Ejecting Rocket Fuel

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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.
 
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  • #2
Baluncore
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Because if you eject it all at once it destroys the rocket structure, and the payload.
 
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  • #3
Filip Larsen
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Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
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).
 
  • #4
jbriggs444
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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.
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.

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.
 
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  • #5
anorlunda
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Amateur rockets come close. They expend their fuel in a small fraction of the flight duration.

 
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  • #6
Vanadium 50
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Amateur rockets come close. They expend their fuel in a small fraction of the flight duration.
You don't want them to be really efficient.

You need them to be about a foot long +/- half an order of magnitude, so they can be easily built. You also don't want them flying more than a few hundred feet (say 500 feet +/- a half order of magnitude) because then recovery can prove difficult. So they need short burns.
 
  • #7
Dale
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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.
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.
 
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  • #8
berkeman
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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...
 
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  • #9
vanhees71
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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.
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?

The derivation of the Tsiolkovsky equation works as follows: You assume an arbitrary exhaust ##\mu(t)## (exhausted mass of fuel per unit time). Then assuming the relative velocity ##v_f## of the fuel to the rocket being constant yields the momentum balance
$$\mathrm{d}_t (m v)=-\mu(v_f-v)=\dot{m} (v-v_g),$$
i.e.,
$$m \dot{v}=-\dot{m} v_g$$
or
$$m \frac{\mathrm{d} v}{\mathrm{d} m}=-v_f$$
from which by separation of of the variables the Tsiolkovsky equation follows.

The main oversimplication seems to be the assumption that ##v_f=\text{const}##.
 
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  • #10
russ_watters
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Faster acceleration means less thrust(percentagewise) wasted supporting the weight of the rocket(when launching vertically).
 
  • #11
Dale
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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?
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.
 
  • #12
vanhees71
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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, but the standard Tsiolkovsky equation doesn't depend on the time needed to exhaust a given amount of fuel, ##\Delta m=m_{\text{ini}}-m_{\text{end}}##. You simply get
$$\Delta v=v_f \ln(m_{\text{ini}}/m_{\text{end}})=v_f \ln [(m_{\text{ini}}/(m_{\text{ini}}-\Delta m)],$$
as you stated yourself above.
 
  • #13
A.T.
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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.
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?

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.
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?
 
  • #14
ergospherical
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My two cents - if a mass ##q## is ejected instantaneously at ##t=0##, i.e. ##m(t) = m(0) - q H(t)##, and at constant velocity ##v_e## relative to the rocket, then the differential equation describing the motion is\begin{align*}
m \dfrac{dv}{dt} = v_e q \delta(t)
\end{align*}where ##\delta(t)## is the Dirac delta. Thus the change in velocity is\begin{align*}
\int dv = v_e q \int \dfrac{\delta (t)}{m(t)} dt = \dfrac{v_e q}{m(0)}
\end{align*}where the integral is taken over any interval containing ##t=0##. (The value of ##m(0)## is that of the rocket plus the initial fuel).

[This result is slightly different from the expression for an instantaneous burn given by @Dale].
 
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  • #15
Filip Larsen
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Allow me to expand on my first reply above:
a) The ideal rocket equation indicates it doesn't matter how quickly you eject your propellant (as long as you do at same relative speed), the rocket will end up with the same total delta-V.
b) In situations where you have drag terms (i.e. atmosphere) for sections of the trajectory, maintaining a lower speed at those sections will (everything else being equal) give higher total delta-V. Add to this the practical limitation that even without atmosphere its not possible to enter a closed orbit around a planet from its surface using only one impulsive launch maneuver (but that only relates to the "shape" of the orbit, not the total speed).
c) For some orbital maneuvers the rocket will obtain the highest total delta-V if the maneuver is done impulsively (i.e. all at once) at the periapsis (see Olberth's effect for details).

So depending on the actual context the original question may be either misleading, true or false.
 
  • #16
DaveC426913
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The really short answer:

... ejecting all at once?
Because this is better known as 'the rocket esploded'.

It is left as an exercise for the OP to determine why this is inefficient.
 
  • #17
Filip Larsen
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Because this is better known as 'the rocket esploded'.
I think this is covered in reply #2
 
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  • #18
DaveC426913
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I think this is covered in reply #2
I actually scanned for such replies before posting. I guess I wasn't as diligent as I should have been.
 
  • #19
Dale
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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,
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.
 
  • #20
ergospherical
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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.
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)}##.
 
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  • #21
Dale
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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)}##.
That is interesting. I hadn’t looked at that.
 
  • #22
Halc
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The 2nd formula is an instantaneous burn, meaning that all of the exhaust has the same final velocity.
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.

In other words, it seems that given X amount of fuel and M rocket mass, in deep space already, you get exactly the same final speed regardless of the rate of burn, which is why burn time doesn't appear in that rocket equation. So the reason for the chosen rate of burn has more to do with engine efficiency and getting off the planet and other considerations.
Burn too slow and too much of the energy is used just countering gravity (a problem only if your rocket is pointed up, hardly the most efficient orientation). A sufficiently low power rocket will never leave the pad at all, at least not until its weight drops below its thrust. Hence the need for high G force at takeoff, but not so high when say leaving Earth orbit.

Faster fuel expenditure means larger engines, and that adds mass to the engines and, as has been pointed out in prior posts, greater structural mass needed to withstand the greater forces involved. Burns also are throttled back at first to prevent excessive speed while still in dense atmosphere. You always hear them talk about throttling up after a minute or so. No need to hold back if there's no atmosphere with which to contend.

Slow expenditure of fuel is indicated when already in space, which is why ion engines are optimal when time is not of the essence. If there's people on board, there's the mass of the life support to think about. Get there faster just so you don't have to cart along so many resources to keep your perishable payload intact.
 
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  • #23
Dale
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so it should be viewed as a limit as the burn time decreases to zero
That is explicitly not what I assumed because then there is no difference. The OP can (should) clarify.
 
  • #24
ergospherical
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Nonetheless either approach is still valid, since as the burn time ##\delta t \rightarrow 0##, then mass ejected ##q \rightarrow 0## for any finite mass ejection rate ##r(t)## and the change in velocity given by the rocket equation\begin{align*}
v_e \ln \left( \dfrac{m(0)}{m(0)-q} \right) = - v_e \ln \left( 1 - \dfrac{q}{m(0)} \right) = \dfrac{v_e q}{m(0)} + O(q^2)
\end{align*}reduces to the formulae in #14 and #20. Clearly a mass ejection rate ##r(t)## which does not remain finite is unphysical.
 

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