Answer:
A) 21.1 m
B) 17.3 m
C) 3.267x10^7 m/s
Explanation:
This is a case of special relativity.
Let the relative speed of astronauts ship to my ship be v.
According to my observation,
My craft is 21.1 m long, according to my observation, astronauts craft is 17.3 m long.
If we fix the reference frame as my ship, then the rest lenght of our identical crafts is 21.1 m and the relativistic lenght is 17.3 m
l' = 21.1 m
l = 17.3 m
From l = l'(1 - p^2)^0.5
Where p is c/v, and c is the speed of light
17.3 = 21.1 x (1 - p^2)^0.5
0.82 = (1 - p^2)^0.5
Square both sides
0.67 = 1 - p^2
P^2 = 0.33
P = 0.1089
Revall p = v/c
v/c = 0.1089
But c = speed of light = 3x10^8 m/s
Therefore,
v = 3x10^8 x 0.1089 = 3.267x10^7 m/s
Following are the response to the given points:
a) Its ship travels to the distance of [tex]21.1\ m[/tex]
b) The astronaut's craft would be at a range of [tex]17.3\ m[/tex]
c) Relativity's use of length contraction:
[tex]\to L=L_0(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to \frac{L}{L_0}=(\sqrt{1-\frac{v^2}{c^2}})[/tex]
Here,
[tex]\to \frac{L}{L_0}=\frac{17.3}{21.1}=0.81[/tex]
Hence
[tex]\to 0.81=(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to 0.6561=1-\frac{v^2}{c^2}\\\\\to \frac{v^2}{c^2} =1- 0.6561\\\\\to \frac{v^2}{c^2} =0.3439\\\\\to \frac{v}{c} =0.58\\\\\to v= 0.58 c[/tex]
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