calcuating time taken for interstellar travel

[snas (Stuart)]

I have just been reading an article from a US skeptic site regarding UFO's and the constraints put on them by the laws of physics. But while the answer to the question appears, at first fairly obvious (or at least approximately so), the equation given, which I have to assume is correct, produces an answer that at first appears wildly incorrect.

The apprently obvious answer to the highlighted equation below is a bit more than 20 years. ie: once the spacecraft reaches max velocity of 0.5c, it will take 2 years to cover each light year.So how do they come up with the answer below?

"For my model I have chosen a spacecraft with a crew of six that will leave its planet for a planet in the habitable zone of a star 10 light years away. It will be accelerated at a rate of 10 m/s2 (10 meters per second squared) to a velocity of 0.5 times the velocity of light (0.5c, where c is the velocity of light). The time for it to reach this velocity is given by this equation:
t = v/a = 1.5×108/10 = 3.06×107 s = 174 days
(a = acceleration in meters/second squared; v = velocity in meters/second; s = time seconds)"

If 174 days is the correct answer, then clearly I am missing something.

Stuart

[mithrandir (Andrew)]

Originally Posted by snas

"For my model I have chosen a spacecraft with a crew of six that will leave its planet for a planet in the habitable zone of a star 10 light years away. It will be accelerated at a rate of 10 m/s2 (10 meters per second squared) to a velocity of 0.5 times the velocity of light (0.5c, where c is the velocity of light). The time for it to reach this velocity is given by this equation:
t = v/a = 1.5×108/10 = 3.06×107 s = 174 days
(a = acceleration in meters/second squared; v = velocity in meters/second; s = time seconds)"

From an external frame of reference, to get from 0 to 0.5c (~150000km/sec) with acceleration 10m/sec
150000000/10/86400 ~= 174 earth days

Did they explain their magic numbers, or take into consideration the relativistic effects, or the mass of propellant required?

Neglecting practical issues and relativity, by that time they will have covered 1.125e12 km or about 0.2 LY.
The next 9.6 LY will take 19.2 years.
Decelerating at the same 10m/sec will then take another 174 days and cover another 0.2 LY
So the total is 19.2 + 2*174/365 or about 20.15 years

[snas (Stuart)]

OOPS! Ignore me. What I didn't pick up on was my critical mistake.

The time for it to reach this velocity (0.5c) is given by this equation:
t = v/a = 1.5×108/10 = 3.06×107 s = 174 days

ie: what they were saying was that the time taken to reach v= 0.5c was 174 days, NOT the time for the entire trip

Sorry!
Stuart