How Does Speed Affect Space Measurements in Special Relativity?

[capslock]

A rocket travels from Earth to the moon (distance measured from the Earth 384000km) at a speed v = 0.8c.

(i) How long does it the trip take according to an observer on earth?

Just used straightforward Newtonian mechanics,

t = 384000*10^3/(0.8*3*10^8) = 1.6s.

(ii) How long does the trip take according to the astronaut?

Used the lorentz transformation,
t' = \gamma (t - u/c^2 x)
= (0.36)^(-1/2) * (1.6 - 0.8/c^2 * 384000*10^3)
= 0.96s

I believe this part is correct? Now the part I am stuck on is:

(iii) What is the earth-moon distance measured by an astronaut on the rocket?

Could someone please explain to me the principles behind the calculation? I am finding it difficult to get my head around SR!

Many thanks, James.

 

[Pengwuino]

I don't have a calculator on hand so i can't verify the first two for you. The third one has to do with a phenomenon called length contraction. As you approach the speed of light, the length measured by the person in the rocket ship of the distance between the Earth and moon becomes contracted according to the equation L' = \frac{{L & & _0 }}{\gamma } where L' is the distance measured by the person traveling in the rocket ship.