Rocket Motion

The motion of a rocket is caused by the reaction force (or thrust) produced by the ejection of combustible fuel; this fuel forms a part of the rocket until it is ejected out. Hence, the motion of rocket is equivalent to motion of a system of variable mass, the thrust being produced by ejecting a part of the mass of the system.

Suppose a system (e.g. a rocket) has mass m and velocity v at some instant of time t, as seen from an inertial frame S (say, earth). In next interval of time Δt, suppose it ejects a mass Δm, which moves with velocity u relative to frame S. The parent system, now with mass (m – Δm) moves with velocity (v + Δv) relative to frame S.
 
In frame S, the initial momentum of the total system is,

P_i = mv.

and final momentum, P_f = (m – Δm) (v + Δv) + (Δm) u

where we have taken the direction of u along v (for v + Δv). According to Newton’s law of motion, in an inertial frame, the rate of change of momentum of a system of change of momentum of a system of fixed (constant) mass is equal to net external force acting on the system; hence we find,

where u_rel is the velocity of the parent system (rocket) relative to ejected mass, in the direction of v.

If the mass of the system increases, we replace dm/dt by – dm/dt. Then, we write



Remember, for decrease in mass dm/dt is a –ve number, while for increase in mass, it is a +ve number. The above equation is rewritten as,



where, F_reac = u_rel dm/dt is the force of reaction or thrust applied by ejected (or added) mass on the parent system.

Consider the motion of a rocket going vertically upwards in earth’s field of gravity, so that

F_ext = mg

Let us assume g to be a constant in the range of motion under consideration. The rocket is also assumed to eject gases with a constant velocity with respect to itself, i.e. u_rel is constant. Hence, for the rectilinear (vertical) motion, we have


Integrating above equation, we find



Constant of integration C₁ is fixed by initial conditions: at t = 0, m (t = 0) = m₀ = m_S + m_F (mass of space-ship + fuel) and let v (t = 0) = v₀. Hence, C₁ = v₀ + u_rel In m₀, and we get



The above equation can further be integrated if we know the variation of mass with time. Assuming that the fuel is ejected at constant rate , we have

dm/dt = 1 –

Or, m (t) = m0 – t

The fuel is ejected for a finite time till all of it is finished. That is, at t = τ, m (t) = ms, mass of space-ship. Hence, we have

m_S = (m_S + m_F) – τ



That is, m (t) = mS + mF (1 – t/ τ)     for t ≤ τ

Substituting m (t), we get

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