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The Rocket Equation


The rocket equation is the governing equation for all of rocketry, it is derived from the conservation of momentum in a time varying mass system. That means that as the rocket uses more fuel, it gets lighter and therefore can accelerate faster. The rocket equation in its basic form is a differential equation, this equation can be manipulated and changed to include all possible forces; like gravity, drag, and many others. By including all of these forces one needs a computer to solve the problem, for our uses including no external forces, or just gravity is a very good approximation. The formula below is the rocket equation with no external forces.

Δv=b ln(m0/m)

Here Δv is the change in velocity, b is the gas velocity, m0 is the initial mass, and m is the mass without fuel. The gas velocity is related to the thrust, which is the basic measurement of a propellant. This says that the smaller m gets the bigger the change in velocity is, and therefore the faster the rocket moves. What matters here is the gas velocity and the size of m0/m. This can be rearranged as follows:

1-m/m0=1-exp(-Δv/b)

This is a more useful equation because we can find a percentage of the rockets mass that must be fuel for the rocket to reach a desired velocity. An example: To make it to low earth orbit a rocket must reach 9.7km/s Δv, so starting from rest and assuming a gas velocity of 4.5km/s, we get 1-m/m0=0.884, this says that 88.4% of the ships mass must be fuel, because 88.4% of the initial mass of the rocket was used up to achieve 9.7km/s and low earth orbit. This leaves 11.6% for payload and the rest of the ship.

To improve efficiency and the amount of payload a rocket can bring into orbit rockets use stages, this means that instead of one big fuel tank and 1 engine there are separate fuel tanks with their own engines and when the lowest tank is empty it detaches from the rocket. This is a typical setup, and a perfect example is the Saturn V rocket, which had 3 stages and an engine onboard the actual spacecraft. The rocket equation holds exactly the same as for a single stage but when the old stage is jettisoned the initial mass is reset to the total mass of the rocket minus the jettisoned stage. Here is an example of a 2 stage rocket with the same values as the single stage rocket. Suppose that the first stage should provide a Δv of 5.0 km/s; 1-Δv=0.671, therefore 67.1% of the initial total mass has to be propellant. The remaining mass is 32.9 %. After deposing of the first stage, a mass remains equal to this 32.9%, minus the mass of the tank and engines of the first stage. Assume that this is 8% of the initial total mass, then 24.9% remains. The second stage should provide a Δv of 4.7 km/s; 1-Δv=0.648, therefore 64.8% of the remaining mass has to be propellant, which is 16.2%, and 8.7% remains for the tank and engines of the second stage, the payload, and in the case of a space shuttle, also the orbiter. Thus together 16.7% is available for the payload and the rest of the ship. As you can see using two stages allows us to bring 5.1% more mass into space.

The form and function of the rocket equation under the effects of constant gravity are the same as for free space, so it will not be discussed other than to show it and explain the meaning of it. The equation is Δv=bln(m0/m)-gt. As you can see there is the term due to gravity added, which effectively takes a constant off the Δv from the free space equation.



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