A spacecraft flies away from the earth with a speed of 4.80×106m/s relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days (1 year) later, as measured by the clock that remained on earth. What is the difference in elapsed time on the two clocks, measured in hour?

1.1215 h

Explanation:

Δt=Δt_o/√(1-u^2/c^2)

(Δt° is the proper time in the rest frame,

Δt is the time intent measured in the second frame of reference u is the speed of the second frame with respect to the rest frame c is the speed of light )  

a) the proper time is the time on spacecraft (rest frame);

Δt = 365 days = 8760 h  

Δt=Δt_o/√(1-u^2/c^2)= 8761.1215 h

the difference = 8761.1215 h — 8760 h   = 1.1215 h

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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