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.

 

[kyle8921]

In special relativity, the concept of length contraction is introduced. This means that, as an object moves at high speeds, its length in the direction of motion appears shorter to an observer. In this case, the astronaut on the rocket is moving at a speed of 0.8c relative to the Earth, so the distance between the Earth and the Moon will appear shorter to them.

The formula for length contraction is given by L' = L/\gamma, where L is the length measured by the stationary observer and L' is the length measured by the moving observer. In this case, the astronaut is the moving observer and the distance between the Earth and the Moon (384000km) is the length measured by the stationary observer.

Using the Lorentz factor \gamma = (1 - v^2/c^2)^(-1/2), we can calculate the distance measured by the astronaut:

L' = L/\gamma = 384000km/(0.6)^(-1/2) = 320000km

So, according to the astronaut on the rocket, the distance between the Earth and the Moon is 320000km. This is shorter than the distance measured by the stationary observer on Earth, due to the effects of length contraction.