2021-09-11 20:10:25
THE ROCKET EQUATION
Derivation & Explanation
History and Introduction
In this post we are going to look at something known as the Tsiolkovsky Rocket Equation. As its name suggests, it applies to something called a "rocket”. This is what we mean by it:
We refer to a “rocket” as anything that can accerate itself due to the conservation of momentum using thrust by expelling part of its mass with high velocity.
The vast range of different things that the word "rocket” can refer to made it especially easy for people to derive well before the invention of the first chemically propelled rocket.
Chemistry of Rockets
Modern rockets burn chemical fuel which relies on redox reactions such as that between hydrazine (N2H4) and dinitrogen tetroxide (N204):
2N2H4(l) + N2O4(g) —> 3N2(g) + 4H2O(g) + Heat
In this process, very large amounds of heat and gases are produced and ejected to produce thrust. Suppose the fuel expelled is made up of pellets. At one moment in time, a pellet of mass dm is expelled at speed 126. By Newton's 3rd Law:
F₁₂ = -F₂₁ ⇔ m₁v͘₁ = -m₂v͘₂
What this means is that there exists a force between the pellet and the remaining rocket. From this, we can figure out a conservation law that will come in handy when deriving the rocket equation.
Conservation of momentum
The trick is to integrate Newton’s Third Law:
∫ m₁v͘₁dt = ∫ -m₂v͘₂dt
We can rearrange it to have m₁v͘₁ + m₂v͘₂ = C. The familiar mv terms are the expression for momenta and so the expression we obtained above is a statement of conservation of momentum —— This is the secret behind rocket flight.
In the example above, initially, a block connected to a compressed spring is placed on a cart at rest. Moment later the spring rebounds and ejects the block, resulting in the cart moving in the opposite direction of the block. This is conservation of momentum in action and everything behind rocket propulsion. Now let’s move onto the real deal.
Derivation
For a moment let’s suppose that the amount of fuel left in the rocket decreases linearly. This way the exhaust velocity 128 in the roket’s frame of reference remains constant throughout the flight.
Initially, the system has the momentum P₀ = mv . When it exhausts a pellet of fuel of mass dm and speed v - vₑ in the frame of an outside observer, the system momentum becomes
Pf = (m - dm)(v + dv) + dm(v - vₑ)
In the absense of any external force, by the conservation of momentum we equate the initial and final momenta of the system to have:
mdv = -vₑdm
This is negative because the mass of the rocket decreases.
As the mass of the rocket changes from m₀ to mf, its speed changes from v₀ to vf. Setting these as the bounds of integration:
∫ᵥ₀ᵛᶠ dv = -vₑ∫ₘ₀ᵐᶠ dm/m
Performing the integration gives us the result:
vf - v₀ ≡ ∆v = vₑ ln (m₀/mf)
This known as the Tsiolkovsky Rocket Equation, or the rocket equation for short. It first was introduced in 1896 by Konstantin Tsiolkovsky, who applied it in order to explore the possibilities of space travel. Plotting the equation on a graph.
Some Analysis
From the plot, we notice that for the ∆v, aka the change in velocity of the vehicle to be greater, the demand for mass ratio m₀/mf increases exponentially. All this is when we neglect external forces.
What this means for us is that we’d always want the wet mass mg to be as high as possible. Usually, this portion is mostly comprised of propellants.
With Gravitation
Due to the rocket's weight in the gravitational field,
dJ = —mg dt
Since impulse equals to the change in momentum,
dP= Pf —P₀ = —mgdt
The expressions for initial and final momenta are the same as before, so we have:
mdv+vₑdm= -mgdt
The expressions for initial and final momenta are the same as before, so we have:
vf - v₀ ≡ ∆v = vₑ ln (m₀/mf) - g∆t
This is the rocket equation for a rocket inside a gravitational field. The new ∆t term tells us that the longer the burn time is the smalled ∆v will be. This is the reason why rockets carry so much fuel.
#Физика
#الفيزياء
#Mohanned_Qasim
443 viewsالأستاذ مهند قاسم, 17:10
𝑴
