As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 m. (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of 4.00×107m/s relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

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Manetho Manetho · Answer 1 · 2020-02-12T07:23:46+01:00

Answer:

a) L= 3568 m

b) 9×10^{-5} sec

c) 8.92×10^{-5} sec

Explanation:

runway length on the Earth L_o = 3600 m

Speed of the spacecraft flying past v= 4.00×10^7 m/s

then Einstein's length contraction theory,

[tex]L =L_o \sqrt{1- \frac{v^2}{c^2} }[/tex]

[tex]L =3600\sqrt{1- \frac{(4\times10^7)^2}{(3\times10^8)^2}[/tex]

Solving we get L= 3568 m

b) Time interval for an observer on earth when the spacecraft is directly over one end of the runway until it is directly over the other end.

L_o= 3600 m

speed v= 4.00×10^7 m/s

Then time interval Δt = L_o/v= [tex]\frac{3600}{4\times10^7}[/tex]= 9×10^{-5} sec

c) for the pilot the time interval will be

Δt_o = L/v = [tex]\frac{3568}{4\times10^7}[/tex] = 8.92×10^{-5} sec