Rocket Propulsion

In a rocket, the fuel burns and produces gases at high temperatures. These gases are ejected out of the rocket from a nozzle at backside of the rocket. The ejecting gas exerts a forward force on the rocket which helps it in accelerating.

Suppose, a rocket together with its fuel has a mass M₀ at t = 0. Let the gas be ejected at a constant rate r = - dM/dt. Also suppose, the gas is ejected at a constant rate r = - dM/dt. Also suppose, the gas is ejected at a constant velocity with respect to the rocket.

At time t, the mass of the rocket together with the remaining fuel is

M = M₀ - rt.

If the velocity of the rocket at time t is v, the linear momentum of this mass M is

P = Mv (M₀ - rt) v … 1

Consider a small time interval  A mass Δm = rΔt of the gas is ejected in this time and the velocity of the rocket becomes v + Δv. The velocity of the gas with respect to ground is

= - u + v

In the forward direction,

The linear momentum of the mass M at t + Δt is,

(M - ΔM) (v + Δv) + ΔM (v - u) … (2)

Assuming no external force on the rocket fuel system from (1) & (2),

(M - ΔM) (v + Δv) + ΔM (v - u) = Mv

Or,

(M - ΔM) (Δv) = (ΔM) u

Or,

Δv = (ΔM) u / M – ΔM

Or,

Δv/Δt = (ΔM/Δt) (u/M - ΔM) = ru/M – r Δt

Taking the limit as Δt →0,

dv/dt = ru/M = ru/M₀ - rt

This gives the acceleration of the rocket. We see that the acceleration keeps on increasing as time passes. If the rocket starts at t = 0 and we neglect any external force such as gravity,

∫^ᵛ_₀ dv = ru ∫^ᵗ_₀ dt/M₀ - rt

Or,

v = ru (- 1/r) ln (M₀ - rt)/M₀

Or,

v = u ln (M₀ - rt)/M₀