Irodov 1.181: Sideways Thrust Rotates Without Slowing

Problem Irodov 1.181

A spaceship of initial mass $m_0$ travels at constant velocity $\vect v_0$ with no external forces acting. A jet engine switches on, continuously ejecting gas at a speed $u$ relative to the ship, always directed perpendicular to the ship’s own instantaneous velocity. The engine runs until the ship’s mass has fallen to $m$ , then shuts off. Find the angle $\alpha$ through which the ship’s velocity direction has turned by the time the engine shuts off.

A snapshot during the burn: the ship, current mass m, moving at v along its own axis while it ejects an element of mass dm through its flank at speed u relative to itself, square to v. — A snapshot during the burn: the ship, current mass $m$ , moving at $\vect v$ along its own axis while it ejects an element of mass $dm$ through its flank at speed $u$ relative to itself, square to $\vect v$ .

Given

$m_0$
$\vect v_0$
$u$
$m$ (mass when engine shuts off)

Find

\alpha

Solution

Step 1The differential rocket equation, no external force

Throughout, $dm$ is the mass that leaves during $dt$ , a positive quantity, and the ship’s own mass falls from $m$ to $m-dm$ . For such a body ejecting $dm$ at velocity $\vect u$ relative to itself, with no external force, the change in its own velocity is

d\vect v = -\vect u\,\dfrac{dm}{m}

(1)

Step 2The perpendicular thrust cannot change the speed

The velocity triangle. v_0 and v are drawn equal because the speed never changes, so the deviation α is the whole effect of the burn, and d v closes the triangle from the tip of v_0 to the tip of v. — The velocity triangle. $\vect v_0$ and $\vect v$ are drawn equal because the speed never changes, so the deviation $\alpha$ is the whole effect of the burn, and $d\vect v$ closes the triangle from the tip of $\vect v_0$ to the tip of $\vect v$ .

The engine is aimed so that its exhaust runs square to the ship’s own instantaneous velocity at every moment of the burn, which is the statement’s one non-obvious given.

\vect u\cdot\vect v = 0 \qquad \text{at every instant} \tag{2}

Every increment the engine delivers is therefore sideways: by (1), $d\vect v$ is along $-\vect u$ and so is perpendicular to $\vect v$ too, which can turn the velocity but not lengthen or shorten it. The identity that turns that picture into an equation is that a vector dotted with its own increment is half the increment of its square.

\vect v\cdot d\vect v = \frac{1}{2}\,d(\vect v\cdot\vect v) = \frac{1}{2}\,d(v^2) \tag{3}

Dotting (1) with $\vect v$ , reading the left side through (3) and the right side through (2),

\begin{align*}\frac{1}{2}\,d(v^2) &= -\bigl(\vect u\cdot\vect v\bigr)\frac{dm}{m} = 0 \\ v^2 &= \text{const}\end{align*}

The ship entered the burn at speed $v_0$ , so that constant is $v_0$ and stays $v_0$ until the engine shuts down.

v = v_0

(4)

Step 3The rotation rate from the perpendicular push

Taking the magnitude of (1), with $|\vect u|=u$ fixed and $dm>0$ , gives the angle turned per unit of mass ejected.

\begin{align*} v_0\,d\alpha &= u\,\frac{dm}{m} \\ d\alpha &= \frac{u}{v_0}\,\frac{dm}{m} \tag{5}\end{align*}

Step 4Integrating over the whole burn

One bridge is needed before (5) can be integrated. Its $dm$ is the mass that leaves, always positive, while the integration runs over the mass the ship still carries, which falls; the two are the same quantity with opposite signs. Writing $m'$ for the running mass and $dm'$ for its change,

dm = -\,dm'. \tag{6}

Substituting (6) into (5) and integrating from the initial mass $m_0$ down to the final mass $m$ accumulates the total deviation.

\begin{align*}\int_0^{\alpha} d\alpha' &= -\frac{u}{v_0}\int_{m_0}^{m}\frac{dm'}{m'} \\ \alpha &= -\frac{u}{v_0}\bigl(\ln m - \ln m_0\bigr)\end{align*}

\alpha = \dfrac{u}{v_0}\ln\dfrac{m_0}{m}

(7)

Check

The logarithm’s argument is a ratio of two masses, so it is a pure number and (7) carries the units of $u/v_0$ , which is an angle in radians as required. Since the engine only ejects mass, $m<m_0$ , so $m_0/m>1$ and the logarithm is positive: the deviation angle comes out positive, as an angle actually turned must. If $m\to m_0$ (the engine fires for a negligible time), $\alpha\to0$ , as it must when no mass has left.

Remark

Because $|\vect v|$ never changes, by (4), this maneuver is a pure rotation of the velocity direction at constant speed, exactly analogous to circular motion under a purely centripetal force, except that here the “centripetal” agent is the sideways thrust and the turning is driven by mass loss rather than by time: the natural parameter for the whole problem is $\ln(m_0/m)$ , not $t$ , which is why the elapsed time never had to enter the derivation at all.