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Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :Richard Hackel uses 10 m/s² = 1.052 ly/y²>
We can now review the journey to Tau Ceti.
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Both Earth and Tau Ceti are considered to be inertial.
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A rocket is stationary on Earth, When its clock show τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².
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a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
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According to your equations v = a⋅t and vₘ = a⋅t/2:
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d = ∫a⋅t⋅dt + 0 ly = a⋅t²/2 => t = √(2⋅d/a)
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The proper time of the rocket when it passes Tau Ceti is:
τ = √(2⋅d/a) = 4.7764 y
Absolutely.
My typo: v = a⋅t = 5.0279 ly/yThe speed of the rocket in the terrestrial frame
when it passes Tau Ceti is: v = a⋅t = 5.2860 ly/y
Yes, Vr=5.0245cSo your "theory" is identical to NM and predicts
BUT: Vo=0.980c:-D
Observable speeds Vo is not real speeds Vr,
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