The physics and mathematics of rocket propulsion
Image from Steve Bowers
A rocket is a device that moves in one direction by squirting part of itself in the other direction. Due to the conservation of momentum, if a rocket craft at rest shoots out a mass m of propellant with velocity Vex (the exhaust velocity), the remainder of the rocket, with mass M, will end up moving at a speed (m/M) × Vex. From this, it is immediately obvious that the more propellant a rocket carries, and the faster it squirts its propellant away, the faster the rocket can go.
One complication is that when a rocket squirts away some of its propellant, this will not just accelerate the rocket's structure and payload, but also all the propellant that has not been used yet. If the total amount of propellant is small compared to the mass of the payload, this will not make much of a difference, but the final velocity will be limited to much less than the exhaust velocity. If the rocket wants to go nearly as fast or faster than it is shooting away its propellant, it will need to carry a mass of propellant much larger than its payload mass. This, in turn, lessens the speed at which initial burns of the propellant can get the rocket going since these initial burns are pushing the mass of the extra propellant along with the rocket. The total amount by which a rocket can change its speed is given by
ΔV = Vex ln(M0/M)
where M0 is the initial mass of the rocket payload + structure + propellant, and M is the total mass remaining after it has burnt its propellant. The function ln is the natural logarithm, or logarithm to the base e ≈ 2.71828. Thus, for a rocket to get moving at the speed of its exhaust velocity, it will need to have an initial mass 2.71828 times larger than its final mass. To go twice as fast as its exhaust velocity, its initial mass will need to be 2.718282 ≈ 7.38906 times larger than its final mass.
Note that the ΔV found for M with empty propellant tanks is the total amount by which a rocket can change its velocity. If it uses all of its ΔV to speed up, it will have no propellant left to slow down again, and will be left forever drifting at its burnout velocity for the rest of time. Rockets without auxilliary forms of propulsion will need to carefully budget their ΔV for their mission.
The force being exerted is the rate at which the momentum changes with time. For a rocket, this is
F = m' × Vex,
where m' is the rate at which the rocket is losing propellant mass (for example, m' might be measured in kilograms of propellant per second).
The acceleration a rocket will experience for a given force is inversely proportional to the rocket's mass,
a = F/M.
Note that since the mass of a rocket will change as it spends propellant, the rocket can accelerate faster near the end of its mission than near the beginning.
| As an example, consider a rocket that shoots out 10 kilograms of propellant per second at an exhaust velocity of 5000 meters per second (typical of high performance chemfuel rockets). This gives a force of 50,000 newtons (1 N = 1 kg/s × 1 m/s = 1 kg × 1 m/s2). If the rocket has a mass of 10,000 kg, it will be accelerated at 5 m/s2, or about half the acceleration due to gravity on the surface of old Earth.|
It takes energy to get propellant moving. The kinetic energy of a mass m moving at a speed Vex is E = ½ m Vex2. Power is the rate at which energy is gained or lost, so the power required to squirt out propellant at a mass flow of m' and an exhaust velocity of Vex is
P = ½ m' Vex2.
Note that the force F = m' × Vex can be substituted into this relationship for power, to relate the power to the force and exhaust velocity
P = ½ F Vex
or the mass flow rate
P = ½ F2/m'.
All this assumes that the rocket shoots its propellant straight backwards, and that all the power goes into moving the propellant and not into residual heat in the exhaust or waste heat in the rocket. If some of the propellant goes to the side (perhaps the exhaust is shot out in a cone instead of straight back, so that some of the propellant goes a bit to one side or the other) it will not get quite as much thrust. This is the nozzle momentum efficiency. If the propellant exhaust stream exits the rocket hotter than it started out in the propellant tanks, it will take more power. This is the nozzle energy efficiency. With well engineered rockets, these are not usually large effects. For example, a rocket that shoots its propellant backward in a cone that is 1/10 as wide as it is long will still have 99.75% as much thrust as if it shot its propellant straight back. Rockets using solid nozzles have efficiencies as high as 70%. Magnetic nozzles can give efficiencies of up to 85% for thermal plasmas, or higher for some athermal plasmas. If the nozzle energy efficiency is denoted ζ and the nozzle momentum efficiency is denoted ξ then
F = ξ m' Vex,
P = ½ m' Vex2 / ζ
P = ½ F Vex / (ζ ξ)
P = ½ F2/(ζ ξ2 m').
Often, a rocket works by heating its propellant until it is a gas or plasma, and then allowing the hot, high pressure propellant to expand away from the rocket in the backward direction, pushing the rocket forward. As the propellant expands, its temperature drops until most of its initial thermal energy has been turned into kinetic energy. If you know the temperature to which the propellant is heated, and the mass of a known quantity of particles of propellant, you can find the exhaust velocity.
If you know the temperature T in kelvin, the efficiency ζ of the rocket, and the molar mass ρ (amount of mass in one mole, and 1 mole = 6.02×1023 particles) of the expelled propellant, the exhaust velocity will be
Vex = √ 3 ζ R T / ρ
where R is the gas constant R = 8.314 J/(mol K). The molar mass of a type of atom is called its atomic weight, and is usually listed in descriptions of that atom or isotope. Be careful - the value for R given above assumes you are measuring ρ in kg/mole, and the molar mass is usually given in grams per mole, so be sure to convert your units.
Often, the temperature will be expressed as an energy τ. If you know the rest mass of a single particle in the spent propellant expressed as an energy ε, then
Vex = c × √ 3 ζ τ / ε ,
where c is the speed of light in vacuum, c = 299,792,458 m/s.
For example, one atom of normal hydrogen has an energy of 938,791 keV due to its rest mass. If a fusion reactor heats hydrogen up to a temperature of 50 keV, and the hydrogen plasma is shot out through a magnetic nozzle at 90% efficiency to form a rocket plume, the exhaust velocity will be
c × √ 3 × 0.9 × 50 / 938,791 = 0.01199 c = 3,600,000 m/s.
Very commonly, a rocket's exhaust will be made of more than one kind of particle. In this case, use the weighted average mass of all the particles in the gas.
| For example, a fusion plasma might consist of 80% helium-3 with a mass of 2,809,440 keV per particle or 0.003016029 kg/mole, and 20% deuterium with a mass of 1,876,140 keV per particle or 0.002014102 kg/mole. The effective mass per particle is then 0.8 × 2,809,440 keV + 0.2 × 1,876,140 keV = 2,622,780 keV, and the effective mass per mole is 0.8 × 0.003016029 kg + 0.2 × 0.002014102 kg = 0.002815644 kg.|
The previous descriptions assume that the exhaust velocity and ΔV are both much less than the speed of light in vacuum. When this assumption breaks down, the analysis will need to be modified.
First calculate a quantity β = Vex/c. The total change in rapidity η available to the rocket is
Δη = β ln(M0/M).
Unlike velocities, at relativistic speeds rapidities do add. So a rocket can accelerate up to any rapidity givin by its Δη, and use any remaining Δη to slow down or change direction. For a given rapidity, the velocity is
V = c × tanh(&eta)
where tanh is the hyperbolic tangent,
and the time dilation factor for the rocket is
γ = cosh(η)
where cosh is the hyperbolic cosine
If you know the total energy E of a particle in the exhaust (including its rest mass-energy), then you can calculate a quantity
γex = E/ε
or, if you know the kinetic energy EK,
γex = 1 + EK/ε
where, as before, ε is the rest mass-energy of one particle. Once you know γex you can find β
| β = √|| |
1 - 1/γex2 .
For a photon rocket, this method works perfectly well - just use β=1; that is, Vex=c.
| For example, protons and antiprotons have rest masses of 938,280 keV each. When they annihilate each other, they produce 1,876,560 keV of energy. This annihilation produces, on average, 1.5 π+ mesons, 1.5 π- mesons, and 2.0 π0 mesons. A π+ or π- meson has a rest mass of 139,570 keV and a π0 has a rest mass of 134,980 keV. In total, then, proton-antiproton annihilation produces 1.5 × 139,570 keV + 1.5 × 139,570 keV + 2.0 × 134,980 keV = 690,000 keV worth of rest mass. Therefore, |
γex = 1,876,560 keV / 690,000 keV = 2.7.
β can be solved for,
β = √ 1 - (1/2.7)² = 0.93.
In actual practice, only the π+ and π- contribute to thrust (the π0 immediately decays into two gamma rays, that cannot be magnetically deflected). This reduces the available Δη by a factor of (1.5 × 139,570 keV + 1.5 × 139,570 keV)/690,000 keV = 61%.