hmvince Messages 44 Reaction score 0 Apr 24, 2012 #31 which once graphed, shows that optimum time is to go as fast as possible.
Ich Science Advisor Messages 1,931 Reaction score 1 Apr 25, 2012 #32 is this correct? No. You can use Earth distance D, then t=D/v t'=t/γ. Or you use spaceship distance D', then D'=D/γ t'=D'/v t=t' γ. Either way, t'=(D/v)/γ.
is this correct? No. You can use Earth distance D, then t=D/v t'=t/γ. Or you use spaceship distance D', then D'=D/γ t'=D'/v t=t' γ. Either way, t'=(D/v)/γ.