Accelerating to the Speed of Light

[ruko]

Hypothetically, how long would it take to get to the speed of light at 1 G of acceleration? Is this simple math or does relativity some how get into the calculation?



























Einstein:
You do not really understand something unless you can explain it to your grandmother.

 

[guerom00]

It would take an infinite time
It is not possible to accelerate and reach the speed of light

 

[flatmaster]

Let's do it classically assuming no relativity.

v=vo+at
c=gt
t=c/g
t=(3*10^8 m/s) / (10 m/s^2)

t = 3*10^7 s

That's just about one Earth year.

 

[mgb_phys]

Relativity is involved, and the reason you can never reach the speed of light, is that as you go faster your mass increases by 1/sqrt(1-v^2/c^2)

You can use this with the normal acceleration formulae to work out, with a constant force, how you will accelerate.

 

[James Leighe]

If you specify a constant acceleration you 'could' accelerate to the speed of light. The only problem is that the energy (force) needed to maintain that acceleration as you near a good clip will diverge to infinity (since your mass does the same).

 

[DaveC426913]

If you specify a constant acceleration you 'could' accelerate to the speed of light.

You would need to specify from whose frame of reference that constant acceleration is measured.

 

[zeromodz]

In theory, nothing with mass can travel faster than the speed of light, but I will answer your question mathematically anyways. Assuming you start from rest and we use the acceleration of 9.8 m/s^2

Vf = Vi + AT
T = Vf - Vi / A
T = ((299792458) - (0)) / (9.8)
T = 30591067.14 seconds or 354 days.

 

[DaveC426913]

In theory, nothing with mass can travel faster than the speed of light, but I will answer your question mathematically anyways. Assuming you start from rest and we use the acceleration of 9.8 m/s^2

Vf = Vi + AT
T = Vf - Vi / A
T = ((299792458) - (0)) / (9.8)
T = 30591067.14 seconds or 354 days.

No. This is NOT correct.

It is only true at speeds well below c. Nearer c, the acceleration levels off. The correct formula is 1/sqrt(1-v^2/c^2).



Here is what you can expect:

After x years of 1g acceleration you will be traveling at a velocity of y:

years accel     velocity
    1                0.77c
    2                0.97c
    5                0.99993c
    8                0.9999998c 
   12                0.99999999996c

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

Note that, after one year, you are only doing 3/4ths of c. It's not until 2 years that you near c.

 

[pnhcafe]

Not to be a total nerd about this, but the mass doesn't increase, the momentum does. And in response to the original question, the reality of the matter is that no one is entirely sure what happens because examples of particles with mass that move at or near the speed of light are very uncommon and rarely weigh more than an electron. Most answers will reflect the theoretical opinion of the person answering. In short, it seems like the 'one Earth year' answer is what you were looking for, and I registered on this forum primarily because I thought it was a really good question.

 

[DaveC426913]

And in response to the original question, the reality of the matter is that no one is entirely sure what happens because examples of particles with mass that move at or near the speed of light are very uncommon and rarely weigh more than an electron.

Not sure where you get your information from but we are very sure what happens. SR is one of the most thoroughly-verified theories in the history of science.

 

[russ_watters]

And in response to the original question, the reality of the matter is that no one is entirely sure what happens because examples of particles with mass that move at or near the speed of light are very uncommon and rarely weigh more than an electron.

Satellites go plenty fast enough for the effects of SR to be measured.

 

[ruko]

No. This is NOT correct.

It is only true at speeds well below c. Nearer c, the acceleration levels off. The correct formula is 1/sqrt(1-v^2/c^2).


c.

One year is mathematically correct if you could somehow maintain one G throughout the significant relativistic phase of the trip.

 

[guerom00]

What does 1G even means when both 1 meter and 1 second are affected by GR effects ?

 

[DaveC426913]

One year is mathematically correct if you could somehow maintain one G throughout the significant relativistic phase of the trip.

The occupant will maintain 1g throughout the entire trip, no matter if the trip lasts 2 years or 62 years.

 

[DaveC426913]

What does 1G even means when both 1 meter and 1 second are affected by GR effects ?

Well, since the OP never mentioned we were doing this within a deep gravity well, we can ignore GR effects.