Our example rocket has a mass of M kilograms and carries m0 kilos of fuel. As the fuel burns, it produces an equal mass of gas, which shoots out of the back of the rocket, causing the rocket to accelerate. But what velocity will the rocket reach once all the fuel has been used up?
The principle of calculus is to consider a large change - the burning of all the rocket's fuel - as the sum of lots of small changes. Consider burning a small quantity of fuel, of mass dm. The gas produced by the burning shoots out of the back of the rocket at velocity u. The momentum of the gas is given by its mass (dm) times its velocity (u).
Change in momentum of gas when it is expelled = u dm
Before the gas was expelled, the rocket was travelling at velocity v. The thrust from the gas causes it to accelerate, increasing its velocity by an amount dv. The total mass of the rocket, including the fuel that still remains in the tank, is (M+m).
Change in momentum of rocket = (M+m) dv
By the law of momentum conservation, the change in momentum of the rocket is equal to the change in momentum of the gas. That means we can combine the two previous equations to relate the tiny amount of fuel that was burned, dm, to the resulting increase in velocity, dv.
u dm = (M+m) dv
The increase in velocity that comes from burning all the fuel can be worked out by summing all of the small increases, remembering to adjust the value of m at each stage as the mass of unburned fuel that remains on the rocket decreases. The more steps that you split the total change into, the more accurate the answer will be.
This is where the real power of calculus comes in: it allows a completely accurate answer to be worked out. Calculus splits the total mass of fuel into infinitely small parts, so that the value of m changes continuously between each step. The sum can then be written as an integral.
Total increase in velocity = ∫ dv = u ∫ dm/(m+M)
This expression means "sum up all the terms, dm/(m+M), for values of m between m0 and 0." When evaluated, the result is
Increase in velocity = u log(1 + m0/M)
The astronomers can now use this formula to work out how much fuel the rocket needs to accelerate to the correct velocity.