A direct calculation would yield 9.8x3600x24x365=309052800 m/s.

However, this is faster than the speed of light. If you were to attempt this rate of acceleration for one year you would find that it is impossible as you approach the speed of light to provide sufficient energy to maintain that rate of acceleration.

Richard

First, put the inside of the spaceship in the observation reference frame, and calculate the acceleration x time = velocity.

　　v₊ = 1G x 1 year (365.25 days) = 9.8m/s² x 31,557,600 seconds = 309,264,480 m/s .

Since the observer is moving, this relative motion (v₊) is called backward motion, in contrast to the forward motion (v₋) in which the observer is stationary and the object is moving. In hierarchical relativity, the progress of time and the speed of light are covariant (the time in the observational frame of reference always changes with the speed of light), so the invariant speed of light in the observational frame of reference (c: invariant speed of light) is

　　c = 299,792,458 m/s.

Since the observer's time progress of the constant speed of light (c) in this observation reference frame and the space movement speed of the backward motion (v₊) are orthogonal, from the Pythagorean theorem, the retreat of the stationary system outside the spacecraft is The speed of light (w₊: recession speed of light) is

　　w₊ = √(c² + v₊²) ≒ 430,720,369 m/s .

This backward speed of light (w₊) is the acceleration from the constant speed of light (c), which is the way the observer's time progresses, and the backward motion (0 to unlimited) is added, so it always reaches superluminal speed (c < w₊). Become.

✕ Absolute time stationary coordinate system: 0 → v Galilean transformation

△ Boundary invariant light velocity system: 0 → v & c Lorentz transformation

〇 Bounded light velocity invariant system: c → √(c² ± v²) Optical scale conversion

Absolute stationary coordinate system (0) → Invariant light speed reference system (c) means that relative motion is nothing more than a parameter for co-varying relative time and light speed, which is (0 → v) still wrong. There is no such global inertial frame invariant to the speed of light. For example, when an observer running on the ground sees the speed of light around him become faster than the speed of light, the theory of special relativity based on Einstein's constancy of the speed of light (the Lorentz transformation does not change the speed of light, only changes the way time progresses) ) is different.

Since the ratio of backward motion (v₊) and backward light speed (w₊) seen from this moving spacecraft is the same as the ratio of forward motion (v₋) and constant light speed (c) seen from the Earth's stationary system,

　　v₋ = (v₊ / w₊)c ≒ 0.718c ≒ 215,256,034 m/s .

With respect to the initial observers at 'launch' you'd be travelling at 77% of the speed of light.

https://www.omnicalculator.com/physics/space-travel

Dear James Garry,

Thank you for introducing a meaningful page.

Please refer to the page of calculation of HR(hierarchical relativity) below.

Take that into account on your web.

ⓧkgの探査機を１Gで加減速し一番近い恒星へ行くのに、ⓨkgの燃料が必要か？｜Hyama Natural Science Research Institute｜note

https://note.com/s_hyama/n/n852bca8cc00a