Faster than the speed of light

[Zigzagcom]

How long does it take to reach the speed of light at 1G acceleration?

The equation for acceleration is a = (v-u) divided by t.

a = acceleration = g = 9.8 m/s/s
v = final velocity, which in this case is the speed of light=300,000,000 m/s
u = initial velocity=0
t = time


9.8= (300000000 - 0) / 't'
9.8 X 't'= 300000000
't'= 300000000 / 9.8
time to reach the speed of light at 1G = 30612244.9 seconds or about 353 days...

I have been experiencing 1G acceleration all my life, well over 15,000 days...so I feel I am entitled to the distinction of being in a state faster than the speed of light. I don't know about you folks.

So, maybe someone can enlighten me, please.

 

[CompuChip]

Welcome to PF.

First of all, you are not constantly being accelerated at 1 g. Normally, when standing on the ground, the ground applies a force upwards (normal force) which prevents you from falling through it.

Secondly, which is your greatest mistake, you are applying non-relativistic ("Newtonian") mechanics to relativistic situations. If you want to speak about velocities near the speed of light, you should use the equations from the theory of special relativity. They will tell you that the closer you get to the speed of light, the more energy you need to pump into reach the same increase in velocity. When the velocity reaches the speed of light, this energy becomes infinite, i.e. you cannot actually attain the speed of light.

 

[sylas]

How long does it take to reach the speed of light at 1G acceleration?

The equation for acceleration is a = (v-u) divided by t.

a = acceleration = g = 9.8 m/s/s
v = final velocity, which in this case is the speed of light=300,000,000 m/s
u = initial velocity=0
t = time9.8= (300000000 - 0) / 't'
9.8 X 't'= 300000000
't'= 300000000 / 9.8
time to reach the speed of light at 1G = 30612244.9 seconds or about 353 days...

I have been experiencing 1G acceleration all my life, well over 15,000 days...so I feel I am entitled to the distinction of being in a state faster than the speed of light. I don't know about you folks.

So, maybe someone can enlighten me, please.

Bear in mind that the acceleration you experience is different from the rate of change of velocity to an external observer. If you actually experience 1G acceleration (9.8 m/s² for 15,000 days (1.296*10⁹ seconds) then your velocity relative to your initial starting point will be close to the speed of light, but just a bit less. Specifically:

• You'll be fast. You will be about 99.99999999999999999999999999% of the speed of light. (I calculated this using logs to get the right number of digits).
• You'll off the planet. (About 1.2 billion billion light years away).
• You'll be behind the times. (In your 15,000 days, the rest of humanity has advanced some 1.2 billion billion years.

(Someone check my numbers...)

Cheers -- sylas

 

[BAnders1]

An object with mass can never reach the speed of light, so at 1g acceleration, you will only get CLOSER to the speed of light (never really reaching it).

As long as you have been on Earth, you have felt a downward acceleration of 1g. Unless you have been free falling for 15,000 days, however, your velocity has not increased due to the 1g acceleration. The reason this acceleration has not increased your velocity to the speed of light is due to the surface of the Earth "pushing" you upward at 1g (exactly canceling out your net acceleration).

The theory of Special Relativity describes speeds near the speed of light, and it is found that at higher speeds, "normal" physics is much different. The speed of light is that same in ALL inertial reference frames.

If you are wondering why nothing with mass can travel faster or equal to the speed of light, then I can only tell you that the reason for this is because "that's just how nature is." Special relativity leads to some very interesting things, and it isn't too hard to understand as long as you have a good means of learning the theory.

 

[Zigzagcom]

Thanks for the explanations. So being at rest (yet experiencing a 1G acceleration) is not the same as being in a rocket and accelerating at 1G...
I would not know in which of the two frames of reference I am in, until I run out of fuel in the rocket, that much I understand, and also the E=m*C^2 part. Weird stuff...

 

[CompuChip]

Thanks for the explanations. So being at rest (yet experiencing a 1G acceleration) is not the same as being in a rocket and accelerating at 1G...

Yes. In the first case, there is no net force. The gravitational force is canceled by the normal force of whatever it is you're standing on. There is no net force acting on you, hence there is no acceleration by Newton's second law (even in special relativity, the functional form F ~ a remains valid). The problem is caused by the colloquial "experiencing an acceleration" by which we (you) mean that if you were to temporarily cancel the normal force, for example by dropping a ball or jumping in the air, the acceleration is always the same and equal to 1g.

I would not know in which of the two frames of reference I am in, until I run out of fuel in the rocket, that much I understand, and also the E=m*C^2 part. Weird stuff...

Yes. In special relativity there is no conceivable experiment where you can decide between two observers moving at constant relative velocities. As soon as accelerations come into play, special relativity is no longer sufficient (just like ordinary mechanics is no longer sufficient at high speeds) and you need general relativity, which is again a little different (and considerably more complicated to conceptually and mathematically grasp). If your rocket is decelerating because it runs out of fuel (and there is some sort of friction) or you are near a body - such as a planet or star - which produces a non-negligible gravitational field, you can unambiguously identify the accelerating and non-accelerating observers.

 

[coverme]

How long does it take to reach the speed of light at 1G acceleration?

The equation for acceleration is a = (v-u) divided by t.

a = acceleration = g = 9.8 m/s/s
v = final velocity, which in this case is the speed of light=300,000,000 m/s
u = initial velocity=0
t = time


9.8= (300000000 - 0) / 't'
9.8 X 't'= 300000000
't'= 300000000 / 9.8
time to reach the speed of light at 1G = 30612244.9 seconds or about 353 days...

I have been experiencing 1G acceleration all my life, well over 15,000 days...so I feel I am entitled to the distinction of being in a state faster than the speed of light. I don't know about you folks.

So, maybe someone can enlighten me, please.

The Earth surface provides you a reaction force which cancels the effect of Gravitational acceleration,making net effect zero,

 

[QuantumPion]

Bear in mind that the acceleration you experience is different from the rate of change of velocity to an external observer. If you actually experience 1G acceleration (9.8 m/s² for 15,000 days (1.296*10⁹ seconds) then your velocity relative to your initial starting point will be close to the speed of light, but just a bit less. Specifically:

• You'll be fast. You will be about 99.99999999999999999999999999% of the speed of light. (I calculated this using logs to get the right number of digits).
• You'll off the planet. (About 1.2 billion billion light years away).
• You'll be behind the times. (In your 15,000 days, the rest of humanity has advanced some 1.2 billion billion years.

(Someone check my numbers...)

Cheers -- sylas

Also I'd imagine the world would be run by apes by that point.

 

[HallsofIvy]

You mean it isn't already?

To emphasize what others have said: you have not been "experiencing 1G acceleration all my life". You have been experiencing the gravitational force which would result in 1G acceleration if it were not offset by other forces: the floor acting on your shoes, the earth, the chair on your backside.

 

[Sloodin]

Thanks for the solution of the question:
"How long does it take to reach the speed of light at 1G acceleration?"
Now, here is the follow-up to that question...
How much time will elapse on Earth as this space traveler speeds away at this rate to the speed of light? Assuming that the time elapsed is just that time period until the speed of light is 'almost' reached?

TRhanks You

 

[jtbell]

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

[Grep]

As soon as accelerations come into play, special relativity is no longer sufficient (just like ordinary mechanics is no longer sufficient at high speeds) and you need general relativity, which is again a little different (and considerably more complicated to conceptually and mathematically grasp).

Actually, it's not correct to say SR can't handle accelerations. Common misconception. They talk about it in Physics FAQ here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

 

[K^2]

Yes. SR can't handle accelerated reference frames, but if you correctly identified an inertial reference frame, you can deal with objects accelerating within that frame. Note, for example that Twin Paradox gives you correct results in SR as long as you chose the system that does not accelerate.