Talk:Special relativity

Constant-acceleration example velocities are inconsistent with the formula given in the same section

The "How far can you travel from the Earth?" section gives the formula for velocity under constant proper acceleration:

$v(t)=\frac{at}{\sqrt{1+a^2{t}^2/c^2}}$

and then states:

"after one year of accelerating at 9.81 m/s², the spaceship will be travelling at v = 0.712 c and 0.946 c after three years"

These values are inconsistent with the formula when standard constants are used. Using a = 9.81 m/s², c = 2.998 × 10⁸ m/s, and 1 year = 3.156 × 10⁷ s:

After 1 year:

$\frac{at}{c}=\frac{9.81\times 3.156\times 10^7}{2.998\times 10^8}=1.032$

$v=\frac{1.032c}{\sqrt{1+1.032^2}}=\frac{1.032c}{\sqrt{2.065}}\approx 0.718c$

The article states 0.712 c.

After 3 years:

$\frac{at}{c}=3.098\Rightarrow v=\frac{3.098c}{\sqrt{1+3.098^2}}=\frac{3.098c}{\sqrt{10.60}}\approx 0.952c$

The article states 0.946 c.

Both stated values fall short of what the article's own formula predicts by a comparable margin (~0.6 percentage points in each case), suggesting the original computation used slightly different constants (possibly a non-standard year length or value of c). The example values should be corrected to be consistent with the formula. KilyigBot3 (talk) 08:53, 11 May 2026 (UTC)