Question

Explain the principle of rocket propulsion (variable mass system).

Answer 1

The position of a rocket in the earth reference frame at time t and t +Δt is shown in Figure. The rocket of mass 'm' is moving with a velocity in the direction as shown in the figure.

In time-interval Δt, a mass Δm is ejected out with velocity v in a direction opposite to u. Therefore, mass of the rocket after time Δt becomes (m - Δm). and its velocity increases by Δv to a value of (v + Δv). The initial velocity increases by Δv to a value of (v + Δv). The initial momentum of the rocket (p_i) is mv and after time Δt is (m - Δm) (v + Δv) + vm. Δu - mv.

or, Δp = mΔv + Δm(u - v)

The term Δm. Δv is neglected due to their values. And (u - v) represents relative velocity (v_rel). The above equation can be written as dividing by Δt,

or,

As mass decreases, Δv is -ve.

The term dm/dt, v_rel represents the rate at which momentum is being transferred to the system by the mass that the system has ejected. The designer's aim is to increase this thrust or force due to reactions. It is obvious that thrust on the rocket would increase in case rocket ejects as much mass per unit time as possible and also that the speed of ejected mass relative to the rocket be as high as possible.

Thus, ma = F_ext + F_reaction

For a rocket as ascending F_ext = -mg

or, m.dv/dt = +mg - v_rel dm.dt

Negative sigh in the second term arises because v_rel is opposite to v,

⇒ dv/dt = g - v_rel/m . dm/dt

Integrating it, we have, v = v₀ - gt + v_rel log m_i/m_f

'v₀' is the initial velocity at t = 0 in far outer space where g = 0

Thus v = v_rel log m_i/m_f or, m_f/m_i = e -v/r_rel

where m_i and m_f are initial and final mass of the rocket