# On two requirements in order for an equation describing the propulsion of a rocket to hold

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?

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.

• 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? Nov 1, 2019 at 10:51
• Hey @Hilbert, check my edit. 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.