Irodov 1.366: A Year at Ten Gravities – Amit & Deepali Sinha Foundation

Problem Irodov 1.366

A rocket leaves the Earth from rest and travels in a straight line. Its motor is run so that at every moment of the flight the rocket’s acceleration, measured in the inertial frame that momentarily travels with the rocket and in which the rocket is momentarily at rest, has the same value $w' = 10g$ , with $g = 9.8$ m/s $^2$ . The boost lasts $\tau = 1.0$ year as measured by clocks of the geocentric frame, which is treated as inertial throughout. Taking $c = 3.00\times 10^{8}$ m/s, find (a) the fractional difference $(c-v)/c$ , expressed as a percentage, between the speed of light and the rocket’s speed $v$ at the end of the boost, and (b) the distance $l$ the rocket has covered in the geocentric frame by that moment.

The geocentric frame K and the co-moving frame K' at one moment of the flight. The rocket P has the speed v in K; K' moves at that same speed, so the rocket is at rest there and its acceleration is w'. — The geocentric frame $K$ and the co-moving frame $K'$ at one moment of the flight. The rocket $P$ has the speed $\vect v$ in $K$ ; $K'$ moves at that same speed, so the rocket is at rest there and its acceleration is $\vect w'$ .

Given

$w' = 10g$
$g = 9.8$ m/s $^2$
$\tau = 1.0$ year $=3.156\times10^{7}$ s
$c = 3.00\times10^{8}$ m/s

Find

(a)

(c-v)/c

in per cent at

t=\tau

, (b) the distance

l

covered

Solution

Step 1Turn the co-moving acceleration into an equation in the Earth frame

The stated $w'$ is a proper acceleration, measured in a frame that changes from instant to instant, so it is useless until it is converted into a statement about $dv/dt$ in the one frame that lasts, the geocentric frame $K$ . Take the moment when $K$ measures the speed $v$ , and let $K'$ be the co-moving frame, which therefore travels at $V = v$ . In a short interval the rocket picks up a speed $du'$ in $K'$ , and the velocity composition formula converts that into the speed $K$ sees, remembering that the rocket already carries $v$ .

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

Expanding the denominator to first order in the small quantity $du'$ and keeping only first-order terms gives the increment as $K$ measures it.

\begin{aligned} v+dv &= \left(v+du'\right)\left(1-\frac{v\,du'}{c^2}\right) \\[3pt] &= v+du'\left(1-\frac{v^2}{c^2}\right) \end{aligned}

The two moments are events at the same place in $K'$ , since the rocket is momentarily at rest there, so the time between them dilates in the ordinary way. Dividing the velocity increment by the time increment gives the equation of motion in $K$ .

dt = \frac{dt'}{\sqrt{1-\dfrac{v^2}{c^2}}}, \qquad \frac{dv}{dt} = \frac{du'\left(1-\dfrac{v^2}{c^2}\right)\sqrt{1-\dfrac{v^2}{c^2}}}{dt'}

The ratio $du'/dt'$ is the acceleration measured in the co-moving frame, which is the stated constant $w'$ , and the two roots combine into a single power.

\frac{dv}{dt} = w'\left(1-\frac{v^2}{c^2}\right)^{3/2}

(1)

Step 2Integrate to the speed at any Earth time

Equation (1) separates, with all the $v$ dependence on one side and the constant $w'$ on the other. The integral on the left is a standard one, which the following identity makes visible: differentiating the combination $v/\sqrt{1-v^2/c^2}$ reproduces exactly the integrand.

\frac{d}{dv}\left(\frac{v}{\sqrt{1-\dfrac{v^2}{c^2}}}\right) = \frac{\sqrt{1-\dfrac{v^2}{c^2}}+\dfrac{v^2}{c^2}\Big/\sqrt{1-\dfrac{v^2}{c^2}}} {1-\dfrac{v^2}{c^2}} = \left(1-\frac{v^2}{c^2}\right)^{-3/2}

Integrating both sides of the equation of motion from the launch, where the rocket is at rest, up to the time $t$ therefore gives

\begin{aligned} \int_0^{v}\left(1-\frac{v_1^2}{c^2}\right)^{-3/2}dv_1 &= \int_0^{t}w'\,dt_1 \\[3pt] \frac{v}{\sqrt{1-\dfrac{v^2}{c^2}}} &= w't \end{aligned}

Squaring that relation and collecting the terms in $v^2$ solves it for the speed.

\begin{aligned} v^2 &= w'^2t^2\left(1-\frac{v^2}{c^2}\right) \\[3pt] v^2\left(1+\frac{w'^2t^2}{c^2}\right) &= w'^2t^2 \end{aligned}

v = \frac{w't}{\sqrt{1+\left(\dfrac{w't}{c}\right)^{2}}}

(2)

The boost against the dimensionless time w't/c: the gold curve is , the dotted gold line at v/c = 1 the speed of light it approaches without reaching. The marked point ends this boost, at w'τ/c = 10.3. — The boost against the dimensionless time $w't/c$ : the gold curve is (2), the dotted gold line at $v/c = 1$ the speed of light it approaches without reaching. The marked point ends this boost, at $w'\tau/c = 10.3$ .

Step 3How far below

c

the speed has come

Write $a$ for the dimensionless combination $w'\tau/c$ that governs the whole result. The fraction asked for follows from (2) evaluated at $t = \tau$ .

\frac{c-v}{c} = 1-\frac{v}{c} = 1-\frac{a}{\sqrt{1+a^2}}

The number $a$ is large here, so pull $a$ out of the root and expand the remaining bracket by the binomial series in the small quantity $1/a^2$ .

\begin{aligned} \frac{a}{\sqrt{1+a^2}} &= \frac{1}{\sqrt{1+\dfrac{1}{a^2}}} = \left(1+\frac{1}{a^2}\right)^{-1/2} \\[3pt] &\approx 1-\frac{1}{2a^2} \end{aligned}

The data fix $a$ : the acceleration is $w' = 10g$ , which is $98$ m/s $^2$ , and the boost lasts $\tau = 3.156\times10^{7}$ s.

a = \frac{w'\tau}{c} = \frac{98\times3.156\times10^{7}}{3.00\times10^{8}} = 10.31

With that number the fraction follows. Numerically,

\frac{c-v}{c} \approx \frac{1}{2\left(10.31\right)^{2}}

\frac{c-v}{c} \approx \frac{1}{2}\left(\frac{c}{w'\tau}\right)^{2} = 4.7\times10^{-3} = 0.47\,\%

(3)

Step 4The distance covered during the boost

The distance is the time integral of the speed from launch to the end of the boost. The integrand is the derivative of a square root in disguise, so the substitution $u = 1+\left(w't/c\right)^2$ , for which $du = 2w'^2t\,dt/c^2$ , carries it through with the limits transformed as well.

\begin{aligned} l &= \int_0^{\tau}v\,dt = \int_0^{\tau}\frac{w't\,dt}{\sqrt{1+\left(\dfrac{w't}{c}\right)^{2}}} \\[3pt] &= \frac{c^2}{2w'}\int_{1}^{1+a^2}\frac{du}{\sqrt{u}} \\[3pt] &= \frac{c^2}{w'}\Bigl[\sqrt{u}\,\Bigr]_{1}^{1+a^2} \end{aligned}

Evaluating the bracket gives the distance in closed form. Since $\tau$ is one year, the light-year $c\tau = 9.47\times10^{15}$ m is the natural unit, and dividing the result by it turns the answer into light-years directly.

l = \frac{c^2}{w'}\left(\sqrt{1+a^2}-1\right), \qquad \frac{l}{c\tau} = \frac{\sqrt{1+a^2}-1}{a} = \frac{10.358-1}{10.31}

\begin{aligned} l &= \left(\sqrt{1+\left(\frac{w'\tau}{c}\right)^{2}}-1\right)\frac{c^2}{w'} \\ &= 8.6\times10^{15}\ \mathrm{m} = 0.91\ \text{light-year} \end{aligned}

(4)

Check

Both formulas must collapse to the Newtonian ones when the boost is short, $w't \ll c$ . Expanding (2) for small $w't/c$ gives $v \approx w't$ , and expanding the boxed distance the same way gives $l \approx \left(c^2/w'\right)\left(a^2/2\right)$ , which is the Galilean $w'\tau^2/2$ ; both are correct for a constant acceleration $w'$ . The combination $c^2/w'$ is a speed squared over an acceleration and so is a length, as the boxed $l$ needs. Finally the exact value of the first answer, $1-10.31/\sqrt{1+10.31^2} = 4.67\times10^{-3}$ , differs from the expanded $4.70\times10^{-3}$ by well under the rounding, which is the license the expansion needed.

Remark

Compare the distance with the $1.00$ light-year that a pulse of light released at launch covers in the same year of terrestrial time. For large $w't/c$ the boxed distance tends to $l \to ct-c^2/w'$ , so the rocket does not merely fall behind the pulse, it falls behind by an amount that settles on the fixed length $c^2/w' = 9.2\times10^{14}$ m $= 0.097$ light-year. At the end of this boost the actual shortfall is $0.092$ light-year, already $95$ percent of that ceiling. The lag is a permanent debt run up in the first few weeks, while the speed was still well below $c$ , and no amount of further burning at $10g$ can pay it back: once $v$ is within a fraction of a percent of $c$ the rocket gains distance at essentially the same rate as the light does, and the gap between them stops growing.