0
$\begingroup$

Suppose that you are in a vehicle in space with a power source generating a constant power of $P_0$ that you can use to fire iron pellets. The iron pellets compose 99% of the ship's mass, $m$. The rate at which you can fire the pellets is controllable, but is limited by the vehicle's power plant.

I have tried solving this, but I seem to get an answer for the optimal rate $\dot{m}$ that is non physical. I think I've done something conceptually wrong, here is my attempt:

We know that for our mass we have, over the time interval $t_0 < t < t_f$: \begin{equation} m(t) = m_0 - \frac{dm}{dt}(t-t_0) \end{equation}

And that the power is: \begin{equation} P = \frac{1}{2\Delta t} m(t)v^2(t)=P_0 \end{equation}

This implies that: \begin{equation} v(t) = \sqrt\frac{2P_0 \Delta t}{m(t)} = \sqrt\frac{2P_0 \Delta t}{m_0 - \frac{dm}{dt}(t-t_0)} \end{equation}

Where $\Delta t = t_f - t_o$. So now I assume that maximizing the distance traveled over the time period $\Delta t$ is equivalent to maximizing the final velocity $v(t_f)$. And we have: \begin{equation} x(t_f) - x_0 = \int_{t_0}^{t_f} v(t)dt = \sqrt{2P_0 \Delta t} \int_{t_0}^{t_f}\frac{dt}{\sqrt{m_0 - \frac{dm}{dt}(t - t_0)}} \end{equation}

The Euler-Lagrange equation then becomes: \begin{equation} \frac{\partial f}{\partial m} - \frac{d}{dt}\frac{\partial f}{\partial \dot{m}} = 0 \implies \end{equation} \begin{equation} -\frac{d}{dt}[\frac{1}{2(m_0 - \dot{m}(t-t_0))^{\frac{3}{2}}}] = 0 \end{equation} Or that: \begin{equation} \frac{1}{2(m_0 - \dot{m}(t-t_0))^{\frac{3}{2}}} =C \end{equation} Then we have that (with some algebra): \begin{equation} \dot{m}(t) = \frac{1}{t-t_0}(m_0 - (\frac{1}{2C})^\frac{2}{3}) \end{equation}

This solution doesn't make sense in that near $t = t_0$ it implies that all of the iron pellets are expelled. Over all I'm just having a difficult time understanding how to setup the fundamental physics equations to get to the correct functional. Please advise.

$\endgroup$
  • $\begingroup$ There is no dependence of the final velocity on the available power. Take a look at the rocket equation. That's the correct solution to the problem. $\endgroup$ – CuriousOne Sep 15 '15 at 20:10
  • $\begingroup$ I'm not sure if I understand what you mean, the problem states the rate at which pellets can be fired is dependent upon the power source. I know of the thrust equation for a rocket but I'm still not really sure how to express the correct functional that needs to be maximized. Also, since the problem statement says the firing rate, and therefore the change in mass is dependent on the fixed power supply I thought it would need to be included in the solution. $\endgroup$ – Captainj2001 Sep 15 '15 at 20:16
0
$\begingroup$

@CuriousOne is correct: the ship's final velocity only depends on the amount of mass the ship is discarding, and the discarded mass's velocity. Changing the rate will change where the ship is when it runs out of fuel, but not the final velocity.

To help your intuition, imagine two identical ships, except that one fires its pellets once every other second, and the other fires its pellets once every third second. After each fires its first pellet, they will both be gliding at the same velocity. The second pellet will change the velocity of each ship by the same amount, only depending on the proportion of mass fired and the firing velocity. The fact that one ship waits a bit longer before firing makes no difference in the final velocity of each. In the end, the first ship will run out of fuel sooner, but when the second ship runs out of fuel they will be traveling at the same velocity (although the first ship will have travelled further since its average velocity was higher).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.