Question 20.2: Sample Exercise: A Contraction in Length A spaceship traveli...
Sample Exercise: A Contraction in Length
A spaceship traveling at the velocity of 1.8 × 10^{8} m/s (0.6c) covers a distance of 900 km as measured by an observer on the Earth.
a. What is the distance traveled in this time as measured by the pilot of the spaceship?
v = 0.6c
c = 3 × 10^{8} m/s
L_{0} = 900 km
L = ?
b. How much time does it take to cover this distance as measured by the observer on Earth and as measured by the pilot?
t = ?
t_{0} = ?From table 20.1,
| table 20.1 | |
|
Values of γ for Different Values of Relative Speed v |
|
| v = 0.01c | γ = 1.000 05 |
| v = 0.1c | γ = 1.005 |
| v = 0.5c | γ = 1.155 |
| v = 0.6c | γ = 1.250 |
| v = 0.8c | γ = 1.667 |
| v = 0.9c | γ = 2.294 |
| v = 0.99c | γ = 7.088 |
γ = 1.25
\frac{1}{γ }= \frac{1}{1.25 }\frac{1}{γ }= 0.8
L = (\frac{1}{γ }) L_{0}
= (0.8)(900 km)
= 720 km
b.
As seen by the observer on Earth: L_{0} = vt
t = \frac{L_{0}}{v}
= \frac{9 × 10^{5} m}{1.8 × 10^{8} m/s}
= 5 × 10^{-3} s = 5 ms
As seen by the spaceship pilot: L = v t_{0}
t_{0} = \frac{L}{v}
= \frac{7.2 × 10^{5} m/s}{1.8 × 10^{5} m/s}
= 4 × 10^{-3} s = 4 ms
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