When the spacecraft's fusion efficiency is 0.01%, the spacecraft's fusion fuel is 1t, and the total

When the spacecraft's fusion efficiency is 0.01%, the spacecraft's fusion fuel is 1t, and the total mass of the spacecraft is 1.2t, the friend bet the final speed of the spacecraft is 80% of the speed of light. The state of the fuel at nuclear fusion is assumed to be D + T and 0.01% of the maximum energy of the fusion of the fuel, although the fusion method has not yet been firmly determined. I am curious about the answer of the final speed of this space ship.
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Superina0xb4i
Restating the numbers to make sure they're right:
- After burning the fuel, the remaining mass is 200kg
- Of the 1000kg of fuel burned, 0.1kg (0.01%) is turned into kinetic energy of the remaining mass.
If so, here's how you calculate the result:
$\gamma =E/m=200.1/200=1.0005$
From $\gamma =1/\sqrt{1-{v}^{2}/{c}^{2}}$, we get ${v}^{2}/{c}^{2}=1-1/{\gamma }^{2}$
Plugging in the numbers:
$v=c\sqrt{1-1/{1.0005}^{2}}=0.032c$ or about 9,500 km/sec.
Fast, but not 0.8c. That would require $\gamma =1.67$, which in turn requires a much higher efficiency and more fuel.