Answer:
A) The time of the observer on the surface of the mars is the proper time.
B) Duration = 434.65 μs
Explanation:
A) From the question, we see that the signal light blinked on the Martian surface which is the same frame of reference of the observer at rest. Hence, we can say that the time of the observer on mars is the proper time.
B) time dilation equation is given as;
Δt =γ•Δt(o)
Where Δt is the duration of the light pulse measured by the pilot of the spaceship.
γ = 1/(√(1-β²))
Now,
β is expressed as β = u/c
Thus;
γ = 1/(√(1-(u/c)²)) = 1/(√(1-(u²/c²)))
Where, u is the speed of the spaceship relative to the surface of planet mars.
From the question, u = 0.985c
Thus, plugging in the relevant values, we obtain;
Δt = [1/(√(1-(0.985²c²/c²))] x 75
Δt = [1/(√(1-(0.985²))] x 75 = [1/(√(0.029775)]75= 5.7953 x 75 =
434.65μs