As you pilot your space utility vehicle at a constant speed toward the moon, a r

Question

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.840 c , relative to you. At the instant thespace-racer passes you, both of you start timers at zero.

Part A

At the instant when you measure that the spaceracer has traveled 1.25×108 m past you, what does the race pilot read on her timer?

Part B

When the race pilot reads the value calculated in part A on her timer, what does she measure to be your distance from her?

Part C

At the instant when the race pilot reads the value calculated in part A on her timer, what do you read on yours?

Explanation / Answer

A) In her frame she has traveled on a shorter path than the actual path distance. Therefore divide by gamma.

Gamma = 1/sqrt[1 – (v/c)^2] = 1.84

Same as just looking at the equation L/ = 1.25*10^8/ (1.84*0.84*3*10^8) = 0.27 seconds

B) From part "a" her measured length has to be smaller than the original. It can't be greater nor the same (so it cant be 1.25 x 10^8 meter). You have to remember that the length traveled in the moving frame will be smaller. So you should be dividing 1.25*10^8/1.84 = 0.68*10^8 m

C) For time contraction think of twin paradox. You know that the person traveling in the spaceship ages slower which means less times passes by for them. So the person going at 0.84c had better be reading shorter times on there watch. In your reference frame that person should be taking a long time to return (meaning that if she were your twin you would be older when she returned) so you watch had better have a longer time read.

So, time = 0.27/1.84 = 0.147 sec

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