Irodov 1.388: The Rocket Adds Rapidity Not Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.388

A rocket moves along a straight line in free space, with no external force acting on it. It drives itself forward by throwing gas out behind it: the exhaust leaves with the velocity $\vect u$ relative to the rocket, the same at every instant, and its magnitude $u$ is small enough compared with the speed of light $c$ that the jet is non-relativistic in the rocket’s own frame. The rocket starts at rest in an inertial reference frame, with rest mass $m_0$ , and its speed $v$ is measured in that same frame throughout. Find the speed $v$ reached by the time the rocket’s rest mass has fallen to $m$ .

The rocket of rest mass m in the launch frame, moving at v while its exhaust leaves the tail at u relative to it. — The rocket of rest mass $m$ in the launch frame, moving at $\vect v$ while its exhaust leaves the tail at $\vect u$ relative to it.

Given

exhaust velocity $\vect u$ constant relative to the rocket
$u\ll c$
initial rest mass $m_0$
current rest mass $m$
$v=0$ when $m=m_0$

Find

the speed

v

as a function of

m

Solution

Step 1The recoil an element gives in the rocket’s own frame

One instant of the burn, twice, to one scale. In K the rocket has rest mass m and speed v; in the comoving K' it is at rest, sheds the rest mass dm from its tail at speed u, and recoils by d v'. — One instant of the burn, twice, to one scale. In $K$ the rocket has rest mass $m$ and speed $v$ ; in the comoving $K'$ it is at rest, sheds the rest mass $dm$ from its tail at speed $u$ , and recoils by $d\vect v'$ .

The jet speed $u$ is quoted relative to the rocket, so the one frame in which it may be used directly is the frame $K'$ that moves with the rocket at this instant. In that frame the rocket is momentarily at rest and everything moves slowly, so ordinary momentum bookkeeping is exact to the order that survives. Let the rocket’s rest mass fall from $m$ to $m-dm$ during the interval, so that $dm$ , the rest mass that leaves, is a positive quantity, and let the ejected element have rest mass $\mu$ . Balancing rest energy in $K'$ , the rocket’s recoil energy being second order and the jet’s kinetic energy smaller than $\mu c^2$ by $u^2/c^2$ , fixes $\mu$ :

\begin{aligned} mc^2 &= (m-dm)c^2+\mu c^2 \\[2pt] \mu &= dm . \end{aligned}

The total momentum in $K'$ was zero before the element left and is zero after, the element going backwards at $u$ and the rocket forwards at $dv'$ . Dropping the product $dm\,dv'$ , which is second order,

\begin{aligned} 0 &= (m-dm)\,dv'-\mu u \\[2pt] &= (m-dm)\,dv'-u\,dm \\[2pt] m\,dv' &= u\,dm . \end{aligned}

dv' = \dfrac{u\,dm}{m}

(1)

Both $u$ and $dm$ are positive, so $dv'$ is positive: the rocket recoils forward, away from its exhaust.

One velocity read in two frames. K' moves through K at V= v, and the rocket P has d v' in K', v+d v in K. The triangle deliberately does not close: the two combine relativistically, not by a Galilean sum. — One velocity read in two frames. $K'$ moves through $K$ at $\vect V=\vect v$ , and the rocket $P$ has $d\vect v'$ in $K'$ , $\vect v+d\vect v$ in $K$ . The triangle deliberately does not close: the two combine relativistically, not by a Galilean sum.

Step 2Carry the increment back to the launch frame

The increment $dv'$ of (1) is measured in $K'$ , but the question asks for the speed in the launch frame $K$ , through which $K'$ is itself moving at $v$ . The two are related by the relativistic velocity composition law for motion along one line:

v+dv = \frac{v+dv'}{1+\dfrac{v\,dv'}{c^2}} .

Expand the denominator to first order in the small quantity $dv'$ and multiply out, keeping only first-order terms:

\begin{aligned} v+dv &= \left(v+dv'\right) \left(1-\frac{v\,dv'}{c^2}+\cdots\right) \\[2pt] &= v+dv'-\frac{v^2\,dv'}{c^2}+O\!\left(dv'^2\right) . \end{aligned}

Subtracting the $v$ that the rocket already had leaves the increment the launch frame actually sees:

dv = \left(1-\dfrac{v^2}{c^2}\right)dv'

(2)

Step 3One differential equation in

v

and

m

Equations (1) and (2) between them relate the launch-frame increment $dv$ to the rest mass lost, and the variables separate at once, all the $v$ dependence on the left and all the $m$ dependence on the right:

\frac{dv}{1-\dfrac{v^2}{c^2}} = u\,\frac{dm}{m} .

One bridge is needed before this can be integrated. Its $dm$ is the rest mass that leaves, always positive, while the integration runs over the rest mass the rocket still has, which falls. Writing $m'$ for the rocket’s running rest mass and $dm'$ for its change, every $dm$ that departs is rest mass the rocket no longer carries:

dm = -\,dm' . \tag{3}

Substituting (3) and integrating from the start of the burn, where the rocket is at rest with rest mass $m_0$ , to the instant of interest, where its rest mass is $m$ :

\begin{aligned} \frac{dv}{1-\dfrac{v^2}{c^2}} &= -u\,\frac{dm'}{m'} \\[4pt] \int_0^{v}\frac{dv}{1-\dfrac{v^2}{c^2}} &= -u\int_{m_0}^{m}\frac{dm'}{m'} . \end{aligned}

The left integrand is split by the standard partial fractions, written out on its own line before it is used:

\frac{1}{1-\dfrac{v^2}{c^2}} = \frac{1}{2}\left(\frac{1}{1-\dfrac{v}{c}}+\frac{1}{1+\dfrac{v}{c}}\right).

Each piece integrates to a logarithm, and the factor $c$ comes out of the substitution $d(v/c)=dv/c$ :

\begin{aligned} \frac{c}{2}\ln\frac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}} &= -u\ln\frac{m}{m_0} \\[4pt] &= u\ln\frac{m_0}{m} . \end{aligned}

Dividing through by $c$ puts both sides in dimensionless form:

\frac{1}{2}\ln\frac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}} = \frac{u}{c}\ln\frac{m_0}{m} .

The left side is the inverse hyperbolic tangent of $v/c$ :

\operatorname{artanh}\dfrac{v}{c} = \dfrac{u}{c}\,\ln\dfrac{m_0}{m}

(4)

Step 4Solve for the speed itself

Exponentiating (4) turns the logarithm of the mass ratio into a power of it, and the exponent $2u/c$ is the one the answer carries:

\begin{aligned} \frac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}} &= \exp\!\left(\frac{2u}{c}\ln\frac{m_0}{m}\right) \\[4pt] &= \left(\frac{m_0}{m}\right)^{2u/c} . \end{aligned}

Abbreviate the reciprocal power by $X=\left(m/m_0\right)^{2u/c}$ , cross multiply and collect the terms in $v/c$ :

\begin{aligned} X\left(1+\frac{v}{c}\right) &= 1-\frac{v}{c} \\[2pt] \frac{v}{c}\left(1+X\right) &= 1-X . \end{aligned}

Dividing by $1+X$ and putting the mass ratio back in place of $X$ gives the dependence asked for:

\dfrac{v}{c} = \dfrac{1-\left(\dfrac{m}{m_0}\right)^{2u/c}} {1+\left(\dfrac{m}{m_0}\right)^{2u/c}}

(5)

The answer against the mass ratio m_0/m, for two exhaust speeds: gold u/c=0.2, royal u/c=0.1. Dashed gold is the classical v=u(m_0/m) at the same u; the two part company as the speed grows. — The answer against the mass ratio $m_0/m$ , for two exhaust speeds: gold $u/c=0.2$ , royal $u/c=0.1$ . Dashed gold is the classical $v=u\ln(m_0/m)$ at the same $u$ ; the two part company as the speed grows.

Check

Take the slow limit first. When the exponent $(u/c)\ln(m_0/m)$ is small, $\operatorname{artanh}(v/c)\approx v/c$ in (4), which returns $v=u\ln(m_0/m)$ , Tsiolkovsky’s classical rocket formula, as it must. Numerically, at $u/c=0.2$ and $m_0/m=5$ the boxed result gives $5^{0.4}=1.904$ and hence $v/c=0.904/2.904=0.311$ , against the classical $0.2\ln5=0.322$ : close, and lower, which is the gap the plotted curves show. The two ends behave too: at $m=m_0$ the numerator vanishes and $v=0$ , and as $m\to0$ the power $\left(m/m_0\right)^{2u/c}$ tends to zero and $v\to c$ , so no finite mass ratio ever reaches the speed of light. The exponent $2u/c$ is a pure number, which keeps both sides dimensionless.

Remark

Equation (4) is the result worth keeping, more than the boxed form it leads to. The quantity $\theta=\operatorname{artanh}(v/c)$ is the rapidity of the rocket in the launch frame, and what the burn does is add rapidity: each element thrown off contributes $d\theta=(u/c)\,dm/m$ , an amount that does not depend on how fast the rocket is already going, so in rapidity the relativistic rocket obeys Tsiolkovsky’s law exactly and only the map $v=c\tanh\theta$ from rapidity to speed is nonlinear. That additivity holds only because every boost here is along the one line the rocket travels; once the motion leaves a single line, successive boosts no longer add this way. Read backwards the same equation prices a target speed: the mass ratio needed grows exponentially in $c/u$ , which is what puts relativistic speeds beyond a chemical exhaust.