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Given a rocket has a minimum acceleration with full tanks of say 1 g and a max acceleration with empty tanks of 3g, what's the average acceleration? Is it just 2

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  • $\begingroup$ It depends on the definition of average acceleration. $\endgroup$ – Uwe Apr 26 '19 at 15:57
  • $\begingroup$ What's your use case for the average acceleration figure? $\endgroup$ – Russell Borogove Apr 26 '19 at 18:20
  • $\begingroup$ I assume that it's going to be the time-average of acceleration. It's obvious physically, and I can't imagine another definition that could make any sense. @Uwe can you think of anything else it could be beside that? The problem with this kind of random comment is that it attracts down votes, and doesn't really help in any way. If you have a constructive comment for the OP, why not communicate to them directly in order to improve something that you believe needs improving. $\endgroup$ – uhoh Apr 27 '19 at 2:47
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For a more thorough discussion, see @Michael Stachowsky's answer.

However, given some assumptions, it's quite manageable:

  • The rocket is at full, constant thrust from full to empty tank
  • It uses fuel at a constant rate
  • It's not affected by any external forces
  • "average" is the change in velocity from full to empty tank, divided by how long it takes to burn all the fuel.

The integral is then pretty simple, and simplifies down to this expressed using just the start acceleration ($a_0$) and the end acceleration ($a_1$):

$$a_{avg} = \frac{a_0 \cdot a_1\cdot ln\left(\frac{a_0}{a_1}\right)}{a_0 - a_1}$$

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As @Uwe said, it depends on your definition of average. The acceleration is not as relevant as you think here. A rocket is governed by the rocket equation, which tells us how much change in velocity we can expect from a rocket, and is given by:

$$\Delta v = v_e\ln{\frac{m_0}{m_f}}$$

Where $v_e$ is the exhaust velocity, $m_0$ is the fully fueled mass, and $m_f$ is the final mass (after all of our fuel is expended). If we let $\Delta t$ be the time it takes to fully run out of fuel, then one way to compute the average acceleration is:

$$a_{avg} = \frac{\Delta v}{\Delta t}$$

Now, to your question: presuming instead we have the initial and final accelerations. What then? The rocket equation is derived according to:

$$\frac{dv}{dt} = \frac{-v_e}{m(t)}\frac{dm}{dt}$$

It is traditionally integrated with respect to $m$, but there is no reason not to integrate with respect to $t$. Assuming that $t_0 = 0$ and the rocket runs out of fuel at some $t = t_f$, then we might define average acceleration in a different way:

$$a_{avg,I} = \frac{1}{t_f}\int_0^{t_f}\frac{dv}{dt}dt = \frac{1}{t_f}\int_0^{t_f}\frac{-v_e}{m(t)}\frac{dm}{dt}dt$$

Presuming you knew how fuel was being expended in your rocket (this would depend on how you throttle it, if it's always in vacuum compared to fighting an ever decreasing atmosphere etc), then you would know the function $\frac{dm}{dt}$ and could carry out the integration.

However, it would depend more on the rocket's specific properties, which aren't specified in the question.

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    $\begingroup$ Thank you for the very detailed answer, you obviously have a far better understanding of this than I have. The specific problem I'm trying to solve is: Given min and max acceleration I want to calculate the average acceleration so I can then calculate the duration of a brachistochrone trajectory. So the throttle would be set at 100% throughout and there would be no external forces such as air resistance. $\endgroup$ – nAUTILUS Apr 26 '19 at 16:53
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    $\begingroup$ Although it has been a long time since I had to try to compute that kind of trajectory (and I've long since forgotten), if your throttle is 100% all the time and you make some simplifying assumptions about how the rocket uses fuel, the $\frac{dm}{dt}$ is just constant, which makes your life a BIT easier. Then $m(t)$ is linear. $\endgroup$ – Michael Stachowsky Apr 26 '19 at 17:04
  • $\begingroup$ Very nice answer! When you say "it depends on your definition of average" can you think of any other possible definition besides the time-average? I think the comments under the question are misleading because they imply that there are other reasonable kinds of averaging. But I can't think of any remote possibilities. $\endgroup$ – uhoh Apr 27 '19 at 2:49
  • $\begingroup$ The integral and the pure average of Delta v/ Delta t may not evaluate to the same value. I'd have to check it to see $\endgroup$ – Michael Stachowsky Apr 27 '19 at 12:07

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