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A rocket moving with a velocity $v$ releases a mass $\Delta m$ (fuel exhaustion) at a speed $v_{0}$ in a time interval $t$ and $t+\Delta t$, and increases its velocity by $\Delta v$.

If we apply conservation of momentum in the reference frame of the rocket, we get (in magnitudes) $$\Delta m \times v_{0} = m \times \Delta v$$

or $$\Delta v = v_{0}\frac{\Delta m}{m}.$$

In A.P. French's Newtonian Mechanics, page $328$, the author says about the above equation,

This equation is not quite exact. But as we let $\Delta t$ approach zero, the error approaches zero. As long as $\Delta v/ v_{0}$ is much smaller than unity, the equation above is an excellent approximation.

Question1: In deriving the equality, we did not impose any condition condition upon $\Delta t$, so where does this said error arise from?

Question2: We obtained the relation using conservation of momentum only; so where did the requirement $\Delta v/ v_{0} \ll 1$ come from? Is it an empirical observation?


2 Answers 2


Let's look at the conservation of momentum with respect to the ground frame (as done in French's book). $$(m+\Delta m)v = m(v+\Delta v) + \underbrace{\Delta m (v-v_0)}_{\text{not exact}} \tag{1}$$

There's a nuance here because $v_0$ is the exit velocity of the burned-up fuel particles with respect to the rocket's frame of reference. And we know that the rocket's speed is changing with time. So, the term with the underbrace in equation $(1)$ is an approximation and can't be valid when the mass $\Delta m$ is being released continuously with exit velocity $v_0$ with respect to the frame of the rocket.

The assumption : In the time interval $\Delta t$, the rocket's speed $v$ hasn't changed by a very large amount ($\Rightarrow \Delta v/v_0 << 1$). As the time interval $\Delta t$ becomes smaller and smaller, the situation becomes more and more equivalent to the case of a mass-chunk $\Delta m$ (as a whole) detaching from the rocket with speed $(v-v_0)$ at time $t+\Delta t$, for which we know equation $(1)$ is exact.

$\mathbf{\text{EDIT (Response to comments are made here)}} :$

If we apply conservation of momentum in the reference frame of the rocket, we get (in magnitudes) $$\Delta m v_0 = m \Delta v \tag{2}$$

Let's be precise here. Equation $(2)$ (which comes from momentum conservation) was calculated in an inertial frame (call it $S$) that was comoving with the rocket at time $t$ : the rocket was at rest with respect to this frame at time $t$ (and time $t$ alone). At time $t+\Delta t$, we observe the rocket to be moving at speed $\Delta v$ in this frame. This frame is different from the non-inertial frame that's attached to the rocket at all times (call it $S'$). If this distinction is clear, you will realize that it's not appropriate to call $S$ as the "rocket frame". Now, let's get to your question.

Equation $(2)$ is exactly true only when the entire $\Delta m$ is ejected continuously from the rocket with speed $v_0$ in frame $S$. But that is not the case for the same reason as I explained above (rocket's speed is changing). Fuel is ejected continuously at speed $v_0$ only in frame $S'$ that is attached to the rocket.

  • $\begingroup$ So the problem is $v$ not being constant between $t$ and $t + \Delta t$. But why did this discrepancy not appear when we were in the rocket reference frame? $\endgroup$
    – Hilbert
    Nov 1, 2019 at 10:51
  • $\begingroup$ Hey @Hilbert, check my edit. $\endgroup$
    – Ajay Mohan
    Nov 1, 2019 at 11:16

If all the exhaust is released at once ($\Delta t$ approaches zero) then all of the rearward momentum given to the exhaust ($\Delta m$) is countered by the increased velocity of the mass $m$.

But over an extended period of time, the portion that is burned first pushes back not on $m$, but on $m$ plus the not-yet-burned $\Delta m$. The extra mass means the velocity increase is not as large.


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