Now, the mass of fuel the rocket initially has on board is equal to m 0 – m f. For the constant mass flow rate R it will therefore take a time T = (m 0 – m f)/ R to burn all this fuel. Integrating both sides of the equation with respect to time from 0 to T (and noting that R = dm/dt allows a substitution on the right) obtains: Δ Δ v = v f − − v 0 = − − v e [ ln ⁡ ⁡ m f − − ln ⁡ ⁡ m 0 ] = v e ln ⁡ ⁡ ( m 0 m f ) .