Assume 4.37 light years away. Obviously the fastest possible trip would be for light and would take 8.74 years round trip. But I’m more curious on two things:
How much time passes on earth while the human (below) makes the trip?
How much time passes for that human since they are travelling so fast?
So, far in the future, mankind has a spaceship with an unprecedented amount of power. The spaceship accelerates at the earths gravitational rate. It does this until it reaches the halfway mark, then the ship rotates 180º and it accelerates in the opposite direction at earths gravitational rate. When it makes a stop, it’s at Alpha Centauri (or it’s close enough!).
The answer to the first one is 8.74 years.
The answer to the second, if they are truly moving at the speed of light, no time will pass at all (but they wouldn’t be able to do that). Since they aren’t moving at the speed of light, since they can’t, it would be necessary to know how close they are to doing so.
Not sure what you’re asking for in the next part (how long the trip will take? How fast they’ll end up going?)
Except the question isn’t, “How long would it seem to take if you could go at the speed of light”, or “How long would it seem to take if you could go at 99% of the speed of light”, but rather: “How long would it seem to take (from Earth’s reference point and from the ship’s reference point) if you could accelerate at 9.8 m/s^2 halfway there, flip and decellerate at 9.8 m/s^2, flip and do the whole thing over again to get back?”
To figure this out, you just need to calculate the observed times for one of the quarters of the trip, and multiply by four. Some math person will have to set up the equation for me, I used to be able to do that kind of thing…
Basic Newton Physics: (with rounding to simple numbers)
c=186,000miles/sec = 3x10^8 m/sec
1 yr = 3.2x10^7 sec
1 ly = (3.2x3) x 10^15 m = 9.6x10^15 m
a=g=10 m/sec
4.3ly = (4.3x9.6)x10^15m = 4.1x10^16 m
Half-way - 2.1ly = 2.0x10^16 m
Time to halfway - d=(1/2)at^2
2.0x10^16=(0.5)(10)t^2
t=sqrt[(2.0/5.0) 10^16] = 6.3x10^7 sec or about 729 days.
speed at turn-over
v=at= (10)(6.3x10^7)=6.3x10^8m/sec
Darn!!
This is twice the speed of light, so I would have to look up the relativistic formula instead.
If it had been at about 1/10 the speed of light or less, a Newtonian calculation would be fairly close.
According to it, at 1g accel/decel it will take 3.6 years for you to get to the star 4.3 light years away. On Earth, about 5.2 years will pass, if I calculated it correctly.
If you could very quickly jump to say 99.9% of the speed of light and do the same thing in reverse at the other end, it would only take a couple of months. If you could make the trip at 99.99% of light speed it would only take about 4 weeks ship time. Of course do to that requires an incredible source of energy (not to mention power) and even if you had that there would still be the crush gee forces from the acceleration so you would need one of those Star Trek inertial damper thingamajigs.
The formulae in the link you’ve provided are correct, but you’ve misapplied them. I used the same formulae (thoguh I already knew what sort of order of magnitudes to expect).
Using the Newtonian formula I hit the speed of light fairly quickly. I did the same math as MD2000 and realized I’d need to use relativity with it which is when I gave up.
I know from one of Cecil’s own columns that for a person travelling at speed u who adds speed v to their total, the final output is actually (u+v)/(1+(uv/c^2)). But that really doesn’t help me take time dilation into account. Also, that’s a distinct change, and in reality I’m always accelerating in the OP. (Sort of the "interest compounded monthly vs. interest compounded continuously [pert]).
@Terr
They’re wrong because if they were travelling at the speed of light to begin with and stopped instantly it would seem to take 4.3 years from earth.
I used exactly the same formula as Terr did from John Baez’s site.
Like I said though I knew what to expect as I’ve doen the calculation before and alsoI believe the figures are presented (but the calculations not shown) in A First Course in General Relatvity by Bernard Schutz for either this exact problem (or a very simlair one).
In case someone does find an error in my calculations, I have to say I haven’t chekced them, but I’m reasonably confident they’re correct as they’re the same as what I remember they should be.
The link says that the trip to 4.3 light years away, with 1 g accel/decel both ways will take 3.6 years. Did they misapply the formulas that they presented as well?