Length contraction and time dilation for relativistic spacecraft

[philadelphia]

I have two problems that I got confused with and I am not sure if I am doing right

Homework Statement

1) The starship ENTERPRISE 1200m long when at rest. How long would it seem to be to someone watching the ship pass by them at 0.75c?

Homework Equations

length contraction
$$\Delta$$ $$t{}$$0 = Find $$\gamma$$ $$\Delta$$t' ?

The Attempt at a Solution

1200/0.75c

Homework Statement

2) A spacecraft 500m long is traveling at 60% of the speed of light (3 x 10^8 m/s)
a) Find $$\gamma$$

b) If 10 years go by one earth, how many years go by according to people on board the spacecraft ?

c)how long is the spacecraft according to Earth observers

Homework Equations

length contraction
$$\Delta$$ $$t{}$$0 = Find $$\gamma$$ $$\Delta$$t' ?

The Attempt at a Solution

I only tried finding a:

gamma = L(sub 0)/ L'
60% of speed of light is 180000000 = L (sub 0)
180000000 / 500 = 36000000 years?
is this correct?

 

[Vanadium 50]

Those aren't the right equations. Gamma has a specific definition and it's not speed.

You also seem to mix up length and time; this may work for these problems, but it will cause you grief in the future.

 

[Goddar]

In special relativity, the gamma factor basically gives you the ratio of one frame of reference to the other (space and time): it's particularly convenient when you express the speeds in terms of c.
So: when an object is moving at 0.75c relative to another, this ratio reduces to 4/sqrt(7), as you should verify..
- In your first problem, the length of the "moving" object will appear contracted relative to a "fixed" frame of reference (earth)
- In the second one, time is "dilatated" in the spaceship
Just use the gamma factor in both cases to convert from one frame of reference to the other.