A spaceship flies past Mars with a speed of 0.985c relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for 75.0μs. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?

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

the results is a decrease in air temperature with an increase of altitude.

explanation:

the answer is a)

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