rocket engines and fuel tanks but no rocket fuel) weighs 100 metric tons (a Saturn V rocket weighs 130t empty), and optimistically selecting an extremely efficient rocket fuel (liquid hydrogen-oxygen, netting an exhaust velocity of ~4.5km/s), we plug our numbers into the Tsiolkovsky Rocket Equation and find that we need a bit over 40,000 tons of rocket fuel onboard at launch in order to perform that acceleration. For comparison, the aforementioned Saturn V carried ~2800t across three stages, which was sufficient to reach the moon, performing most of the transit at a relatively paltry ~2.5km/s.

Now, we go back to that rail launcher. It needs to accelerate 40,000 tons and change to 27km/sec over 2400km. Assuming linear acceleration, that means we need to perform the acceleration over the course of roughly three minutes.

This task will require the linear accelerator to deliver 14,580,000,000,000,000 joules (14.58 petajoules) of kinetic energy into the vehicle. The best railguns are about 66% efficient, so the power source needs to supply a total of 22.1PJ. To do so in the three-minute timeframe will require a continuous supply of 123 terawatts of electrical power -- enough to illuminate more than two trillion 60W light bulbs. The combined electrical generation power of the entire United States is approximately one terawatt; the entire world can generate about 7.7TW at theoretical peak capacity (which is never attained).

So, before you can operate your rail gun, you're going to need to install some serious power generation and delivery equipment on the moon. I'm envisioning nuclear power plants (because most power generation technologies we know today wouldn't work on the moon, although I'd be interested to work out what you could accomplish with solar) and millions of supercapacitors spread out along the railgun to deliver the energy.

But wait, I'm not done! Recall that our launch rail is two-thirds the Moon's diameter and needs to be straight. We'll need to bore a hole through the moon. I'll spare you the math, but if we drill a chord through the moon from surface to surface of the appropriate length, we luckily avoid the molten iron outer core, and we gain some serviceability and safety by having two entrances! Maximum distance of this tunnel from the nearest surface point is about 462km, and, as stated before, the total length of the tunnel is 2400km. The deepest borehole we've ever created on Earth is about 12km deep and only 12" wide. As the hole gets wider, more depth gets more difficult, and I hope I don't need to remind you that our 40 kiloton vehicle will be a bit wider than 12". Needless to say, we will need to make significant strides in our boring technology before this becomes feasible! Maybe Elon Musk can help.

As a bonus to this through-the-moon approach, however, is that we get a light assist from the Moon's gravity during the first half of the acceleration phase. It's completely offset by the second half as we move farther from the center of the moon, but it's better than fighting gravity the entire time, as is typical of most launch systems.

Factors still not considered:

• How to get the launch vehicle, its fuel, the railgun, the power generation equipment, and all supporting infrastructure to the moon
• How to build a 2400km-long railgun
• Additional energy required to escape the Moon's and Earth's gravity wells
• Since we're boring a hole through the moon, we can only use it at one specific moment, when it's aimed straight at Mars. Otherwise, we're wasting a significant portion of our launch energy and need to pack even more rocket fuel in order to adjust course. I'm not sure how often the hole would be in alignment and not obstructed by Earth, nor how wide the launch windows would be.
• Note that both Earth and Mars are moving quite fast (relative to the Sun). As such, you can't just make a straight-line path from one to the other. Instead, you need to plan a Transfer orbit, which takes a much longer path -- usually roughly 2x the direct line.

Between the borehole, the launch power requirements, the vehicle size, and the logistics, I think that we can conclude two things: 1. We as a species are nowhere close to being ready to implement this. 2. Maybe just go a bit slower.

Note: I used Saturn V for reference because it's the heaviest-lift-capable launch vehicle I'm aware of that was actually put into use and therefore has significant data available about it. Ares V was, and SLS is, slated to be more capable, and I'm guessing somewhat more efficient, but detailed figures are not readily available.

Finally, just to address a couple misconceptions in the question and its comments:

• The speed of sound on Earth and in space have nothing to do with this. On Earth, the speed of sound is relevant because you have to deal with shockwaves when you exceed it. The moon's lack of an atmosphere renders this moot.
• From comments: "if there is no resistance against the moving object, how do you determine the speed it can achieve?" When there is no resistance, there is no speed limit -- you are only limited by your ability to apply force in the correct direction to accelerate. However, there is in fact some resistance -- the vehicle will be in the Sun's gravity well the entire time. It's a relatively small force, but it will cause the coasting vehicle to decelerate slightly beginning the moment the railgun stops accelerating the vehicle.