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I have been developing a science fiction novel, and something akin to the following mathematics problem has arisen: (Assume Newtonian mechanics for the moment.)

Alice is a medium-distance space-hauler, and has to move her cargo in a straight line from point $A$ to point $B$, a distance of $9\times 10^{12}$ meters. Because high accelerations are uncomfortable for humans over long intervals, Alice will be accelerating towards point B for the first half of her journey at $+10 \ m/s^2$, and switching to $-10 \ m/s^2$ to slow down halfway through.

Blorb, a freaky alien space pirate, has plans to plunder Alice's booty, and due to his alien physiology can tolerate accelerations up to $\pm 25 \ m/s^2$. If he must leave $24$ hours after Alice to avoid making his intentions obvious, what is the soonest he can intercept Alice, assuming he must match not only her position, but also her velocity to properly board her ship?

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  • $\begingroup$ This is better suited for Physics SE $\endgroup$ Jun 11, 2020 at 5:35
  • $\begingroup$ I would have posted it to Physics SE, but it's heavily an optimization problem, though someone has removed that tag. Not only must the velocity and position both be matched SIMULTANEOUSLY, as matching one or the other is easy, but they must simultaneously matched at the earliest possible point. $\endgroup$ Jun 11, 2020 at 5:42
  • $\begingroup$ It's more a constraint problem than an optimization problem, I think, though I concede the distinction is a fine one. The easiest solution is probably trial and error, but maybe someone will come up with an elegant approach. $\endgroup$
    – Brian Tung
    Jun 11, 2020 at 5:45
  • $\begingroup$ Just as a back-of-the-envelope estimate, Alice will take about a week-and-a-half to get to the halfway point. A one-day headstart just isn't that much, given the disparity in accelerations. My guess is that Blorb can match position and velocity in three or four days, tops. (I.e., she will still be accelerating when Blorb catches up.) $\endgroup$
    – Brian Tung
    Jun 11, 2020 at 5:49
  • $\begingroup$ Just to give you an idea of the distances involved, $9 \times 10^{12}$ meters is about twice the distance to Neptune. It's really not that far. $\endgroup$
    – Brian Tung
    Jun 11, 2020 at 5:52

2 Answers 2

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The only flight plans that matter are: accelerate towards $B$ for time $t_1$, then accelerate towards $A$ for the rest of time (say until $t_2$). Doing anything else wastes time.

Proof: suppose for time $t_a$ through $t_b$ you didn't accelerate (or did so less than the maximum). Now consider accelerating towards $B$ for some small period of time, then decelerate for some period of time such that your start and end speed are unchanged. You then have a faster average speed so you use less time to cover that same distance.

We simply give the total distance and ending velocity of such a flight plan, and you can get simultaneous equations for Alice and Blorb matching at true-clock time $t$.

From a standing start, accelerating continuously gives you $$\begin{align*}v(t)&=a(t-t_0)\\x(t)&=\frac{1}{2}a(t-t_0)^2\end{align*}$$ by directly integrating $v(t)$. Therefore at the turnover point, your current physical coordinates are $v_1=v(t_1)=a(t_1-t_0), x_1=x(t_1)=\frac{1}{2}(t_1-t_0)^2$.

Next, consider the retrograde acceleration phase which gives you $$\begin{align*}v(t)&=v_1-a(t-t_1)\\x(t)&=x_1+v_1(t-t_1)-\frac{1}{2}a(t-t_1)^2\end{align*}$$ again by direct integration $\int_{t_1}^tv(t)\,\mathrm{d}t$.

Let Alice's head start be $s$, to match velocities at unknown intercept time $t$ and Blorb turnaround time $t_1$, $0\leq t_1\leq t$ we have $$a_Bt_1-a_B(t-t_1)=a_A(t+s)\quad\Rightarrow\quad t_1=\frac{1}{2}\left((1+a)t+as\right)$$ where we're going to nondimensionalize by letting $a=\frac{a_A}{a_B}$.

Plug that into the constraint matching positions at time $t$: $$\begin{align*} \frac{1}{2}t_1^2+t_1(t-t_1)-\frac{1}{2}(t-t_1)^2&=\frac{1}{2}a(t+s)^2\\ -t_1^2+2t_1t-\frac{1}{2}t^2&=\frac{1}{2}a(t+s)^2\\ (1-a^2)t^2-2as(1+a)t-as^2(2+a)&=0\\ t&=\frac{s}{1-a}\left(a+\sqrt{\frac{2a}{1+a}}\right) \end{align*}$$ Tedious but straightforward.

With the actual numbers in the question, the results are $$\begin{align*} t&=46\!:\!14\!:\!13.7\text{ (+1 day head start)}\\ t_1&=37\!:\!09\!:\!57.6\\ d&=3.1968\times10^{11}\text{m} \end{align*}$$ so Alice gets caught just under 3 days out, about 3.5% of the way there.

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The intercept happens well before Alice reaches halfway, simplifying calculations. The optimal solution is a bang-bang control.

Alice's velocity at time $t$ seconds after Blorb departs will be $v_A(t)=10(t+D)$ m/s (where $D=86400$s), and their position will be $$ x_A(t) = 5(t+D)^2\text{ m}. $$ (Integrate $v(t')$ from $t'=-D$ to $t'=t$)

So Alice reaches the halfway point roughly 14.5 ($=\frac{\sqrt{9/5\times10^{12}}-D}{D}$) days after Blorb departs. This is plenty of time for Blorb to catch up.

Blorb's strategy is to accelerate at 25m/s for as long as possible, and just at the right moment start to decelerate at 25m/s until the positions and velocities are equal. We have two unknowns, $t_1$, the time at which Blorb starts decelerating, and $t_2$ the time at which the ships meet. We require $x_A(t_2)=x_B(t_2)$ and $v_A(t_2)=v_B(t_2)$ where the A subscript refers to Alice's ship and the B subscript to Blorb's.

For Blorb's ship, the acceleration is $a_B(t)=25$ if $t<t_1$ and $-25$ is $t>t_1$. The velocity therefore is $$v_B(t) = \begin{cases} 25t &\text{if $t<t_1$} \\ 25t_1-25(t-t_1) & \text{if $t>t_1$} \end{cases}=\begin{cases} 25t &\text{if $t<t_1$} \\ 50t_1-25t & \text{if $t>t_1$} \end{cases} $$ To get the position we integrate again, $$x_B(t) =\begin{cases} 12.5t^2 &\text{if $t<t_1$} \\ 12.5t_1^2+50t_1(t-t_1)-12.5(t^2-t_1^2) & \text{if $t>t_1$} \end{cases} $$

Now solve the equations matching velocities and positions to find deceleration time $t_1$ and intercept time $t_2$. The velocities give $$50t_1-25t_2=10(t_2+D)$$ and positions give $$12.5t_1^2+50t_1(t_2-t_1)-12.5(t_2^2-t_1^2)=5(t_2+D)^2. $$

The first equation gives $$t_1=\frac{35t_2+10D}{50}=\frac{7t_2+2D}{10}$$ and then the second gives a complicated quadratic for $t_2$, $$12.5\left(\frac{7t_2+2D}{10}\right)^2+50\frac{7t_2+2D}{10}\left(t_2-\frac{7t_2+2D}{10}\right)-12.5\left(t_2^2-\left(\frac{7t_2+2D}{10}\right)^2\right)=5(t_2+D)^2$$ with solution $$t_2=\frac{14+10\sqrt{7}}{21}D,$$ so $$ t_1 = \frac{2+\sqrt{7}}{3}D. $$

This gives the deceleration time as approximately 133798s or 1 day 13 hours and 9 minutes, and the intercept time as approximately 166454s or 1 day 22 hours and 14 minutes (these times are after Blorb departs, not Alice). Add 24 hours for the times for Alice (she gets caught in just under 3 days).

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  • $\begingroup$ I'm glad we actually got the same answers, that arithmetic was real tedious (I pulled WolframAlpha simplify) $\endgroup$
    – obscurans
    Jun 11, 2020 at 8:45
  • $\begingroup$ @obscurans Yes I was glad to have confirmation too! $\endgroup$
    – David
    Jun 11, 2020 at 11:30

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