Lookin' for loopholes allowing for superluminal speeds
Posted Aug19-12 at 01:29 AM by qhyperbola
Updated Mar1-13 at 06:35 PM by qhyperbola (to remove tangential rambling)
Updated Mar1-13 at 06:35 PM by qhyperbola (to remove tangential rambling)
Trying to find loopholes in the laws of physics is something alot of us amateur hacks spend an embarassing amount of effort at trying to do. For instance, trying to get around the speed of light limit either to allow for superluminal travel in space - or for time travel of some kind. In my 1983 physics book there is a section on the rocket equation - describing the motion of systems with varying mass. I was able to reckon out from the equations that for a rocket burning fuel at a steady rate (given as mass burned per second 200kg/s) and exhausting gas at a given relative velocity ie 2 km/s one can find the final velocity of said rocket from the ratio of (fuel + payload)/(payload) by a logarithmic equation:
(exhaust speed)ln(initial mass/final mass) = final velocity. So for a rocket at rest (with initial velocity of zero) with 4000kg payload and 16000kg fuel so total mass initially is 20000kg. Burning fuel at a rate 200kg per second to produce a jet of gas exhausting at relative velocity of 2km/s and we get final velocity from Uexln ( massinitial/ massfinal) which 2 km/s(ln 5 ) which is 3.22 km/s and it takes 80 seconds to go from 0 to 3.22 km/s assuming no other force is acting on rocket.
Interestingly the logarithmic nature of the equation does impose a limit on the final velocities obtainable For instance if we lessened the payload to 1000kg and added an extra 3000kg of fuel (for an extra 15 seconds burn time we end up with final velocity of 5.99km/s.
I reckoned it out so as to find out howmuch fuel at a given exhaust velocity would be needed to reach a target velocity: payload*evelocityfinal/uex so for a one kilogram payload to break the light barrier using conventional rocket fuel which I assumed could be made to exhaust velocities like the ones in the book's examples - one could use e^(300,000/2) which is a huge number. According to that same physics book 1052kg is the mass of the universe and e^(150,000) exceeds that number by quite a bit.- a real number that should give a real result that puts the payload's final velocity at more than the speed of light.
Does anyone else have ? I've never seen the rocket equation mentioned in relation to relativity.
(exhaust speed)ln(initial mass/final mass) = final velocity. So for a rocket at rest (with initial velocity of zero) with 4000kg payload and 16000kg fuel so total mass initially is 20000kg. Burning fuel at a rate 200kg per second to produce a jet of gas exhausting at relative velocity of 2km/s and we get final velocity from Uexln ( massinitial/ massfinal) which 2 km/s(ln 5 ) which is 3.22 km/s and it takes 80 seconds to go from 0 to 3.22 km/s assuming no other force is acting on rocket.
Interestingly the logarithmic nature of the equation does impose a limit on the final velocities obtainable For instance if we lessened the payload to 1000kg and added an extra 3000kg of fuel (for an extra 15 seconds burn time we end up with final velocity of 5.99km/s.
I reckoned it out so as to find out howmuch fuel at a given exhaust velocity would be needed to reach a target velocity: payload*evelocityfinal/uex so for a one kilogram payload to break the light barrier using conventional rocket fuel which I assumed could be made to exhaust velocities like the ones in the book's examples - one could use e^(300,000/2) which is a huge number. According to that same physics book 1052kg is the mass of the universe and e^(150,000) exceeds that number by quite a bit.- a real number that should give a real result that puts the payload's final velocity at more than the speed of light.
Does anyone else have ? I've never seen the rocket equation mentioned in relation to relativity.
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