- #1
BlackyTheCat
- 4
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A spaceship leaves earth. One twin stays back, the other is on the ship. The ship accelerates for 5 years with a constant acceleration (the 5 years are in the reference frame of the ship), then it decelerates for 5 years. Then, it turns around and does the same thing again. All the accelerations and decelerations are at the same rate. When he returns, he is 40 years old. How old is the twin who stayed on Earth?I know the following equations that could be relevant:
a = (1-v2/c2)3/2/(1+v*u/c2)*a'
Where a is the acceleration in the resting frame, a' in the moving frame, v is the velocity of the frame that is moving, and u is the velocity of the object moving within the moving frame.
Also, time dilatation formula.I know how to solve the normal twin paradox with time dilatation. Here, I assume that I should use the acceleration formula above somehow. However, I am at a loss at how. I do not know any velocities. Also, even if I could for example calculate a, I don't see how it would help. I could get gamma from it I guess, but for that I would already need gamma before, so it would be a bit redundant.
I see that if I solve the problem for one step (one acceleration or deceleration), I am done, since it is symmetric though.
a = (1-v2/c2)3/2/(1+v*u/c2)*a'
Where a is the acceleration in the resting frame, a' in the moving frame, v is the velocity of the frame that is moving, and u is the velocity of the object moving within the moving frame.
Also, time dilatation formula.I know how to solve the normal twin paradox with time dilatation. Here, I assume that I should use the acceleration formula above somehow. However, I am at a loss at how. I do not know any velocities. Also, even if I could for example calculate a, I don't see how it would help. I could get gamma from it I guess, but for that I would already need gamma before, so it would be a bit redundant.
I see that if I solve the problem for one step (one acceleration or deceleration), I am done, since it is symmetric though.