= Solution
The spacecraft clock measures <proper time> $\tau$, while a clock at rest with Earth measures the launch-to-emission interval $t$. Neglect Earth's subsequent motion and gravity, as in the intended <special relativity> model. The <Lorentz factor> and <time dilation> give
$$
\gamma=\frac1{\sqrt{1-v^2/c^2}}=\frac3{\sqrt8},\qquad
t=\gamma\tau=\frac3{\sqrt8}(12\,\mathrm h)=12.7279\,\mathrm h.
$$
In Earth's <inertial frame>, the spacecraft's <displacement> at emission is $d=vt$. Therefore
$$
\boxed{d=\frac{c\tau}{\sqrt8}=4.24264\ \text{light-hours}
\simeq4.58\times10^{12}\,\mathrm m.}
$$
Here $\tau=43200\,\mathrm s$ and the <speed of light> is $c=2.99792458\times10^8\,\mathrm{m\,s^{-1}}$. The distance is evaluated at the emission event, using simultaneous Earth-frame endpoints, as required by <relativity of simultaneity>; the return signal has not yet reached Earth. A distance assigned on the spacecraft's simultaneous time slice is different: Earth has moved away at $c/3$ for $12\,\mathrm h$ in that frame, giving $c\tau/3=4$ light-hours. Specifying the <reference frame> is essential to the numerical answer.
Back to article page
