- 3033 - ROCKET REACTIONS - what are the physics, the math? - Rockets are very interesting if you want to learn physics. Rocket designs get right down to the fundamentals.
----------- 3033 - ROCKET REACTIONS - what are the physics, the math?
- It all starts with the law of motion that says:
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------------------- “Action = Reaction”.
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------------------- Action in physics means the product of energy and time. (A=E*dt)
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- To learn more let’s put a rocket in outer space with no other forces acting on it. We will burn the fuel and determine the velocity of the rocket. The rocket fuel burns at a constant rate. It exit’s the exhaust at a certain rate that is constant in kilograms per second.
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- How can the rocket go faster that the speed of the exiting gas? Can we use a heavier gas and get more thrust from the rocket? Can we use ion propulsion to generate higher velocity from the exiting gas and get a faster rocket? Can we use photons exiting the rocket traveling at the speed of light and get light speed from our rocket?
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- Because Kinetic Energy is ½ m*v^2, which is ½ m*v*v, which is momentum times velocity, like action = reaction, the Conservation of Momentum is basically , again, the Conservation of Energy.
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- The Conservation of Momentum means that mass of the rocket * the velocity of the rocket in one direction = mass of the gas * the velocity of the gas in the opposite direction.
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- Initially the rocket has an original mass, “Mr“, and we light the fuel and the gas exits out the back at a velocity, “Vg“, at a rate of so much mass per second, “dm/dt“, the amount of mass of gas per unit time. Over the next instant of time, “dt“, the rocket increases its velocity, “dv” by ejecting the mass of gas, “dm“, which is equal to “dm/dt * dt“.
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--------- Rocket momentum = Exhaust gas momentum
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---------- Mr * dVr = Vg * dm/dt * dt
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----------- dVr = Vg * dm/Mr
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---------- The differential of the velocity of the rocket = the ratio of the velocity of the gas * the constant rate mass of gas exiting per second to the mass of the rocket.
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- Another way to say this is the rate of change of the velocity of the rocket = the ratio of the velocity of the gas * the constant rate mass of gas exiting per second to the mass of the rocket.
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-------- We have to use calculus to solve this problem because over every instant of time the rocket is loosing mass and getting lighter. If we integrate the differential of the velocity of the rocket we can get the velocity of the rocket at each instant.
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- To get the final velocity we have to integrate, or sum up, the rate of changes in velocity over the entire range from the initial mass of the rocket to the final mass of the rocket, which is much less after the fuel is all burned up.
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--------------------- Integrate: dVr = Vg * Integrate: dm/Mr
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--------------------- Vr = Vg * Integrate: (1/Mr) *dm
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---------- We can put “Vg” outside the integral because the velocity of the exiting gas is a constant.
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- Now we have to go to the Calculus books to learn the Integral of the function (1/m) dm: Calculus is the math you use for something that is continuously changing.
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- Calculus is the math of change. It there is change in the product of two quantities that is area ( length times width). The summation of areas is Integration. Example: Action = Energy * dt, kilogram*m^2/sec.)
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- If there is a ratio of two quantities, this is a slope or rate of change, and that is differentiation. Example: Velocity = ds / dt, meters per second.
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--------- The Integral of 1/x dx = natural log of x
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---------- Therefore: Vr = Vg * natural log of m
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---------- If we have a rocket mass of 1000 than the natural log of 1000 = 6.9 so the final velocity of the rocket will be 6.9 times the velocity
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- The Integral of dm/m can be looked at two ways, as a ratio or as a product of (1/m) dm. The Intergral of dm/m is the natural log of “m”. The natural log is the exponent to the base “e”, whare “e” = 2.71828.….. The Intergral of e^x*dx is e^x.
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- The differential of e^x is e^x*dx. e^x is the only function that has this of the gas exiting the rocket. If the mass of the rocket was 10 then the velocity of the rocket would reach the natural log of 10 = 2.3 times the velocity of gas exiting the rocket.
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- The more rocket fuel you have the faster the rocket will go. But, there is a trade-off here. It takes more energy to lift a heavier rocket. It takes more thrust from the rocket to accelerate it if the rocket is more massive. And, the more energy you use the faster your fuel is used up.
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- The velocity of the rocket is directly proportional to the velocity of the gas, or mass per second, exiting the rocket. Why not make the exiting mass go faster out the tail of the rocket to make the rocket go faster? Let’s design an ion rocket.
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- Chemical reactions are limited to a few electron-volts of energy. We can set up a negative electric charge at the rear of the rocket to 200,000 volts and have positive cesium ions at the front of the rocket. The cesium beam of ions would accelerate out the back of the rocket with an enormous velocity. Let’s recalculate the velocity of the rocket with this design.
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- The formula for the Conservation of Momentum is the same:
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---------------- Momentum of the rocket = Momentum of the ion exhaust.
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------------------ Mr * dVr = Vg * dm/dt * dt = Vg * dm
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------------------ The thrust is the force that moves the rocket and force is the rate of change of momentum with respect to time. Thrust is momentum / second. Or, in Calculus language, it is the first derivative of momentum with time. ( F=dm*v/dt)
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----------------- Trust = Vg * dm/dt
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----------------- Thrust = the velocity of the ions * the rate of mass exhaust / second
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- To find the velocity of the cesium ions we need to use the Conservation of Energy, again. The Kinetic Energy of the fast moving ions must equal the Potential Energy caused by the high electrical voltage, 200,000 volts.
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- Kinetic Energy is ½ m*v^2. Potential Energy is the one electron charge of the cesium ion * the 200,000 volts. A volt is the energy per unit charge, “q”. Therefore, Potential Energy = q* Volts.
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----------------------- Kinetic Energy = Potential Energy
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----------------------- ½ m * v^2 = q * Volts
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---------------------- v^2 = 2 * Volts * q / m
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A mole is a unit of measure of the quantity of atoms. In this case cesium atoms. One mole = 6.02*10^23 atoms. There are 96,500 coulombs of charge per mole. And, cesium’s atomic weight, which is mass per mole is 0.133 kilograms per mole.
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- What we need is the ratio so we will soon get rid of those pesky moles. 96,500 coulombs / mole / 0.133 kilograms / mole = 725,564 coulombs / kilogram = q / m.
A volt is a joule of energy / coulomb. So,
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------------------ v^2 = 2*200,000 joules / coulomb * 725,564 coulombs / kilogram
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------------------- v^2 = 29 * 10^10 joules / kilogram = 29*10^10 kg * m^2/(sec^2 * kg)
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------------------ v = 5.4 * 10^5 m / sec.
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- The velocity of the ions is 540,000 meters per second, or, 1,200,000 miles per hour. That is 200,000 times faster the then the gas exhaust from the chemical rocket. So, with that extreme velocity this ought to be a great rocket.
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- But, before we jump to that conclusion let’s calculate how much energy it takes to get that much thrust from ion propulsion. The ions are a positive charge and, in effect, an electric current exiting the rocket.
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- If the current was one ampere, which is one coulomb of charge per second, that would be the reciprocal of the 96,500 coulombs per mole, which is 1.036*10^-5 moles / second. And, the atomic weight of cesium is 0.133 kilograms per mole. Multiplying these together gives us 1.378*10^-6 kilograms / second / ampere. Multiply this times the velocity, 5.4*10^5 m/sec and we get 0.74 kilogram* m/sec^2 / ampere.
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- In other words we only get ¾ of a newton of force / ampere from these ions. That is not very much thrust per ampere. Electric Current * Voltage = power. Amperes are coulombs per second * volts are joules per coulomb = joules per seconds which is watts of power.
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- One ampere * 200,000 volts = 200,000 watts. In other words, the ion propulsion rocket is giving us ¾ of newton of thrust per 200,000 watts of power. That is 3.7 *10^-6 newtons per watt. This is not a very good rocket because it takes so much energy to get the thrust we need.
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- What if we used photons instead of cesium ions. Photons have the fastest velocities of anything in the Universe. The velocity out the back of the photon rocket would be the speed of light.
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- We know that E = mc^2 = m*c*c = momentum * c. The momentum / energy = 1/c would be the thrust to power ratio for our photon rocket. That is 3.3*10^-9 newtons / watt.
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- That is even worse than our ion rocket. It is 1000 times worse than our ion rocket and 1,000,000 times worse that our chemical rocket. You probably knew that because you do not feel much of a thrust backwards when you turn on a flashlight. It looks like the chemical rocket is our best rocket design after all.
- If the same energy per atom is released, since kinetic energy = ½ mv^2, then, the lighter the atom we use in our chemical fuel the faster the velocity, and energy is velocity squared.
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- Therefore, for the chemical fuels, we should use the lightest atoms possible. That would be hydrogen and helium. Unfortunately helium won’t burn very easily so we are stuck with hydrogen and oxygen as the best choices available.
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- Or, hydrogen peroxide and oxygen. Next you design the rocket nozzle to get the highest percentage of chemical energy transferred into outgoing velocity. There you have it the best rocket design in thrust / energy and the physics behind why that is so.
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- Reactions? What is your reaction? Could you follow the physics? Could you follow the math? My experience is that you have to have a pencil and paper and do it yourself before it starts to make sense. That is the primary reason I write these reviews. I learn better through my fingers than I do through my eyes, or my ears.
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February 6, 2021 ROCKET REACTIONS 887 3033
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--------------------- --- Sunday, February 7, 2021 ---------------------------
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