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The Physics of Space Battles
An anonymous reader writes PBS' It's OK to be Smart made this interesting video showing us what is and isn't physically realistic or possible in the space battles we've watched on TV and the movies. From the article: "You're probably aware that most sci-fi space battles aren't realistic. The original Star Wars' Death Star scene was based on a World War II movie, for example. But have you wondered what it would really be like to duke it out in the void? PBS is more than happy to explain in its latest It's Okay To Be Smart video. As you'll see below, Newtonian physics would dictate battles that are more like Asteroids than the latest summer blockbuster. You'd need to thrust every time you wanted to change direction, and projectiles would trump lasers (which can't focus at long distances); you wouldn't hear any sound, either."
I seem to recall that the Starfury of Babylon 5 got the physics (more or less) right.
...I can tell you another thing about space battles: you don't see anything aside from a few tracks on a computer screen. If you have a telescope pointed in the right direction at the exact right time you see a very unimpressive and quick flash.
The ranges, timing, and velocities involved are far too great for human perception.
I am very small, utmostly microscopic.
Demonstrating the physics of space fighters with Kerbals in them:
https://www.youtube.com/watch?...
Poul Anderson, The Star Fox
Larry Niven, Protector
C.J. Cherryh, Downbelow Station
I read a rebuttal to that which was fairly compelling: http://scienceblogs.com/builto...
The equation given isn’t derived. We have no idea where they’re getting that 13.4 proportionality constant. Dimensionally it’s correct, and it’s pretty easy to derive the equation up to that constant which will depend on the sensitivity of the detector. That equation modulo some uncertainty with respect to that constant is accurate as far as it goes given a spacecraft of hull temperature T and cross-sectional area A.
I would take you through the steps of the derivation, but it would be pointless because the assumption that the hull temperature has anything to do with the interior temperature is simply flat wrong. We can prove this with a potato.
Switch your oven to the “Bake” setting at a temperature of 350 F. After preheating, put in the potato. The interior of the oven, and eventually the potato, are maintained at a constant temperature of 350 degrees. How hot is the exterior surface of the oven? Depends on how well insulated your oven is, but I can guarantee it’s a lot less than 350 degrees.
The key is the understanding the relationship between heat and energy. Put hot coffee in a thermos – the hot coffee is hot because it contains thermal energy. If the energy can’t leave, the coffee will stay hot because the energy stays inside the thermos. The outside of the thermos stays at the temperature of the surroundings. Now neither the thermos nor the oven is a perfect insulator. Some energy leaks out of the oven’s interior, cooling it down. The oven thus has to pump energy into the heating elements to make up for this loss. Equilibrium is reached when the rate of energy being put into the oven equals the rate of loss through the insulation.
For a spacecraft in a vacuum, the pretty much the only way to lose energy from the interior is by radiant heat. The higher the temperature of the outside, the higher the rate of energy loss via radiation. But the temperature itself is irrelevant, since just like the oven and the thermos it’s not necessarily related to the actual temperature inside the cabin at all. It is always and everywhere a function of the total power passing through the hull. If the temperature inside the cabin is constant, the power leaving the hull by radiation is exactly equal to the power being generated inside the hull.
So how far away can we detect a given amount of emitted power? According to Wikipedia, a telescope of 24 aperture can detect stars of magnitude 22 after a half-hour exposure. I think this is a pretty good realistic limit for detection with reasonable equipment in a reasonable time frame. Now we need to compare this magnitude to something of known power output. How about the Sun? The sun has magnitude -26.73 as seen from the Earth’s surface (smaller magnitude is brighter), for a difference in magnitude of 48.73. The exponent used for magnitude is 2.512, so the difference in power per unit area of telescope is 2.512^48.73 = 3.1 x 1019. Since the Sun radiates about 1000 watts per square meter at the distance of the earth, the smallest radiant power we can reasonably detect in our telescope is about 3.123.1 x 10-17 watts per square meter.
Our hypothetical spacecraft is radiating that power into space, evenly distributed over the surface of a sphere of radius r, where r is the distance to the detector. When that power-per-area is the same as the limit of our telescopic capability, that gives us the maximum detection range. Mathematically,
Where rho is the sensitivity of our detector. Solve for r:
So what’s the power? Well, each human on board is going to produce about 100 W just from basic bodily metabolism. Computers, life support, sanitation, and all the rest will contribute more. We might assume 10,000 watts total for a futuri
Not necessarily - if you paint them black and launch them discretely (by rail gun?) a missile could coast unnoticed across the void to your apprxoimate location and only engage its engines for the final approach.
Radar is viable on Earth because the horizon is only a few dozen miles away and the area of interest is typically only a mile or two high - practicaly 2-dimensional. In 3D space you'd be talking about many orders of magnitude more power to run an broad-focus radar system with the sort of range you'd need to be useful. Especially considering projectile speeds are potentially several orders of magnitude faster than is possible to sustain within an atmosphere.
--- Most topics have many sides worth arguing, allow me to take one opposite you.
Shock wave? Sorry - that requires an atmosphere. If you are in one (inside a ship), you could hear debris hitting your ship. You could hear sounds of power transmission (perhaps loss effects from transformers like the hum you hear under certain power lines on earth today). But once in a vacuum there isn't going to be a shock wave. They actually covered that in the video.
Likewise, perhaps *we* can't focus a laser today, but that's not an inherent limitation of lasers even by today's known physics, that's a limitation of our technology
I'm pretty sure the video author is not aware of it, but that's actually a limitation of physics, not of laser technology. The fact that you cannot focus a laser at long distances is related caused by momentum-position duality in quantum physics: Laser is basically a bunch of photons going all in the same direction, with the same color and coherent phases; technically with the same wavevector. However quantum physics dictates that there is going to be a certain spread, uncertainty, in the wavevector of each photon. This uncertainty is inversely proportional to the size of the chamber where the laser was initially pumped, namely the size of the laser gun.
It is really quite similar to projectile weapons: The longer the barrel, the more accurate the shot.
this post contain no useful information, no need to mod it down
Likewise, perhaps *we* can't focus a laser today, but that's not an inherent limitation of lasers even by today's known physics,
In fact we can focus lasers to within the limits imposed by physics. That is diffraction limited optics and Gaussian beam optics. Real lasers and optics get very close to these limits.
Long story short its all about the size of your laser, or rather aperture. Lets assume a 500nm laser with a 1 meter wide aperture and we assume we want a spot size 1 meter or less. From the math that means we can focus that good out to a range of just over 3140km. In the middle the beam size is about 70cm. At a range of 4700km the spot size is about 2meters. These ranges scale with the square of aperture size with the caveat that we only focus to the aperture size. So a 2 meter one has 4 times the range where the spot size is 2 meters or less. It is also proportional to wavelength.
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