# Lorentz Factor / Spacecraft to Star

1. Oct 8, 2011

### ZedCar

1. The problem statement, all variables and given/known data

A star is located 4.2 light years from Earth.

At what constant velocity must a spacecraft travel from Earth if it is to reach the star in 3.0 years time?

2. Relevant equations

I've been working on this for a couple of days, but I'm not entirely sure which equation to use.

Also, the fact that the distance is given in light years. Generally formulas require the distance input in metres (I'm in UK), so what would I input into a formula for distance? If I input 4.2 it may indicate 4.2 metres!

Thank you

2. Oct 8, 2011

### noleguy33

L = (gamma)-1*L0

Where L is the new length and L0 is the initial length. Solve for gamma and extract out the velocity.

3. Oct 8, 2011

### vela

Staff Emeritus
The idea here is that when the traveler is moving, the distance to the star is length-contracted.

You can keep the distances in light years. If you include the units when you plug the various quantities in, you'll see the units cancel out, so it doesn't matter if you convert to meters first or not.

4. Oct 9, 2011

### ZedCar

If I solve for v I get;

v = (L0 * c) / ( (c * t)^2 + L0^2) )^0.5

L0 = 4.2
c = 3 * 10^8
t = 3

I get an answer of v = 1.4

I'm not sure if this is correct or not? If it is correct, what is meant by 1.4? 1.4 what exactly?

5. Oct 9, 2011

### vela

Staff Emeritus
I think you're just plugging the numbers in wrong. You have $$\frac{v}{c} = \frac{L_0}{\sqrt{(ct)^2+L_0^2}}$$Note that the denominator is larger than the numerator, so v/c must be less than 1.

Now you have t=3.0 years, so ct is the distance light travels in 3.0 years, i.e., 3.0 lightyears. So you get$$\frac{v}{c} = \frac{4.2}{\sqrt{3.0^2+4.2^2}} = 0.81$$or v=0.81c.

6. Oct 9, 2011

### ZedCar

That's fantastic, thanks very much vela!

I actually see what I was doing wrong now.

In the denominator of the final eqn which you posted, I was inserting (3 x c)^2 where you have 3.0^2 and thus obtaining an answer of v = 1.4 due to the larger denominator.