Prerequisites:
- Knowledge of vectors and scalars
- Basic kinematics, Newton’s laws, conservation of linear momentum
- Fundamental knowledge of limits, derivatives, exponentials/logarithms
- Integrals as inverses of derivatives and the difference between definite and indefinite integration
- Basic techniques in integration and differential equations
- Newton’s Law of Universal Gravitation (Only for problem 3)
- The quadratic drag equation (Only for problem 3)
The goal of the derivation of the rocket equation is to be able to give the velocity of a rocket at any moment in its trajectory, purely as a function of the mass ejected from take-off or the time elapsed (for the force-less case, other forces may cause velocity to gain dependence on other factors). In addition, we can reflect on what could’ve been changed in the derivation in order to make the solution more precise or to apply it in a different setting.
To use the conservation of momentum, we must compare the initial and final total momentum; however, there is a minor complication in this. Mass is not ejected one-by-one in a rocket, instead it is fired continuously. Therefore, we approximate the continuous process by assuming that the exhaust gases are released with a mass loss of -Δm (Δm is negative) at a constant speed of u (relative to the rocket) for every passage of Δt. The discrete process can be turned continuous by taking the limit as Δt goes to zero of the equation while keeping the ratio Δm/Δt the same.
Slight but important nuances must be realized in the last term in equation 2. Firstly, since the constant exhaust velocity is measured relative to the rocket, a velocity addition must be applied as the frame of reference being considered is the Earth (or wherever you are measuring the rocket’s velocity from). Secondly, u is the constant speed of the exhaust gases which travel downwards, which is represented by the minus sign in front of the u.
By Newton’s 2nd Law, we claim that dP/dt = Fₑₓₜ . We can find the derivative of momentum from our existing equations by using the limit definition of the derivative.
The first fraction in the limit becomes mass times acceleration since Δv goes to zero as Δt does. The same logic applies with the third fraction to give the derivative of the rocket’s mass with respect to time. However, in the second fraction, there are two terms in the numerator which go to zero, allowing us to write the term as a product of a term that goes to zero times a finite derivative, making the whole fraction go to zero. After all this, we obtain the following differential equation.
In order to solve this equation, we must first find the net force on the rocket or determine whether it is a constant or not, as we will have to integrate with respect to time to solve for v or m. For instance, if the rocket is in an empty vacuum and far away from any planets, the net force would be zero. If the rocket is stationed on the surface of the Earth, then air resistance and gravity will be acting against the rocket’s motion.
We will first solve the equation for the case in which the net force is zero:
Using the chain rule, we can see that the equation loses its time-dependence.
Then, using the method of separation of variables and definite integration, using the initial conditions for the limits of integration, we can find the velocity as a function of the rocket’s mass.
The prime symbol is to make the differences between the dummy integration variables and the final velocity v and final mass m clear.
Now for the case with gravity acting on the rocket.
For simplicity, we will assume that the rocket experiences Earth’s gravity as a multiple of its mass mg. However, this is only an approximation which is valid when the distance traveled by the rocket, in the duration in which the thrust is engaged, is too small to have a significant change on the gravitational force experienced by the rocket. This derivation will be omitted as the process is very similar to the Fₑₓₜ = 0 case.
If we assume that dm/dt is constant, we can write the velocity as a function of time only by introducing an additional parameter: γ =-dm/dt.
Now with the differential equation solved, we can explore some interesting properties of the rocket equation. For instance, we can figure out a rocket’s maximum speed based on the total mass of the rocket and it’s fuel, the mass of the rocket without the fuel, and the exhaust speed.
Using the differential equation in terms of Fₑₓₜ, we could also find the velocity for a number of different forces acting on the rocket. However, the complexity of the differential equation can become overwhelming very quickly, which may necessitate the use of approximative methods.
An interesting application of this equation is predicting the height that a rocket could reach. This may be used as a very crude estimation for fuel needed to reach an orbit around Earth. It is crude because it neglects air resistance and the variation of the gravitational field strength with distance from the Earth. Another application of the differential equation is to calculate the acceleration experienced by the rocket, which could be useful in determining whether the rocket — and its potential inhabitants — will be able to withstand it.
If you’d like to, try these problems below that apply the equations in different ways:
- Derive equation 8 or 9 using F = ma, where m is the rocket’s mass and a is the rocket’s acceleration, and thus determine the thrust of the rocket (the reaction force that accelerates the rocket upwards) . (Difficulty: Easy)
- Using equation 16, determine the displacement of the rocket at time t above the surface of the Earth given that the rocket was initially positioned on the surface of the Earth. (Difficulty: Easy)
- Using equation 6, assuming constant γ =-dm/dt , derive the differential equation that describes the motion of a rocket, including drag (F = -kv²). Can this differential equation be solved by separation of variables, integrating factor method, or by a substitution? (Difficulty: Moderately difficult)
- Using equation 16, determine the set of conditions on u and γ such that the maximum acceleration experienced by the rocket does not exceed ng where n is a positive integer and g = 9.81 ms⁻¹ in terms of the total time where the thrust is engaged and n. (Difficulty: Moderate)
Thank you very much for reading my first article! Please let me know what you think and follow me if you enjoyed it or found it informative. There may be a follow up that explores solving the equation with air resistance, which will be analyzed using numerical and approximative methods.
