# Question posed in a science magazine

1. Oct 17, 2005

### jimmies

Hello, recently I saw a poster with a clip from a magazine that had this situation and I was wondering if this is a completely impossible situation? Wouldn't you have to go several times the speed of light ?!

Suppose you wish to visit the red giant star Betelgeuse, which is 430 light years away, and that you want your rocket to move so fast that you age only 20 years during the round trip. How fast must the rocket travel relative to earth?

Last edited: Oct 18, 2005
2. Oct 17, 2005

### pervect

Staff Emeritus
Are you sure this isn't a homework problem?

Anyway, the short non-numerical answer is that because of time dilation, you can travel 4 light years (as measured from Earth) in less than 4 years "ship time", though it will always take you more than 4 years earth-time.

3. Oct 17, 2005

### JesseM

Of course, from your point of view on the ship, the shortness of the trip isn't because of time dilation, it's because of Lorentz contraction of the distance between you and the star, so it's less than 4 light-years away...from your point of view you get there in less than 4 years, but you never exceed the speed of light.

4. Oct 18, 2005

### mitchellmckain

So if you want to get there in 20 years then you need to contract space by the factor 20/430 at least right? So use the length contraction equation to see what velocity gives you the contraction l = (20/430) l0. This will not be an exact answer but it will get you in the ball park.

For an exact answer you need l/v = 20 years, so you need solve,
sqrt(1-v^2/c^2) 430ly / v = 20 years, and that is not hard to do. Same answer as above to 5 decimal places. But when you are talking about traveling near the speed of light, 5 digit accurracy is not saying much. For example there is a BIG difference between .999995 c and .9999999995 c. I wonder if maybe when talking about the accuracy of a velocity calculation like this maybe the 9s on the left when in this form should not count as significant. Then we would say that velocity .999725 match .999723 to only 2 significant figures. Of course the lorentz contraction factor gamma seem to me like the most sensible way of representing velocities this close to the speed of light.

Last edited: Oct 18, 2005
5. Oct 18, 2005

### jimmies

ah i see. i've never worked with such large numbers before so i didn't think that time dilation and length contraction could make such a large trip take so short, in the perspective of the space team. i'll try the quantative method once i have access to a computer or a cell phone...at the time the only calculator i was equipped with is the one built into my cell phone.
edit: just tried this out with smaller numbers. Alpha Centauri is 4.3 light years away and you want to take a 1 way trip, how fast would the ship have to go if they were to age 12.7 years?
What i did was use the length contraction like mitchellmckain said, but the answer I'm getting seems off. Thus, my equation is sqrt(1-v^2/c^2) 4.3ly / v = 12.7 years. I keep getting v = 0.33858.
(I'm using http://webs.morningside.edu/slaven/Physics/relativity/relativity10.html to practice)

6. Oct 20, 2005

### mitchellmckain

Either your algebra is wrong or you are using the wrong number for c. In the units we are using c = 1.00 ly/year. But your answer is only off in the second significant digit.

7. Oct 20, 2005

### jimmies

Ah I see, you were totally correct :). I got the same answer as it gave doing sqrt(1/((20/860)^2+1)) . Using a computer over a calculator to solve to significant digits helps too :). Usually I just used a calc and stared at the poster while waiting for the previous class to finish. Thanks a bunch!