A spacecraft flies away from the earth with a speed of 4.00 106 m/s relative to

Question

A spacecraft flies away from the earth with a speed of 4.00 106 m/s relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully sychronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 730 days (2 years) later, as measured by the clock that remained on earth.

(a) What is the difference of the elapsed times on the two clocks, measured in hours?

h

(b) Which clock, the one in the spacecraft or the one on the earth, shows the smallest elapsed time? the clock on the earth the clock in the spacecraft both clocks show the same elapsed time

Explanation / Answer

A)

B = v/c, where

v = 4 x 10^6 m/s

c = 3 x 10^8 m/s

B = 0.013

So,

Y = 1/(1-B^2)^1/2 = 1.000084

The atomic clock measures the proper time TS on the spaceship, so time on Earth is dilated, i.e. 730 days = Y*TS.

So,

Y*TS - TS = 1.47 Hours

B)

The spaceship clock therefore records the smaller elapsed time.

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