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10.3: The Rocket Equation

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Initially at time t = 0, the mass of the rocket, including fuel, is m0.

We suppose that the rocket is burning fuel at a rate of b kg s-1 so that, at time t, the mass of the rocket-plus-remaining-fuel is m=m0bt. The rate of increase of mass with time is dmdt=b and is supposed constant with time. (The rate of "increase" is, of course, negative.)

We suppose that the speed of the ejected fuel, relative to the rocket, is V. The thrust of the ejected fuel on the rocket is therefore Vb, or Vdmdt. This is equal to the instantaneous mass times acceleration of the rocket:

Vb=mdvdt=(m0bt)dvdt.

Thus

v0dv=Vbt0dtm0bt.

(Don't be tempted to write the right hand side as Vbt0dtbtm0. You are anticipating a logarithm, so keep the denominator positive. We have met this before in Chapter 6.) On integration, we obtain

v=Vlnm0m0bt.

The acceleration is

dvdt=Vbm0bt.

At t = 0, the speed is zero and the acceleration is Vbm0.

At time t=m0b, the remaining mass is zero and the speed and acceleration are both infinite. However, this is so only if the initial mass is 100% fuel and nothing else. This is not realistic. If the fraction of the total mass was initially f, the fuel will be completely expended after a time fm0b at which time the speed will be (which is, of course, positive), and the speed will remain constant thereafter. For example, if f = 99%, the final speed will be 4.6 V.

Equations ??? and ??? are shown in Figures X.1 and X.2. In Figure X.1, the speed of the rocket is plotted in units of V, the ejection speed of the burnt fuel. The time is plotted in units of m0b. The fuel initially comprised 90% of the rocket, so that the rocket runs out of fuel in time 0.9 m0b, at which time its speed is 2.3V. In Figure X.2, the acceleration is plotted in units of the initial acceleration, which is Vbm0. When the fuel is exhausted, the acceleration is ten times this.

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In Equation ???, v is of course dxdt, so the equation can be integrated to obtain the distance:time relation:

x=V[t+(m0bt)ln(1btm0)].

Elimination of t between Equations ??? and ??? gives the relation between speed and distance:

x=Vm0b[1(1+vV)evV].

If f is the fraction of the initial mass that is fuel, the fuel supply will be exhausted after a time fm0b, at which time its speed will be Vln(1f)(this is positive, because 1 f is less than 1), its acceleration will be 1(1f) and it will have travelled a distance Vm0b[f+(1f)ln(1f)]. If the entire initial mass is fuel, so that f = 1, the fuel will burn for a time m0ib, at which time its speed and acceleration will be infinite, it will have travelled a finite distance Vm0b and the mass will have been reduced to zero, This remarkable result is not very believable, for two reasons. In the first place it is not very realistic. More importantly, when the speed becomes comparable to the speed of light, the equations which we have developed for nonrelativistic speeds are no longer approximately valid, and the correct relativistic equations must be used. The speed cannot then reach the speed of light as long as the remaining mass is non-zero.

Equations ??? and ??? are illustrated in Figures X.3 and X.4, in which , the fraction of the initial mass that is fuel, is 0.9. The units for distance, time and speed in these graphs are, respectively, Vm0b, m0b and V.

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This page titled 10.3: The Rocket Equation is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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