The factor increase in the relativistic energy is 0.129.
The relativistic kinetic energy of the spacecraft is determined as follows;
[tex]K = (\frac{1}{\sqrt{1- \frac{v^2}{c^2} } } )mc^2[/tex]
where;
At 0.255 speed of light, the relativistic kinetic energy is calculated as follows;
[tex]K = (\frac{1}{\sqrt{1- \frac{(0.255c)^2}{c^2} } } )mc^2\\\\K = (\frac{1}{\sqrt{1- \frac{0.065c^2}{c^2} } } )mc^2\\\\K = 1.034mc^2[/tex]
When the speed doubles, v = 2(0.255c)
[tex]K = (\frac{1}{\sqrt{1- \frac{(2 \times 0.255c)^2}{c^2} } } )mc^2\\\\K = (\frac{1}{\sqrt{1- \frac{0.261c^2}{c^2} } } )mc^2\\\\K = 1.163mc^2[/tex]
ΔK = 1.163mc² - 1.034mc²
ΔK = 0.129mc²
Thus, the factor increase in the relativistic energy is 0.129.
Learn more about relativistic energy here: https://brainly.com/question/9864983
