Time Travel: Calculating Velocity for 10 Year Trip

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The discussion focuses on calculating the velocity for a hypothetical 10-year time travel scenario, emphasizing the importance of selecting an appropriate reference frame for accurate calculations. It highlights that when all velocities are measured from the same frame of reference, the velocity addition formula is unnecessary. The conversation also touches on the implications of time dilation and how time is perceived differently depending on the observer's frame. Clarification is needed regarding the definition of "Earth years" in the context of the problem. Overall, understanding the reference frame is crucial for solving the time travel velocity calculations accurately.
jselms99
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Homework Statement
One twin leaves earth on his 25th birthday traveling at velocity .87c. After traveling away from earth for 5 Earth-years, he abruptly turns anround and travels back to Earth at the same speed. How old will his twin brother be, whom he left on earth? How old will he be?
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So at first I thought that the time would be 10 years, and that I’d have to consider the outbound motion as v = .87c and inbound motion as v = -.87c but I’m struggling with addition of the velocities and whether or not this is even the right approach?
 
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Why not choose a suitable reference frame in which to do the calculations? That would be a good start.
 
jselms99 said:
I’m struggling with addition of the velocities
If all of the velocities are measured from the same frame of reference, the velocity addition formula does not enter in. Velocity addition gets used when you need to add a velocity measured in one frame to a velocity that has been measured from another.

If a guy on a moving rocket fires a bullet from a rifle, you need the velocity addition formula to find the resulting velocity of the bullet.

If you just want to talk about a guy riding a rocket, you do not need the velocity addition formula.
 
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jselms99 said:
After traveling away from earth for 5 Earth-years
I presume this to mean 'for 5 years as measured in Earth's inertial frame'. If it doesn't mean that, then a more precise statement of the problem is needed.

I mean, by 'Earth years', I don't think they mean 'as opposed to Mars years', which are larger units of time, but that would be the literal interpretation of the quoted bit above.
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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