- 21 Jul '08 05:11 / 1 editYou are on a trip to Alpha Centauri, you have the means to accelerate at 330 G's safely, but it stops short, only accelerates at that rate for 1000 seconds and the engine cuts out. You have the means to go into hibernation for millenia if need be. You make the calculations, find out its going to take a lot longer than the 5 or 6 years you first planned, so you go into hibernation.

How many years does it take to go the 4 light years (we make it exactly 4 for simplicity) to Alpha Centauri and how much different has time flowed on the ship V time flowing on Earth?

In other words, you start out on your journey in the year 2200, jan 1.

You get to AC in what year on your calander, but figured out its not the same on Earth, what year is it on Earth time? Getting it to within one month is fine. - 21 Jul '08 06:47

First step:*Originally posted by sonhouse***You are on a trip to Alpha Centauri, you have the means to accelerate at 330 G's safely, but it stops short, only accelerates at that rate for 1000 seconds and the engine cuts out. You have the means to go into hibernation for millenia if need be. You make the calculations, find out its going to take a lot longer than the 5 or 6 years you first planned, so ...[text shortened]... s not the same on Earth, what year is it on Earth time? Getting it to within one month is fine.**

d = di + vi*t +(1/2)at^2, di = vi = 0

d = (1/2)at^2

d = (1/2)330*32(ft/s^2)(1000s)^2

d = 2.38 * 10^9 ft

By the time the engines cut out, the ship has travelled 2.38*10^9 ft. - 21 Jul '08 06:50

v = vi + at*Originally posted by sonhouse***You are on a trip to Alpha Centauri, you have the means to accelerate at 330 G's safely, but it stops short, only accelerates at that rate for 1000 seconds and the engine cuts out. You have the means to go into hibernation for millenia if need be. You make the calculations, find out its going to take a lot longer than the 5 or 6 years you first planned, so ...[text shortened]... s not the same on Earth, what year is it on Earth time? Getting it to within one month is fine.**

v = 330(ft/s^2)(1000 s)

v = 3.3*10^5 ft/s

The ship after 1000 seconds is moving at 3.3*10^5 ft/s. - 21 Jul '08 06:59 / 1 editAlpha Centauri is 4 LY away. One LY is 9,460,730,472,580.8 km. I will use 9.5 * 10^12 km for simplicity. 4 * 9.5 * 10^12 is

d = 3.8 *10^13 km

d = di + vt + at^2

3.8 * 10^13 km = 2.38*10^9 ft + (3.3*10^5 ft/s)t + 0

(3.8 * 10^13 km - 2.38*10^9 ft)/(3.3*10^5 ft/s) = t

That will be the amount of time the trip takes from the perspective of the people in the ship. - 21 Jul '08 07:00 / 1 edit

t = (3.8*10^13 km - 2.38*10^9 ft)/(3.3*10^5 ft/s)*Originally posted by AThousandYoung***Alpha Centauri is 4 LY away. One LY is 9,460,730,472,580.8 km. I will use 9.5 * 10^12 km for simplicity. 4 * 9.5 * 10^12 is**

d = 3.8 *10^13 km

d = di + vt + at^2

3.8 * 10^13 km = 2.38*10^9 ft + (3.3*10^5 ft/s)t + 0

(3.8 * 10^13 km - 2.38*10^9 ft)/(3.3*10^5 ft/) = t

That will be the amount of time the trip takes from the perspective of the people in the ship.

t = (3.8*10^13 km - 2.38*10^9 ft*[km/3281 ft])/(3.3*10^5 ft/s*[km/3281 ft]

t = (3.8*10^13 km - 7.25*10^5 km)/100 km/s

We can see the distance covered while accelerating is negligable.

t = 3.8*10^11 s

That's how long the trip takes.

t = 3.8*10^11 s * (year/3.15*10^7 s)

t = 1.2*10^4 years.

It's a good thing we can hibernate! We're going to experience 12,000 years of travel! - 21 Jul '08 07:32 / 1 edit

vf = (3.3*10^5 ft/s)(m/3.3 ft)*Originally posted by AThousandYoung***v = vi + at**

v = 330(ft/s^2)(1000 s)

v = 3.3*10^5 ft/s

The ship after 1000 seconds is moving at 3.3*10^5 ft/s.

vf = 10^5 m/s

is the final velocity. The speed of light is three orders of magnitude higher. The ship is not going at relativistic speeds and there will be no appreciable time dilation.

This is not consistent with the statement

"but figured out its not the same on Earth"

Is this an SAT prep problem? - 23 Jul '08 06:02

There you go.*Originally posted by AThousandYoung***Forgot the factor of 32. Still, is that enough?**

3.3x10^5 x 32 is...1.1x10^7. Ahh...

But I also wanted to show there was a bit of time shift too, see if you can figure out what it is, relativistic effects show up even at 0.01C.

So what is the trip time now you spotted your error? - 25 Jul '08 15:45

You multiply the speed I calculated by the Lorenz factor, but I've been too lazy and busy to do that and give you the answer. Maybe later.*Originally posted by sonhouse***There you go.**

But I also wanted to show there was a bit of time shift too, see if you can figure out what it is, relativistic effects show up even at 0.01C.

So what is the trip time now you spotted your error?