Liste des Groupes | Revenir à sp relativity |
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.
>
Both Earth and Tau Ceti are considered to be inertial.
>
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.
>
a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
>
According to your equations v = a⋅t and vₘ = a⋅t/2:
===================================================
>
d = ∫a⋅t⋅dt + 0 ly = a⋅t²/2 => t = √(2⋅d/a)
>
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,
Les messages affichés proviennent d'usenet.