Irodov 1.179: Speed Is the Log of Mass – Amit & Deepali Sinha Foundation

Problem Irodov 1.179

A rocket travels in free space, with no external force acting on it. It ejects a steady jet of gas whose velocity $\vect u$ , measured relative to the rocket, keeps exactly the same magnitude and the same direction throughout the burn. At the initial moment the rocket is at rest and its mass is $m_0$ . Find the rocket’s velocity $\vect v$ at the later moment when its mass has fallen to $m$ .

One interval of the burn: the body has mass m and velocity v at t, mass m-dm and velocity v+d v at t+dt, the element of mass dm having left the tail at u relative to the body. Dimension arrows mark its thickness. — One interval of the burn: the body has mass $m$ and velocity $\vect v$ at $t$ , mass $m-dm$ and velocity $\vect v+d\vect v$ at $t+dt$ , the element of mass $dm$ having left the tail at $\vect u$ relative to the body. Dimension arrows mark its thickness.

Given

$\vect u$ constant relative to the rocket
initial mass $m_0$
initial velocity $\vect 0$
no external force
current mass $m$

Find

the rocket’s velocity

\vect v

at the moment its mass is

m

Solution

Step 1Close the system around the element about to leave

The rocket by itself is not a fixed collection of matter, and a momentum balance is a statement 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 $K$ in which the rocket started at rest. Nothing enters or leaves that system across the interval.

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

Step 2Refer the exhaust velocity 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 measured in $K$ . The definition of a relative velocity, applied at the moment the element separates, connects the two.

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

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 so that the orders can be compared.

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

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

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

d\vect p = m\,d\vect v + \vect u\,dm

(2)

Step 4Set the increment to zero and read off the equation of motion

No external force acts on the rocket, and none acts on the gas element either once it is counted inside the system, so the system of Step 1 is isolated and its momentum cannot change over $dt$ . Setting (2) to zero gives the differential equation the whole problem rests on.

m\,d\vect v + \vect u\,dm = \vect 0

m\,d\vect v = -\vect u\,dm

(3)

A free-body reading of the same instant. The thrust - u dm/dt, the only force, is the sky arrow below the hull; the charcoal arrows are motion, not force: the body's velocity v and the exhaust velocity u. — A free-body reading of the same instant. The thrust $-\vect u\,dm/dt$ , the only force, is the sky arrow below the hull; the charcoal arrows are motion, not force: the body’s velocity $\vect v$ and the exhaust velocity $\vect u$ .

Step 5Reduce the vector equation to a single line

A rocket’s nozzle is on the body’s own axis of symmetry, through the center of mass, so the jet it ejects exerts no torque about that center; and with no external force there is nothing else to turn the body: its axis, and with it the vector $\vect u$ in the frame $K$ , keeps a fixed direction. Writing (3) as an increment of velocity shows what each element of ejected mass does to the motion.

d\vect v = -\vect u\,\frac{dm}{m}, \qquad dm > 0

Every increment is therefore along $-\vect u$ , and the rocket started from rest, so the velocity never acquires a component off that line and the motion is rectilinear. That is what allows $\vect u$ to be carried outside the integration as a constant vector.

Step 6Integrate from the initial mass to the final mass

One bridge is needed before (3) can be integrated. Its $dm$ is the mass that leaves, always positive, while the integration runs over the mass the rocket still has, which falls; the two are the same quantity with opposite signs, since every $dm$ that departs is mass the body no longer carries. Writing $m'$ for the body’s running mass, and $dm'$ for its change,

dm = -\,dm'. \tag{4}

Substituting (4) into (3) turns the balance into an equation in $m'$ alone, with the variables already separated, velocity on one side and mass on the other.

d\vect v = \vect u\,\frac{dm'}{m'}

The burn is then integrated by taking the body from its initial state, at rest with mass $m_0$ , to the state in question, with mass $m$ , the dummy $m'$ running over the intermediate masses.

\begin{align*}\int_{\vect 0}^{\vect v} d\vect v' &= \vect u \int_{m_0}^{m} \frac{dm'}{m'} \\ \vect v &= \vect u\Bigl[\ln m'\Bigr]_{m_0}^{m} \\ &= \vect u\bigl(\ln m - \ln m_0\bigr)\end{align*}

The difference of the two logarithms is the logarithm of the mass ratio, and turning that ratio upside down changes the sign in front of it, which puts the result in the form that reads directly as a speed along $-\vect u$ , since the ratio $m_0/m$ is greater than one.

\vect v = -\vect u\ln\dfrac{m_0}{m}

(5)

Step 7Confirm the ledger: what the exhaust carries away

The system as a whole started at rest, so everything ever thrown off must carry exactly the momentum the rocket gained, with the opposite sign. Each element was shed when the body’s mass was some $m'$ , so its own velocity in $K$ is the boxed answer evaluated at $m'$ plus $\vect u$ , and its mass is the $dm$ of (4), namely $-dm'$ ; summing over the burn gives the exhaust’s total momentum.

\begin{align*}\vect p_{\mathrm{gas}} &= \int_{m_0}^{m}\bigl(\vect v(m')+\vect u\bigr)(-dm') \\ &= \int_{m_0}^{m}\left(\vect u\ln\frac{m'}{m_0}+\vect u\right)(-dm') \\ &= -\vect u\int_{m_0}^{m}\left(1+\ln\frac{m'}{m_0}\right)dm'\end{align*}

The antiderivative of that integrand is the following, as differentiating the right-hand side returns the bracket.

\int\left(1+\ln\frac{m'}{m_0}\right)dm' = m'\ln\frac{m'}{m_0}

Evaluating it between the limits, the lower one contributing nothing because the logarithm vanishes at $m' = m_0$ , leaves the exhaust’s momentum in the same form as the rocket’s own.

\vect p_{\mathrm{gas}} = -\vect u\left[m'\ln\frac{m'}{m_0}\right]_{m_0}^{m} = -\vect u\,m\ln\frac{m}{m_0}

Comparing this with $m$ times the boxed velocity shows the two momenta cancel, which is the isolated system’s own statement about itself.

\vect p_{\mathrm{gas}} = -m\vect v

(6)

The boxed answer plotted: the speed ratio v/u against the mass ratio m_0/m. The logarithm is zero at a mass ratio of one; the point P marks the ratio e, where the rocket's speed equals its exhaust speed. — The boxed answer plotted: the speed ratio $v/u$ against the mass ratio $m_0/m$ . The logarithm is zero at a mass ratio of one; the point $P$ marks the ratio $e$ , where the rocket’s speed equals its exhaust speed.

Check

The logarithm’s argument is a ratio of two masses and so is a pure number, which leaves the boxed $\vect v$ carrying exactly the units of $\vect u$ , a velocity. At the initial moment $m$ equals $m_0$ , the ratio is $1$ , the logarithm vanishes and $\vect v$ is zero, which is the stated starting condition. For any later moment $m$ is smaller than $m_0$ , so $\ln(m_0/m)$ is positive and the boxed velocity points along $-\vect u$ : the rocket goes the way its exhaust does not, as the free body figure requires. And to reach the speed $u$ itself the mass ratio $m_0/m$ must climb to $e$ , about $2.72$ , the point $P$ on the plot, so a rocket must throw away about $63$ percent of its initial mass merely to match its own exhaust speed.

Remark

Neither the discharge rate nor the duration of the burn survives into the answer, and the derivation never needed the jet to be steady: all it asked was that every element leave with the same relative velocity $\vect u$ . The balance (3) ties $d\vect v$ to $dm$ and never to $dt$ , so the same propellant thrown off in a second and in a year buys the same final speed. What restores the clock is an external force. Let one act, and that same balance gains a term $\vect F\,dt$ whose integral depends on how long the mass takes to leave, which is why a real launch loses speed by burning slowly: a rocket that hovers on its jet consumes propellant and gains nothing, and the shortfall against the boxed result is the gravity loss of the ascent.