Irodov 1.178: The System Must Include the Exhaust – Amit & Deepali Sinha Foundation

Problem Irodov 1.178

A rocket flies under the action of a total external force $\vect F$ while its engine ejects a steady jet of gas. The gas leaves with velocity $\vect u$ measured relative to the rocket, and the mass discharged per unit time is $\mu$ , in kilograms per second. Let $m$ be the mass of the rocket at the instant considered and $\vect w$ its acceleration at that same instant. Show that the rocket’s equation of motion is $m\vect w = \vect F - \mu\vect u$ .

The rocket at the two ends of one interval dt, to one scale: mass m and velocity v at t, mass m-μ dt and velocity v+d v at t+dt, the element μ dt gone at u relative to the body. — The rocket at the two ends of one interval $dt$ , to one scale: mass $m$ and velocity $\vect v$ at $t$ , mass $m-\mu\,dt$ and velocity $\vect v+d\vect v$ at $t+dt$ , the element $\mu\,dt$ gone at $\vect u$ relative to the body.

Given

$\vect u$ (gas velocity relative to the rocket)
$\mu$ in kg/s
$m$
$\vect F$

Find

the equation of motion, that is, a demonstration of

m\vect w = \vect F - \mu\vect u

Solution

Step 1Close the system around the element about to leave

The rocket alone is not a fixed collection of matter, and the impulse-momentum theorem speaks only about a fixed one. So take as the system the rocket of mass $m$ at the instant $t$ together with the gas element it is about to eject during the following $dt$ , and refer every velocity to one inertial frame $K$ . That system loses nothing across $dt$ .

m_{\mathrm{sys}}(t) = m, \qquad m_{\mathrm{sys}}(t+dt) = (m-\mu\,dt) + \mu\,dt = m

Step 2Refer the relative velocity

\vect u

to the inertial frame

The datum $\vect u$ is a velocity relative to the rocket, while the momenta of the next step must all be taken in the inertial frame $K$ . The two are connected by the definition of a relative velocity, applied at the moment the element separates.

\begin{align*} \vect u &= \vect v_{\mathrm{gas}} - \vect v_{\mathrm{rocket}} \\ \vect v_{\mathrm{gas}} &= \vect v + \vect u \tag{1}\end{align*}

The rocket’s own velocity drifts from $\vect v$ to $\vect v+d\vect v$ across the interval, so writing $\vect v$ rather than $\vect v+d\vect v$ on the right misplaces a quantity of order $d\vect v$ in a term whose mass factor is already $\mu\,dt$ .

Step 3The momentum increment of the closed system

Add the two momenta at each end of the interval, the body’s and the element’s, the latter through (1), and expand the products.

\begin{align*}\vect p(t) &= m\vect v \\ \vect p(t+dt) &= (m-\mu\,dt)(\vect v+d\vect v) + \mu\,dt\,(\vect v+\vect u) \\ &= m\vect v + m\,d\vect v - \mu\,dt\,\vect v - \mu\,dt\,d\vect v + \mu\,dt\,\vect v + \mu\,dt\,\vect u \\ &= m\vect v + m\,d\vect v + \mu\,\vect u\,dt - \mu\,d\vect v\,dt\end{align*}

Subtracting the first line from the last gives the increment, whose final term carries two differentials and is therefore negligible beside the first-order ones.

d\vect p = \vect p(t+dt) - \vect p(t) = m\,d\vect v + \mu\,\vect u\,dt - \mu\,d\vect v\,dt

d\vect p = m\,d\vect v + \mu\,\vect u\,dt

(2)

Step 4Equate the increment to the impulse of the external force

The system of Step 1 is closed, so the only agency that changes its momentum over $dt$ is the external force, and its impulse is what (2) must equal.

d\vect p = \vect F\,dt

Reading the two expressions for $d\vect p$ against each other gives the equation of motion in its differential form, which is Meshchersky’s equation.

m\,d\vect v + \mu\,\vect u\,dt = \vect F\,dt

(3)

The same instant as a free body. F acts at the center of mass; the only other force is the jet's reaction -μ u, a forward thrust for the rearward jet drawn. The charcoal v and u are motion, not force. — The same instant as a free body. $\vect F$ acts at the center of mass; the only other force is the jet’s reaction $-\mu\vect u$ , a forward thrust for the rearward jet drawn. The charcoal $\vect v$ and $\vect u$ are motion, not force.

Step 5Divide by the interval to reach the equation of motion

Dividing (3) through by $dt$ turns the two impulses into forces, the two the figure shows.

m\,\frac{d\vect v}{dt} + \mu\vect u = \vect F

The rocket’s acceleration is $\vect w = d\vect v/dt$ , and carrying the jet term to the right leaves the announced form.

m\vect w = \vect F - \mu\vect u

(4)

Check

Dimensions first: $\mu$ is a mass per time and $\vect u$ a velocity, so $\mu u$ has the units $(\mathrm{kg/s})(\mathrm{m/s}) = \mathrm{kg\,m/s^2}$ , a force, matching $m\vect w$ and $\vect F$ term for term. Now the case $\vect u = \vect 0$ , matter shed with no velocity relative to the body, such as sand dribbling from a hole in a moving wagon: the boxed equation collapses to $m\vect w = \vect F$ , so a body may shed mass at any rate $\mu$ and get no push at all out of the shedding. The thrust comes from the relative velocity given to the ejecta, never from the loss of mass by itself.

Remark

The tempting alternative, that the external force equals the time derivative of the rocket’s own momentum $m\vect v$ , is not merely a different bookkeeping of the same content: it is not even frame independent. Under a change to another inertial frame $\vect v$ shifts by the constant boost while $m$ , $\mu$ and $\vect w$ do not, so a law written with $\vect v$ in it would assert something different in every frame whenever $\mu \neq 0$ . The boxed equation escapes that because $\vect u$ is a relative velocity and is the same in all inertial frames, which is the deeper reason the jet’s contribution had to be written with $\vect u$ and not with the exhaust’s velocity in the ground frame. The rate of change of the rocket’s own momentum is not $\vect F$ ; it falls short of it by the momentum the gas carries away each second.