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A rocket's launch 
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#19
Dec2312, 11:56 AM

P: 699

Would it make any significant difference to launch a rocket with some initial velocity?
For example suppose I created vertical launch tube which extended 240 meters below ground. My rocket is accelerated by electromagnets at 3g for 4 seconds so that by the time it is at ground level and the engines are started it is already going 120 m/s. Would this be any more or less beneficial then launching from a platform 720 meters above ground level? On the one hand the launch tube would seem to result in 4 seconds less engine burn time to reach orbital velocity which should save fuel. On the other hand, the kinetic energy added by the launch tube would be equal to the gravitational potential energy added by the platform so they should be equivalent. 


#20
Jan913, 06:16 AM

P: 45

Hello BruceW,
Might be you can help? Regarding efficiency of using fuel: K/K+Q, where Q=total kinetic energy of the emitted gases, K=final kinetic energy of rocket. Sometime ago I read in a popular book of Gerard t'Hooft "Playing with Planets" where he says that efficiency can be near 100% if the velocities of emitted gas gradually increases during the flight. I guess he knows what he says... However I cannot get this by calculations. Assume for simplicity no gravity. Differentiating momentum equality we have (similar to your equation): m' Vg = m Vr', where m  mass Vg a is the velocity of the gas relative to the rocket, Vr' is acceleration of rocket. If we assume Vg=Vr, than it looks like that all gas power is transformed to the rocket kinetic power. For every moment the gas kinetic energy is zero in the frame where the rocket has speed Vr. Solving the above diff equation one can get the simplest formula for the Vr: Vr(m)=v0*m0/m m0 initial mass, v0 initial speed. Using this formula one can calculate again the efficiency to 100%, but the formula looks odd because we stay in the reference frame where the rocket has initial velocity v0. If i modify it to be "correct": Vr(m)=v0*m/m0v0 than simple calc gives the efficiency 1m/m0. For the constant gas speed velocity v0 we have Vr(m)=v0*ln(alpha), alpha=m0/m, and efficiency ln(alpha)^2/(alpha1), which is worse than previous. If I am a captain of rocket and I want to use my fuel in a most efficient way, which time (or mass) dependence of gas velocity I should choose? 


#21
Jan913, 06:35 AM

P: 1,018




#22
Jan913, 07:21 AM

P: 45

What strategy of velocity of emitted gases V_g(m) should I use to minimize the factor: K(m)/(K(m)+Q(m0m)) where K(m) is kinetic energy of the rocket when its mass is 'm', Q(m0m) total kinetic energy of the emitted gases. m0 is initial mass when the rocket has zero speed. 


#23
Jan913, 07:21 AM

P: 45




#24
Jan913, 02:22 PM

HW Helper
P: 3,447

Actually, that is a really good question. Using the same old calculation (without gravity) gives:
[tex]v_{ex} \dot{m}=m \dot{v} [/tex] (where [itex]v_{ex}[/itex] is the velocity of the propellant with respect to the rocket, i.e. the exhaust velocity). Then, doing a similar calculation, to find the rate of change of total kinetic energy of the system, I get: [tex]\frac{dQ}{dt}= \frac{1}{2} \dot{m} {v_{ex}}^2 [/tex] (where the left hand side of the equation is the rate of change of the total kinetic energy in the system. So you can relate this to the rate at which chemical energy is being used up, since change in kinetic energy is due to a change in chemical energy). Note: when the rate of change of mass of the rocket is negative, the rate of change of total kinetic energy of the system is positive, as we would expect. Also, we know that the kinetic energy of the rocket is given by: [tex]T=\frac{1}{2}mv^2[/tex] (Where I am using the symbol T to mean the KE of the rocket). Also, the rate of change of the kinetic energy of the rocket is: [tex]\frac{dT}{dt}=\dot{m}v( \frac{1}{2} v+ v_{ex}) [/tex] So now, using our two equations for rate of change of rocket KE with time, and rate of change of total KE with time, we get: [tex]\frac{dT}{dQ}=  {(\frac{v}{v_{ex}})}^2  \frac{v}{v_{ex}} [/tex] Awesome. This equation is telling us the rate of change of KE of the rocket, with respect to change in the total KE of the system (i.e. this equation is the 'instantaneous' efficiency). Now, we want to find what value [itex]v_{ex}[/itex] must take, for this equation to be maximised, so if we take the partial derivative with respect to [itex]v_{ex}[/itex], then we get [itex]v_{ex}=2v[/itex] and if this condition is fulfilled, then the instantaneous efficiency is at its maximum value of 1/2 So, after all that, we find that to get max efficiency for the rocket, we require [itex]v_{ex}=2v[/itex] so this means that the exhaust velocity must always be at twice the velocity of the rocket (and in opposite direction, of course). So, equivalently, this means we require that the actual velocity of the propellant must always be equal and opposite to the velocity of the rocket (as viewed by an inertial reference frame). So, I agree with your book, that initially the propellant must come out slowly, and gradually begin to come out more quickly. But I find that the efficiency is 1/2, not 1. In fact, an efficiency of 1 does not make sense, because this would imply that the total kinetic energy of all the expelled propellant is zero. But this is clearly not possible. Also, I should say, this is for an ideal rocket, and a real rocket is probably much more complicated. But it is nice to work out that the max theoretical efficiency is 1/2, assuming I've done the calculations right ;) 


#25
Jan913, 06:17 PM

P: 588

For example the launches from Merritt Island Florida, at 28.5 degrees north latitude, yields a "free" initial speed of 1,472Km/h before the rocket even leaves the pad. 


#26
Jan913, 09:12 PM

P: 127

Also, the thrust from a "fixed" thrust rocket increases a bit as it gets above the atmosphere. The exhaust doesn't have to fight atmospheric pressure in getting out of the nozzle.



#27
Jan1013, 03:23 AM

P: 45

Q'= m'(V_exV)^2/2 T'=((mV^2)/2)'=mVV' For T' we take derivative only for the speed, not mass, because we are interested only in final gain in the kinetic energy of the rocket with final mass m. Since mV'=m'V_ex, we have T'=m'*V*V_ex Then it is easy to find: efficiency of power (or energy rate) T'/(T'+Q')=2(V_ex/V)/(1+(V_ex/V)^2), i.e. the maximum 100% is achieved at V_ex=V. If we now assume V_ex=V and solve the equation of motion mV'=m'V we will get V(m) = v0*m0/m +const. If const=0 than this would mean that the rocket initially had speed v0 and the engine has been already working so gases had speed v0 with respect to the rocket and we will then calculate the efficiency to 100%. Simply because in the frame where we do the calculation the emitted gases always has velocity==zero, according to our assumption V_ex=V. But this looks extremely odd because we do the calculation with respect to some frame where the rocket has already got initial velocity v0... If we choose v0=0, which is fair, meaning that we start with gas velocity 0 and rocket speed 0, than we will not accelerate. Initially one should consider a rocket that is not accelerating but moving inertially, then we switch on engine. The easiest is to calculate in center of momentum reference frame, i.e. when the rocket has initial speed zero. In this case I cannot get 100% of efficiency. 


#28
Jan1113, 10:24 AM

HW Helper
P: 3,447

[tex]mv = (m+ \Delta m)(v+ \Delta v)  \Delta m(v_{ex} +v)[/tex] Now, cancelling some terms, and dividing by the small time interval [itex]\Delta t[/itex], we get: [tex]m \frac{\Delta v}{\Delta t} + \frac{\Delta m \Delta v}{\Delta t} = v_{ex} \frac{\Delta m}{\Delta t} [/tex] Now, take the limit that these small changes are infinitesimally small changes, then the second term on the left hand side disappears (because it is much smaller than the other terms), so we get left with: [tex]m \frac{dv}{dt} = v_{ex} \frac{dm}{dt} [/tex] hooray! Now, it is possible to find things like the rate of change of the KE of the rocket and the rate of change of total KE, by using a very similar line of reasoning as above. Right, so to find out what will be the small change in KE of the rocket, we simply use the KE of rocket after, minus the KE of the rocket before, so we get: [tex]\Delta T = \frac{1}{2} (m+ \Delta m)(v+ \Delta v )^2  \frac{1}{2}mv^2 [/tex] And straight away, I am going to ignore any terms which have a product of more than one 'delta', since these terms are going to be much smaller, when we take our limit of very small changes, so therefore, we get: [tex]\Delta T = mv \Delta v + \frac{1}{2} v^2 \Delta m [/tex] So, now, again dividing by [itex]\Delta t[/itex] and taking the limit of very small change, we get: [tex]\frac{dT}{dt}=mv \frac{dv}{dt} + \frac{1}{2} v^2 \frac{dm}{dt} [/tex] And now we simply use the relation [itex]\dot{m} v_{ex} = m \dot{v} [/itex] with the first term of our equation for T, to get the nicer expression: [tex]\frac{dT}{dt} = \frac{dm}{dt} v (v_{ex} + \frac{1}{2} v ) [/tex] hooray. And now, to find the change in KE of the system, we use the KE of the rocket after, and plus the KE of the propellant, and minus the KE of the rocket before (since we want to find the total change in KE of the system). So we get: [tex]\Delta Q = \Delta T  \frac{1}{2} \Delta m (v_{ex} + v)^2 [/tex] Note: we are adding the KE of the propellant, but since the mass of the propellant is [itex] \Delta m [/itex], we get a minus sign. So again, discarding any terms where there is a product of delta's, we have: [tex]\Delta Q = mv \Delta v  v_{ex} v \Delta{m}  \frac{1}{2} v_{ex}^2 \Delta m [/tex] So, divide by delta t and take limit of small changes: [tex]\dot{Q} = mv \dot{v}  \dot{m} v_{ex} v  \frac{1}{2} v_{ex}^2 \dot{m} [/tex] But luckily, using our equation [itex]\dot{m} v_{ex} = m \dot{v} [/itex] the first two terms on the righthand side cancel, so we end up with: [tex]\dot{Q} =  \frac{1}{2} \dot{m} v_{ex}^2 [/tex] 


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