The velocity of the exhaust V e → → in the observer frame is related to the velocity of the exhaust in the rocket frame v e by: v e → → = V e → → − − V → → thus, V e → → = V → → + v e → → Solving this yields: P 2 → → − − P 1 → → = m Δ Δ V → → + v e → → Δ Δ m − − Δ Δ m Δ Δ V → → If V → → and v e → → are opposite, F i → → have the same direction as V → → , Δ Δ m Δ Δ V → → are negligible (since d m d v → → → → 0 ), and using d m = − − Δ Δ m (since ejecting a positive Δ Δ m results in a decrease in rocket mass in time), ∑ ∑ i F i = m d V d t + v e d m d t