Easy rocket movement

Today, in a slightly more technical and, above all, mathematical, I want to talk about a problem that seems very simple and whose results, especially if you come from middle studies may be curious. To do this example, obviously, let’s simplify the problem: the movement of a rocket.

As you know, a rocket works very simply; “Burns” fuel by expelling gas which, by Newton’s third law (action and reaction) drives the rocket.

Imagine that the gases that the rocket ejects are at a constant speed (then the combustion is always the same – first simplification) and therefore the mass of the rocket varies by a percentage. Yes, in addition, as a second simplification, we obviate external forces (such as gravity). Can we calculate the speed of this rocket? This is called the variable mass problem and is a typical basic problem of vector differentiation.

As external forces do not act we say that F=0.

We will take into account the moment p or linear moment, p=mv.

With these two things we have enough to solve the problem.

Let us operate with force, which is the differential in the time of the linear moment.

From here, we draw the known equation, as long as the mass is constant, which gives F=ma, but this is not our case since the mass is not constant.

Let us call vg the velocity of the gases ejected by the rocket (to differentiate it from the velocity v of the rocket itself).

As we have said that external forces do not act and therefore F=0 using what we have drawn above about the linear moment, it will remain constant, so that we calculate the moment of all forces. For this we will think that mass is a function of time (because it varies in time) and that velocity, also, for this reason at time t the mass will be m(t) and velocity will be v(t). For now everything very simple and I do not think I have lost anyone along the way since the only thing we are doing is to have what we have, laying the groundwork.

At another later time, an increase in t, mass and speed will have changed:

here, obviously, the mass increase will be negative (or less than 0, depending on how you want to see it) because the rocket loses mass by “releasing” it.

The ejected gases will have the mass that has been lost (the increase of m) and the velocity will be v+vg we will have as speed the speed of the rocket and his own.

And now that we have already defined everything we use that the linear momentum is constant with which, we take the moments of each thing in the increase of t (rocket and gases) and in t (rocket alone):

And now comes the fun part, raised the equation, play with it, for this simplify and divide everything by increasing t and take the limit when said increase tends to 0.

Where, the fun of the matter and that you have to remember is that when an increase tends to zero, we are deriving from all the life.

Now, in order to solve this differentiation, it remains for us, although we already know much about how the velocity of the rocket varies according to the mass of the gases that are expelled, let us imagine and imagine two times, t=0 and t=T.

Where, in all this kernel the only thing that has to be taken into account is that both v and vg are vectors.

Now, if you look, the logarithm is negative… negative!… impossible!. No, it is not impossible, just remember that the mass at t=0 is greater than the mass at t=T and therefore, the value of the logarithm is positive.

That is to say, as you see, in the end, in spite of being vectorial differentiation, in the end, we have not differentiated or integrated any vector but, as I said, we must take into account that the speed (of the rocket and the gases) is a vector. So although it sounds very complicated, it is not as difficult as it seems, although this example is very simplified (we have not put any restrictions on the speed that the rocket can reach and we know that relativity is our friend).

And, the last, I will find how to put LaTeX ecuations here because it will be better for all of us, even if you have friends say that “what is LaTeX for if no one uses it” by shooting with contempt of the things that they do not value them without thinking that there can be other people for whom it is fundamental. 😉

1 Comment

  1. I see a fault on the resolved ecuation where I take t=0 and t=T… is not ln(m(9) it’s ln(m(0)… number 9 is next to 0.



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