Basic question on length contraction.

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SUMMARY

The discussion revolves around the application of Lorentz's length contraction in special relativity, specifically for a body of length L=20m traveling at a speed of 0.8c from Earth to the Moon. The calculations confirm that the apparent length contraction observed from a stationary observer on Earth results in an apparent length of 12 meters. The formula used is L' = Lγ-1, where γ = 1/√(1 - v²/c²). The calculations were verified and deemed correct by participants in the discussion.

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good night,

this is not actually a homework question, this is just plain curiosity.
we've written a basic problem on length contraction and tried to resolve it ourselves. we just want to know if the concept of the problem is right.

Homework Statement



a body of length [tex]L=20m[/tex] travels from the Earth to the Moon at a speed [tex]\vec{v}=0.8c[/tex].
find the apparent length contraction of the body seen from an observer who is at rest in relation to the Earth.
known data: distance from the Earth to the Moon: approx. 480000 km.

Homework Equations



Lorentz's length contraction: [tex]L'=L\gamma^{-1}[/tex], where [tex]\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex].

The Attempt at a Solution



first of all:
c = 300000 km/s;
20m = 0.02 km.
substituting the variables in Lorentz's length contraction,
[tex]L'=L\sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]
[tex]L'=0.02\sqrt{1- \frac{(0.8c)^{2}}{c^{2}} } = 0.02\sqrt{1-(0.8)^{2}}[/tex]
[tex]L'=0.02\sqrt{1-0.64}=0.02\sqrt{0.36}=0.012km[/tex]
[tex]L'=12m[/tex]
therefore, the apparent length of the body will be 12 meters.

are the concept and calculations right?
 
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Looks good to me.
 
Doc Al said:
Looks good to me.

very good, thank you for the confirmation.
 

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