<A HREF="../../movies/Liftoff.mpg"> <IMG src="./images/liftoff.gif"></A><br> Rocket science is nothing more than the application of Newton's Laws using the full power of the calculus. In fact, most aspects of today's rocket flights could have been understood and accurately described by Newton over 300 years ago. Up to this point we have viewed only problems in which the mass of the system remains constant. Newton himself recognized that situations in which the mass is not constant were still covered by his formulation of the Second Law of Motion even though such situations are less common. These cases always demand that we know how to handle differential equations. One example where the mass does change is the case of rockets. Rockets are perfect for space travel because they carry their fuel and oxygen supply with them. In fact, most of the mass of the rocket on the ground is in the form of fuel and oxidizer. In space, the burning fuel is ejected from the rear of the rocket. This action produces a reaction force on the rocket body which propels it forward. Note that there is no need for any air to ``push against'' for the rocket to work. Newton's Third Law assures us that ejection of an object from the system MUST propel the system in the opposite direction. The propulsive force is referred to as the thrust of the rocket. To derive the equation which describes rocket motion, we begin by translating words about the rocket's actions into equations. We begin by noting that the rocket expels gases from burning fuel. If we imagine that the rocket is in outer space, far from the earth, then there are no drag forces or gravity so there are no external forces on the rocket or the gases it expels. Therefore, if we consider the rocket and the exhaust to be a system, then, the momentum of this system is constant with time. Consider the figure below: At initial time t, the rocket is moving with velocity and has a mass M. Thus the momentum of the system is just Mv and will always remain so since momentum is conserved for the system. After a small amount of time , the rocket will have expelled a mass of gas and the rocket body will have gained a small amount of speed . The change in momentum of the rocket body is This equation says that the rocket changes momentum due to its loss in mass (which is just the mass of gas expelled) times its initial velocity plus the mass of the rocket times its change in velocity. To get the momentum of the expelled gas, we need to know its velocity. Typically, rocket engineers measure the speed relative to the rocket body at which gases are ejected as a way of characterizing the engine. If we call the speed relative to the rocket, uex, then the speed of exhaust relative to us as observers is v - uex. The total mass of the exhaust for this time is just , the mass lost by the rocket body, hence the net change in momentum of the exhaust is where the minus sign indicates that the momentum change in the exhaust is opposite to that of the rocket body motion. Since momentum is conserved, the total changes in momentum of the rocket body and exhaust gases must cancel We can develop this last equation into a differential equation by noting that our expression becomes more and more accurate a description of the rocket's motion for any given time if we let . Therefore, the rocket equation that describes the motion for any time t is simply where the minus sign indicates that the rocket moves forward since the rocket loses mass as it accelerates, therefore dM/dt will be negative for any rocket. If we look at the left hand side of the equation, we see the usual expression for force as mass times dv/dt or acceleration. So the equation tells us that the force on the rocket body is given by the amount of mass expelled and the velocity of that mass relative to the rocket body. Since other forces can simply be added to the rocket force, if there is an external force like gravity or wind resistance acting on the rocket, we simply add these effects together to get the net external force and rewrite our equation as where Fext is the sum of the external forces. The force, is what we call the thrust of the rocket. All that is missing for your degree in rocket science is the ability to apply the rocket equation to typical problems. One thing we usually want to know is how much fuel we need to get the rocket body up to a given final velocity. This is trivial to solve in Maple. #First begin by defining the rocket equation for Maple. Let's assume no #external forces for now. rocket := M(t)*diff(v(t), t) = -uex*diff(M(t), t); # #Now ask Maple to solve it for us with the initial condition being that #v(t=0) = 0, i.e. the velocity is initially zero. dsolve({rocket, v(0) = 0}, v(t)); # #Maple delivers the solution in a form which is not the most compact, we #can rewrite in such a way as to plot the result by making a function #dependent on x where x = M(t=0)/M(t), i.e. we rewrite the Maple result #from above as v(t) = uex * ln(M(t=0)/M(t)), then substitute in x to get #the final velocity as vf := x -> uex*ln(x); # #Now put in a sensible value for the exhaust velocity. Chemical rockets #normally cannot generate exhaust speeds above 4000 m/s, so let's put #in that value for uex and plot what we get uex := 4; plot(vf(x), x=1..100); # The logarithmic rise in speed predicted from the rocket equation shows that rocketry is very demanding technically. After about half the fuel is consumed, the rocket is only moving at uexln(2) or 70%of the exhaust velocity. Since x represents the original mass of fuel compared to the final mass at time t, a value of x = 100 means that only 1 percent of the original mass is left. At that time, the rocket is moving at about 18 km/s, or 40,000 miles/hour. This is a high speed, but not very practical since the payload of the rocket must be less than 1%of the original mass even neglecting gravity and air resistance. The limitations on uex come from the maximum temperatures that can be generated in chemical reactions. The most energetic reaction in use today is the combination of hydrogen and oxygen. The typical temperature here is about 10,000 degrees Kelvin. If you want to know more about this reaction, check here. Large values of M(0)/M only help moderately as we saw in our plots. Since the exploration of even the outer planets is very limited by maximum exhaust velocities less than tens of kilometers per second, some futurists seek reactions with much higher temperatures. One possibility is nuclear reactions such as fission or fusion. To do the differential equation with an external force, gravity for example, we simply note that the term -M*g must be added for a rocket which lifts off vertically. Our equation then becomes with solution If M0 is the initial mass of the rocket and Mfuel is the mass of the fuel at liftoff, then M0 - Mfuel represents the mass of the payload. The final burnout velocity of the rocket is then given by This equation ignores air resistance, the weight of the rocket body itself, and does not take into account the fact that the force of gravity reduces with height (we need integration to do that problem), but nevertheless is pretty accurate for characterizing most rocket launches, including the Space Shuttle. To see the rocket equation in action, do exercise 3-13. Next: What's Your Reaction? Up: CONSERVING EQUATIONS Previous: Momentum Conservation in
Up to this point we have viewed only problems in which the mass of the system remains constant. Newton himself recognized that situations in which the mass is not constant were still covered by his formulation of the Second Law of Motion even though such situations are less common. These cases always demand that we know how to handle differential equations.
One example where the mass does change is the case of rockets. Rockets are perfect for space travel because they carry their fuel and oxygen supply with them. In fact, most of the mass of the rocket on the ground is in the form of fuel and oxidizer. In space, the burning fuel is ejected from the rear of the rocket. This action produces a reaction force on the rocket body which propels it forward. Note that there is no need for any air to ``push against'' for the rocket to work. Newton's Third Law assures us that ejection of an object from the system MUST propel the system in the opposite direction. The propulsive force is referred to as the thrust of the rocket.
To derive the equation which describes rocket motion, we begin by translating words about the rocket's actions into equations. We begin by noting that the rocket expels gases from burning fuel. If we imagine that the rocket is in outer space, far from the earth, then there are no drag forces or gravity so there are no external forces on the rocket or the gases it expels. Therefore, if we consider the rocket and the exhaust to be a system, then, the momentum of this system is constant with time. Consider the figure below:
At initial time t, the rocket is moving with velocity and has a mass M. Thus the momentum of the system is just Mv and will always remain so since momentum is conserved for the system. After a small amount of time , the rocket will have expelled a mass of gas and the rocket body will have gained a small amount of speed . The change in momentum of the rocket body is
This equation says that the rocket changes momentum due to its loss in mass (which is just the mass of gas expelled) times its initial velocity plus the mass of the rocket times its change in velocity. To get the momentum of the expelled gas, we need to know its velocity. Typically, rocket engineers measure the speed relative to the rocket body at which gases are ejected as a way of characterizing the engine. If we call the speed relative to the rocket, uex, then the speed of exhaust relative to us as observers is v - uex. The total mass of the exhaust for this time is just , the mass lost by the rocket body, hence the net change in momentum of the exhaust is
where the minus sign indicates that the momentum change in the exhaust is opposite to that of the rocket body motion. Since momentum is conserved, the total changes in momentum of the rocket body and exhaust gases must cancel
We can develop this last equation into a differential equation by noting that our expression becomes more and more accurate a description of the rocket's motion for any given time if we let . Therefore, the rocket equation that describes the motion for any time t is simply
where the minus sign indicates that the rocket moves forward since the rocket loses mass as it accelerates, therefore dM/dt will be negative for any rocket. If we look at the left hand side of the equation, we see the usual expression for force as mass times dv/dt or acceleration. So the equation tells us that the force on the rocket body is given by the amount of mass expelled and the velocity of that mass relative to the rocket body. Since other forces can simply be added to the rocket force, if there is an external force like gravity or wind resistance acting on the rocket, we simply add these effects together to get the net external force and rewrite our equation as
where Fext is the sum of the external forces. The force, is what we call the thrust of the rocket.
All that is missing for your degree in rocket science is the ability to apply the rocket equation to typical problems. One thing we usually want to know is how much fuel we need to get the rocket body up to a given final velocity. This is trivial to solve in Maple.
#First begin by defining the rocket equation for Maple. Let's assume no #external forces for now. rocket := M(t)*diff(v(t), t) = -uex*diff(M(t), t);
# #Now ask Maple to solve it for us with the initial condition being that #v(t=0) = 0, i.e. the velocity is initially zero. dsolve({rocket, v(0) = 0}, v(t));
# #Maple delivers the solution in a form which is not the most compact, we #can rewrite in such a way as to plot the result by making a function #dependent on x where x = M(t=0)/M(t), i.e. we rewrite the Maple result #from above as v(t) = uex * ln(M(t=0)/M(t)), then substitute in x to get #the final velocity as vf := x -> uex*ln(x); # #Now put in a sensible value for the exhaust velocity. Chemical rockets #normally cannot generate exhaust speeds above 4000 m/s, so let's put #in that value for uex and plot what we get uex := 4; plot(vf(x), x=1..100); #
The limitations on uex come from the maximum temperatures that can be generated in chemical reactions. The most energetic reaction in use today is the combination of hydrogen and oxygen. The typical temperature here is about 10,000 degrees Kelvin. If you want to know more about this reaction, check here.
Large values of M(0)/M only help moderately as we saw in our plots. Since the exploration of even the outer planets is very limited by maximum exhaust velocities less than tens of kilometers per second, some futurists seek reactions with much higher temperatures. One possibility is nuclear reactions such as fission or fusion.
To do the differential equation with an external force, gravity for example, we simply note that the term -M*g must be added for a rocket which lifts off vertically. Our equation then becomes
with solution
If M0 is the initial mass of the rocket and Mfuel is the mass of the fuel at liftoff, then M0 - Mfuel represents the mass of the payload. The final burnout velocity of the rocket is then given by
This equation ignores air resistance, the weight of the rocket body itself, and does not take into account the fact that the force of gravity reduces with height (we need integration to do that problem), but nevertheless is pretty accurate for characterizing most rocket launches, including the Space Shuttle.
To see the rocket equation in action, do exercise 3-13.
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