How to go about calculating the size and energy requirements for a space-based linear accelerator or railgun to accelerate small probes? - Space Exploration Stack Exchange most recent 30 from space.stackexchange.com 2019-08-25T20:25:15Z https://space.stackexchange.com/feeds/question/20320 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://space.stackexchange.com/q/20320 2 How to go about calculating the size and energy requirements for a space-based linear accelerator or railgun to accelerate small probes? J.D. Ray https://space.stackexchange.com/users/14104 2017-02-23T17:11:15Z 2017-02-23T18:18:58Z <p>Apologies for the long question.</p> <p>With all the speculation surrounding the newly-discovered Earth-sized planets a "mere" 40 ly distant, I'm revisiting earlier musings on ways to send interstellar probes on a timescale that's useful (e.g. less than thousands of years).</p> <p>If we were going to use some sort of linear accelerator or rail gun (I don't understand the differences well enough to know which is best applied) to "shoot" a sub-kilogram scale probe at another solar system, how do you go about calculating the "muzzle velocity" achieved using factors for energy input and length? I suppose the "sub-kilogram" part isn't that relevant, except to set the scope of the accelerated mass variable, but I wanted to establish the limits around what I'm trying to discover.</p> <p>Associated questions are:</p> <ul> <li>For the amount of energy it would take, do we have a good way to use it in orbit? For instance, if we put a hectare of solar panels in orbit, and use them to charge an ultracapacitor, is that sufficient technology to power our accelerator?</li> <li>Where is a good place to locate such an accelerator?</li> <li>How much acceleration force (gees) is going to be applied to the probe for a given length of "barrel" and "muzzle velocity"?</li> </ul> https://space.stackexchange.com/questions/20320/-/20323#20323 4 Answer by Richard Alexander for How to go about calculating the size and energy requirements for a space-based linear accelerator or railgun to accelerate small probes? Richard Alexander https://space.stackexchange.com/users/18638 2017-02-23T18:18:58Z 2017-02-23T18:18:58Z <p>Start with basic physics. Force = mass times acceleration (F=ma). I find it easiest to start with 1 kg, then scale up or down as necessary. What velocity do you want to achieve? Final velocity is initial velocity plus acceleration times time (v1 = v0 + at). Convert forty light-years into meters (easy to do using online calculators). 40 ly = 3.784292e+17 m. I like to round off to an easy unit, between 3 x 10^17 to 4 x 10^17. Let's take 4 x 10^17, because it will divide nicely with the next value. </p> <p>Time is distance divided by velocity. You want to reach the system in less than 1000 years. Convert years into seconds. Again, that's easy to do with a Web search tool. 1000 years = 3.1556926 x 10^10 seconds. Say 4 x 10^10 seconds for lower velocity (because of more time). So, velocity is 4 x 10^17 / 4 x 10^10, that is, you need to reach a velocity of 1 x 10^7 meters/sec, or 10 x 10^6 m/s in engineering form. </p> <p>Let's say that you build an accelerator that is 10,000 km long, or 10 x 10^6 meters. Average acceleration is change in velocity over two times the distance, (v1^2 - v0^2)/2s . So, (10 x 10^6 m/s) / 2 x 10 x 10^6 m = 5 x 10^6 m/s^2 . That is about 49 million g. </p> <p>At this point, you can see that your accelerator--whether rail gun, coil gun or blow dart--is going have to be much longer than 10 million meters if you want normal materials to survive. If acceleration were at least 1000 times longer, some macroscopic objects might survive. We need a (longer-than) 10 million kilometer accelerator. That's why Stephen Hawking is talking about using laser beams shot into space; he doesn't want to have to build a multi-million kilometer-long tube into space. Theoretical physicists don't make that much money. </p> <p>Acceleration is the change in velocity over the change in time. Rearranging, we find that time is 10 x 10^6 m/s / 5 x 10^6 m/s^2 = 2 seconds</p> <p>Kinetic energy is 0.5 mass x velocity^2 , so, 5 x 10^12 Joules . Power is the energy divided by time, so 2.5 x 10^12 Joules/sec . Let's say 2x10^12 J/s, or 2 TW. That's a lot of solar panels. If sunlight delivers 1500 Watts/m^2, and our solar panels are 30% efficient, we would need more than 4x10^9 m^2 of solar panels to provide that much power in that much time. Of course, we could use a capacitor and a smaller array. </p>