Big spaceship No1 travels at +Hypersonic speed through space... Spaceship No 2 travels from back to front inside No1 at +hypersonic speed. Spaceship No3 travels from back to front inside No2 at +hypersonic speed.... and so on, until spaceship No. X reaches speed of light by adding all its host spaceships speeds plus its own.... and voila!
2007-03-29 19:26:57 · 15 answers · asked by another 2 in Science & Mathematics ➔ Astronomy & Space
Right now we have many theories explaining ,how to achieve the speed of light by an object.But unfortunately ,those are not practical.
2007-03-29 22:18:04 · answer #1 · answered by ⇐DâV£ MaΧiMiÅnO⇒ 6 · 1⤊ 2⤋
Using the field equations for Special relativity theres an interesting cancelation that occurs and shows that even by adding the two velocities (eg. in opposite directions) the final velocity acheived will not be larger of equal to the speed of light. I cannot remember the equation so cannot be any further use there.
(Yes it is just a theory and an equation that has not been fully tested but its withsstanded the test of time up till now - from which we shall see)
"So it is imposible to reach C without new unknown vast sources of energy. " - Or without a new Equation. =P
One other thing, i was wondering, even if spaceship A were going 60%c and another spaceship(B) was going 60%c in th opposite direction, you would normally - if you wasnt dealing with the speed of light - calculate that to be 120% the speed of light. OK, i was wondering, yes you would be going 1.2c in relative to say ship A, but say, ship B would still only be going at a speed of 60%c!. Just like if you were in a car and the same situation was available (on a road this time) and two cars were going at opposite directions to each other, both at say, 60 m/h, then each cars speed-dial would still say 60 m/h even though they are going 120 m/h relative to each other?
Is this correct?
2007-03-30 12:26:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If you want to travel with the speed of light and solve all the equations of relativity, you realise that your mass will become infinite and the energy needed to catch that speed would be infinite too. So it is imposible to reach C without new unknown vast sources of energy.
Another thing is that adding speed in real space isn't quite what it seems. Scientists believe that C is the fastest speed in Universe. But someone could ask that: Imagine I am travelling with my car with 50 mph and another car comes from the opposite side of the road towards me, with the same speed. At that time I am approaching the other car with 50+50=100 mph. If we do the same thing with spaceships moving with 60% of C, would that mean that they are approaching each other with a speed greater than C (60%C + 60%C = 120%C) ? The answer is no. Even in these conditions, we wouldn't calculate a speed greater than C! (even if it is against Newtonian physics)
2007-03-29 20:19:05 · answer #3 · answered by johnny206greece 1 · 1⤊ 0⤋
"Hypersonic" speed technically means 5 times the speed of sound, so you are talking maybe 4000 miles per hour or over one mile per second. You also need to add time to accelerate to speed. Then you want ship #2 to do the same thing INSIDE ship #1? And #3 INSIDE #2 inside #1. Just how long is this outermost ship?
And don't forget that light travels at 186,000 miles per second, so you'll have to nest 186,000 ships (Yes, it's additive, not geometric), and the outermost ship will have to haul ALL the inner ships and their fuel to the base speed. It will need to be a few light years long because of the time and distance it will take to coordinate the acceleration of all the inner ships.
Um, who's paying for this again?
2007-03-29 20:00:24 · answer #4 · answered by skepsis 7 · 1⤊ 0⤋
Think of it this way. A fighter plane is flying at 500 mph. It shoots a gun that has bullets that leave the barrel at 500 mph. The bullet leaves the airplane traveling at 1000 mph. Now, you are in the fighter plane going 500 mph. You turn on the landing gear lights, which is traveling at the speed of light. The light does not leave the airplane going 186,000 miles per sec plus 500 mph - the light leaves at 186,000 miles per second, no matter how fast the plane is going.
2007-03-29 20:23:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Einstein responded this question. right this is a in spite of the undeniable fact that workout: think of you're on a spacecraft in some photograph voltaic device with a huge call this is approximately to bypass great nova (explode). on the 2nd the huge call explodes, you bounce to lightspeed removed from the huge call. in case you seem lower back on the huge call, what do you spot? The seen explosion may be easy, yet you're travelling at easy velocity, so do you spot the great nova in freeze physique? Nope, you word the explosion purely as you may had you no longer jumped to easy velocity. the rationalization is by using the fact easy continually travels on the linked fee of sunshine relative to any physique of reference. This does reason unusual anomalies called time dilation, the place theory of time transformations in the process frames of connection with make specific the the excellent option habit occurs in all frames of reference. transformations in perceived distance is termed suitable length. on your occasion, the practice travelling at easy velocity might appear like being in a typical practice, different than the panorama may be flying via you.
2016-12-15 11:54:12 · answer #6 · answered by deibert 4 · 0⤊ 0⤋
Nope. It won't work because when you add relativistic velocities, 1+1 doesn't =2. It equals a little bit less. So even if you add 99% of the speed of light to another 99%of the speed of light, you still get less than 100%.
2007-03-29 22:14:55 · answer #7 · answered by zee_prime 6 · 0⤊ 0⤋
You enter the room with IQ of 150/ next another guy will enter the room with IQ of 125 , and so on and so forth . last Guy won't have an IQ equaling to all the others before him . got the hint ? either you are trying to apply sling shot theory in the most bizarre way imaginable or you really believe in your technique . By the way I ain't no scientist so don't take my word for granted . you may want to check the source , I guess you will get the most accurate answer there .
2007-03-29 20:04:02 · answer #8 · answered by Anonymous · 0⤊ 0⤋
Sure. But if you can, it's more efficient to just accelerate your smallest spaceship from zero to C without the surrounding aircraft.
Your scheme is much like multi-stage rocketry, where each stage forms a launch platform for the next.
2007-03-29 19:38:03 · answer #9 · answered by 2n2222 6 · 1⤊ 0⤋
The total speed would be the speed of the original ship.
You could throw a rock,all the molecules in the rock are traveling due to the brownian movement..
Adding them together wouldn't make the rock go faster.
2007-03-29 23:37:11 · answer #10 · answered by Billy Butthead 7 · 0⤊ 1⤋