Consider three clocks A,B and C all set at t=0 at t=0(adsbygoogle = window.adsbygoogle || []).push({});

Clock A is stationary on earth

Clock B is on a rocket at a distance of 16.43 (10^14) meters away from earth and accelerates instantaneously to .6c and arrives at earth. Clock A time = 105.6 days.

Clock C is on a rocket at a distance of 4.08 (10^14) meters from earth and at t=0 starts accelerating in the direction of earth with acceleration of 9.81m/s/s

From v= sq rt (2) (acceleration) (distance). And distance = 1/2 (acceleration) (tt). And ds/dt = (acceleration) (t) It takes the Clock C rocket 105.6 days Clock A time to reach earth that is an elapsed time on Clock A of t= 105.6 days when rocket with Clock C arrives.

Clock B which traveled , distance /rate , also arrives at Clock A time of 105.6 days. The elapsed time on Clock B is, (1/gamma )(Clock A) = .8 (105.6) = 84.5 days

So Clock A reads 105.6 days when all three clocks are back at x=0

Clock B reads 84.5 days

Just before Clock C rocket with v= 8.94 (10^7) m/s strikes earth what time does it show?

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# Elapsed time on accelerating clock

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