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fatpanda1
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- TL;DR Summary
- Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
Can you give both intuitive and mathematical explanation please.
In contexts or models where the rocket equation applies this is question is a bit misleading, so I am curious as to the context the question has been posed in? (There are quite common situations in the normal life of a launcher that is not modeled well by the rocket equation and which also gives a good answer to the question, but the question itself is just posed as if applicable to any situation which is not true).fatpanda1 said:Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
Because ejecting the fuel all at once and having a single clump of exhaust gasses then travel at its normal exhaust velocity relative to the rocket is an incorrect expectation.fatpanda1 said:Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
Can you give both intuitive and mathematical explanation please.
You don't want them to be really efficient.anorlunda said:Amateur rockets come close. They expend their fuel in a small fraction of the flight duration.
For the math, the Tsiolkovsky rocket equation is: $$\Delta v = v_e \ln \left( \frac{m_r + m_f}{m_r} \right)$$ where ##m_r## is the mass of the rocket without the fuel and ##m_f## is the mass of the fuel and ##v_e## is the exhaust velocity.fatpanda1 said:Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
Can you give both intuitive and mathematical explanation please.
In addition to the better answers above, as you shorten the duration of the burn, you increase the acceleration for that initial period during the burn. Depending on the payload, there is a practical limit to how high the acceleration can be. Even without humans on board, the electronic assemblies and electro-mechanical components have limited tolerance for high accelerations...fatpanda1 said:Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
Can you give both intuitive and mathematical explanation please.
I don't understand this argument. The Tsiolkovsky equation does not contain the time you take to exhaust the fuel. So in this approximation it doesn't matter how quickly you exhaust the fule. The change in velocity is always the same. So why should your 2nd formula apply for exhausting the fuel instantaneously and why shouldn't then apply the Tsiolkovsky equation as well?Dale said:For the math, the Tsiolkovsky rocket equation is: $$\Delta v = v_e \ln \left( \frac{m_r + m_f}{m_r} \right)$$ where ##m_r## is the mass of the rocket without the fuel and ##m_f## is the mass of the fuel and ##v_e## is the exhaust velocity.
If you simply burn it all instantaneously then conservation of momentum gives $$\Delta v = v_e \frac{m_f}{m_r} $$
A series expansion of gives $$\ln \left( \frac{m_r+ m_f}{m_r} \right) \approx \frac{m_f}{m_r} - \frac{m_f^2}{2 m_r^2} + O(m_f^3)$$
So I get the opposite, that ejecting it all at once makes it go faster.
The 2nd formula is an instantaneous burn, meaning that all of the exhaust has the same final velocity. Even though the rocket equation doesn't explicitly contain the burn time, it is based on the assumption of a non-instantaneous flow, so the exhaust velocity varies as the rocket accelerates. So the final state in the two cases is different. I.e. in my formulation an instantaneous burn is not just a limiting case of a non-instantaneous burn with a very short burn time.vanhees71 said:I don't understand this argument. The Tiolkovsky equation does not contain the time you take to exhaust the fuel. So in this approximation it doesn't matter how quickly you exhaust the fule. The change in velocity is always the same. So why should your 2nd formula apply for exhausting the fuel instantaneously and why shouldn't then apply the Tsiolkovsky equation as well?
Based on that expectation, burning all at once would give maximal final velocity? Because that would give the exhaust the maximal momentum in the opposite direction?jbriggs444 said:Because ejecting the fuel all at once and having a single clump of exhaust gasses then travel at its normal exhaust velocity relative to the rocket is an incorrect expectation.
So the tiny rocket flies off before all fuel was burned, and the fuel energy goes into accelerating the exhaust in all directions, not just backwards?jbriggs444 said:If you eject a huge blob of fuel from a tiny rocket at a particular exhaust velocity relative to the rocket, it takes only a small amount of energy. You might well think of it as ejecting a tiny rocket from a huge blob of fuel. You wind up with a huge unburnt blob of fuel.
Because this is better known as 'the rocket esploded'.fatpanda1 said:... ejecting all at once?
I think this is covered in reply #2DaveC426913 said:Because this is better known as 'the rocket esploded'.
I actually scanned for such replies before posting. I guess I wasn't as diligent as I should have been.Filip Larsen said:I think this is covered in reply #2
Exactly. For an instantaneous burn the velocity of the fuel relative to the rocket is undefined since the velocity of the rocket is discontinuous (of course it is all well defined in the limit, but I am not assuming a limiting case). So for an instantaneous burn (the way I assumed) the ##v_e## is the velocity of the exhaust gas relative to the inertial frame in which the rocket was initially at rest.vanhees71 said:Hm, as I just added to my posting, the assumption underlying the Tsiolkovsky equation is ##v_f=\text{const}## (where ##v_f## is the velocity of the fuel relative to the rocket). If this is not the case you likely get a different result,
In some ways it's more natural to take ##v_e## as the relative velocity after the burn because it gives the correct limiting behaviour as per #14 i.e. ##m(0)v = (m(0)-q)(v+\Delta v) + q(v + \Delta v-v_e) \implies \Delta v = \dfrac{v_e q}{m(0)}##.Dale said:So for an instantaneous burn (the way I assumed) the ##v_e## is the velocity of the exhaust gas relative to the inertial frame in which the rocket was initially at rest.
That is interesting. I hadn’t looked at that.ergospherical said:In some ways it's more natural to take ##v_e## as the relative velocity after the burn because it gives the correct limiting behaviour as per #14 i.e. ##m(0)v = (m(0)-q)(v+\Delta v) + q(v + \Delta v-v_e) \implies \Delta v = \dfrac{v_e q}{m(0)}##.
This seems wrong since the rocket also 'instantaneously' accelerates to its final speed, so it should be viewed as a limit as the burn time decreases to zero, which seems to be the same transfer of momentum to the exhaust regardless of the rate of burn. No matter how short you reduce the burn time, the exhaust velocity never approaches the same value.Dale said:The 2nd formula is an instantaneous burn, meaning that all of the exhaust has the same final velocity.
That is explicitly not what I assumed because then there is no difference. The OP can (should) clarify.Halc said:so it should be viewed as a limit as the burn time decreases to zero
One possible interpretation would be the limit of pulsing ##n## chunks each of mass ##\frac{m_\text{fuel}}{n}## and each at exhaust velocity ##v_e##. [Measuring exhaust velocity relative to the remaining payload velocity not relative to pre-burn craft velocity].Dale said:That is explicitly not what I assumed because then there is no difference. The OP can clarify.
I would say nonsensical too! That the mass ejection rate must vary only between finite limits (on physical grounds) guarantees that ##q \rightarrow 0## in the limit of an instantaneous burn, in which case the rocket equation reduces to the result of the collision problem.Dale said:Unphysical, yes, but not nonsensical.
Yes -- if we were to harvest all of the energy in the fuel and eject it as a single lump.A.T. said:Based on that expectation, burning all at once would give maximal final velocity? Because that would give the exhaust the maximal momentum in the opposite direction?
The fuel energy goes -- somewhere. We do not care. The model just says we did not use it all. Maybe we are spraying unburnt fuel. Maybe the unburnt fuel burns and scatters. Whichever.A.T. said:So the tiny rocket flies off before all fuel was burned, and the fuel energy goes into accelerating the exhaust in all directions, not just backwards?
The collision problem is hardly non-sensical. Sorry, but I disagree.ergospherical said:I would say nonsensical too! That the mass ejection rate must vary only between finite limits (on physical grounds) guarantees that ##q \rightarrow 0## in the limit of an instantaneous burn, in which case the rocket equation reduces to the result of the collision problem.
Unbounded mass ejection rates are perfectly physical.ergospherical said:The non-sensical part is to consider any mass ejection rate ##r(t)## which is unbounded. If one does so then I don't think it should be a surprise when discrepancies arise.
jbriggs444 said:Unbounded mass ejection rates are perfectly physical.
You burn all of the fuel and package it up into a canister ready for expulsion. You use the energy to expel the cannister so as to use up all of the energy harvested from the burning of the fuel. You eject it in a single impulsive *whoomp*. That's an inelastic collision.
This is the part that is not physically realisable - the duration of the release will be always non-zero hence the mass ejection rate always bounded.Dale said:The moment it leaves your hand is the moment where the mass ejection rate is infinite.
Classically there will always be some instant where you stop touching the mass. That can be defined as the instant where the mass is ejected, giving an infinite mass ejection rate.ergospherical said:This is the part that is not physically realisable - the duration of the release will be always non-zero hence the mass ejection rate always bounded.
This isn’t right; by mass ejection rate it is meant that there is some boundary ##\Sigma(t)## through which a mass ##r(t)## flows per unit time. Any clump of matter that you throw is of finite size (i.e. length parallel to velocity ##\sim l##) so no matter how fast you throw it, it still takes ##t \sim l/v## to eject the mass.Dale said:Classically there will always be some instant where you stop touching the mass. That can be defined as the instant where the mass is ejected, giving an infinite mass ejection rate.