- #1
pc2-brazil
- 198
- 3
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.
a body of length L=20m travels from the Earth to the Moon at a speed \vec{v}=0.8c.
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.
Lorentz's length contraction: L'=L\gamma^{-1}, where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}.
first of all:
c = 300000 km/s;
20m = 0.02 km.
substituting the variables in Lorentz's length contraction,
L'=L\sqrt{1-\frac{v^{2}}{c^{2}}}
L'=0.02\sqrt{1- \frac{(0.8c)^{2}}{c^{2}} } = 0.02\sqrt{1-(0.8)^{2}}
L'=0.02\sqrt{1-0.64}=0.02\sqrt{0.36}=0.012km
L'=12m
therefore, the apparent length of the body will be 12 meters.
are the concept and calculations right?
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 L=20m travels from the Earth to the Moon at a speed \vec{v}=0.8c.
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: L'=L\gamma^{-1}, where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}.
The Attempt at a Solution
first of all:
c = 300000 km/s;
20m = 0.02 km.
substituting the variables in Lorentz's length contraction,
L'=L\sqrt{1-\frac{v^{2}}{c^{2}}}
L'=0.02\sqrt{1- \frac{(0.8c)^{2}}{c^{2}} } = 0.02\sqrt{1-(0.8)^{2}}
L'=0.02\sqrt{1-0.64}=0.02\sqrt{0.36}=0.012km
L'=12m
therefore, the apparent length of the body will be 12 meters.
are the concept and calculations right?
