## Comment Actually, 0.043 c (Score 5, Informative) 711

I believe your math is wrong. U235 releases 202.5 MeV per atom undergoing fission, so that means 1 kg can generate 83.14 TJ from fission. Assuming 100% efficiency, a massless drive, and no mass loss from propellants, that means there is enough energy from fission to reach a velocity of 0.043 c relative to the rest frame.

dE = (m - m') c^2 = m' c^2 (gamma - 1) => m' c^2 = m c^2 (1 - dE/(m c^2)) = m c^2 (1 - rho)

rho = dE/(m c^2) = 83.14 TJ / 89.88 PJ = 9.25e-4

rho = (1 - rho) (gamma - 1) => gamma = 1/(1 - rho) = 1/sqrt(1 - beta^2)

(1 - rho)^2 = 1 - beta^2 => beta^2 = rho (2 - rho) = 1.85e-3

beta = sqrt(rho (2 - rho)) = 0.0430