# Toy Rocket Excel Sheet - Instantaneous Rest Frame

1. Mar 8, 2012

### joe_cool2

So, I need to make an Excel spreadsheet that realistically models the motion of a toy rocket. I need to take drag, gravity, and thrust into account, keeping in mind that the latter two will be influenced by the change in mass. The equation for force I must use, then, is:

F = ma + v(dm/dt)

I should just use the rocket’s instantaneous rest frame to say that v=0 and therefore F = ma. D_H, a PF mentor, says that the force inside this instantaneous rest frame is constant. However, because of another post by HallsofIvy, I am convinced that the force when measured from Earth’s reference frame is not constant. This does not make sense intuitively to me because as D_H says, the acceleration is constant regardless of refrence frame. However, it does make sense mathematically.

I will post HallsofIvy’s steps here, with a slight change, because I think there was a small error towards the end. I come to a slightly different conclusion, but the basic idea is the same, that the force when measured from Earth’s reference frame is not constant:

(at+ b)dv/dt+ av= F gives dv/dt= (F- av)/(at+ b) or dv/(F-av)= dt/(at+b).

Here is where I begin to differ a little from Ivy:

Integrating both sides: (-1/a)ln|F- av|+ C = (1/a)ln|at+ b|+ C

Raise both sides to e: e^(-1/a)(F – av) = e^(1/a)|at+b|

e^(2/a)(at+b)=|F-av|

So the question is whether I’m right about this: that I have to use the force that is calculated relative to the Earth’s reference frame. I think this is true because I need to take into account drag and gravity, so I need to add all of the forces together, keeping them in the same reference frame. After all, to calculate drag, I need to know the velocity, and in order to do that, I must consider the acceleration due to the combined effects of weight and force of drag (using the assumption that these forces cause some net acceleration, which causes a velocity, which causes drag.)

Am I wrong about this? Is there some reason that I actually can just use a constant thrust in looking at the net force? And, if I could do this, does a manufacturer saying that the toy rocket's engine exerts 40.9 N mean that, as D_H seems to suggest, the manufacturer is talking about the instantaneous rest frame?

Last edited: Mar 8, 2012
2. Mar 9, 2012

### D H

Staff Emeritus
Why is that the equation you "must" use?

There are three choices for addressing variable mass systems in the context of Newton's laws:
1. $\vec F=m\vec a$, in which case thrust is an external force,
2. $\vec F=\frac{d\vec p}{dt}$, in which case force is not frame invariant, or
3. Newton's laws are valid only for constant mass particles, in which case Newton's laws do not apply (at least not directly) to variable mass systems.

Your definition of force, $\vec F = m\vec a +\frac{dm}{dt}\vec v$ corresponds to choice #2. This yields a force that is neither frame invariant nor equal to thrust. Most people who work with rockets use option #1 or #3.

3. Mar 10, 2012

### joe_cool2

So, essentially what you're saying is that I should use option 1, assuming that the mass changes every step in the excel file, because the thrust is defined from the instantaneous rest frame?

It seems like that would be on par with my level of experience in physics. But I wonder what option #3 entails?

And, I gather from your assessment of option two that the force actually does change from the frame of the Earth, and that it is not equal to thrust. So, in what context would option number 2 be used?

One final question: you're saying that option 1 treats the force as being external. By this, do you mean to imply that option 2 treats the force as an internal one?

4. Mar 10, 2012

### D H

Staff Emeritus
What follows is a slightly out of order set of responses.

That works, at least in the simple case where the location of the center of mass of the rocket is fixed within the rocket. I have more to say on this further on in this post.

Options #1 and #2 are identical in two situations. One is a system of constant mass where dm/dt=0, making the argument of F=ma versus F=dp/dt a case of arguing over how many angels can dance on the point of a pin. The other is the instantaneous rest mass of the object in question. In this case, v=0, so the v*dm/dt term vanishes.

In the general case of a non-constant mass and a non-zero velocity, force defined by F=dp/dt is not equal to thrust. The advantage of using F=dp/dt is the obvious connection with momentum. The disadvantage is that force becomes a frame dependent quantity. The avantages and disadvantages reverse with respect to option #1. That force is no longer the time derivative of momentum means the connection with the conservation laws is a bit tenuous.

No. The forces F in F=ma and the F=dp/dt are net external force acting on the system. Both options come up with the same answer for force in the case that v=0.

There's a deeper problem here in dynamic mass systems: What is this thing we call force?

This is the option where one says that force doesn't quite make sense for systems of dynamic mass. Instead one uses the conservation laws as the basis for analyzing behavior. The conservation laws are derived from Newton's laws in basic Newtonian mechanics. If you proceed in physics you will learn Lagrangian and Hamiltonian mechanics, where it is the conservation laws that are fundamental1.

Here's a simple but somewhat nonsensical example that illustrates why the concept of force doesn't quite make sense for a non-constant mass system. Imagine a non-rotating rod in empty space, far, far removed from any external influences. Because the rod isn't rotating and isn't subject to any external forces, the rod will be stationary with respect to some inertial observer.

Now let's do something stupid: We'll construct a system boundary such that only part of the rod is inside the system boundary. To make matters worse, we'll make this system boundary vary with time. For example, suppose the system boundary is a cylinder that starts at one end of the rod and encompasses the circumference of the rod, but that the height of the cylinder varies sinusoidally with time between 0 and the length of the rod (e.g., h=l/2*(1+sin(ωt)), where l is the length of the rod).

By construction, there are no forces acting on the rod. There aren't even any internal forces anywhere inside the rod, either (i.e., the rod isn't under stress). Yet with this intentionally silly system boundary, the center of mass of the portion of the rod that lies inside the boundary behaves as a simple harmonic oscillator. One can use the conservation laws to explain why the tip of the rocket that is always inside the system boundary is not accelerating.

Now imagine replacing this rod with a rocket, with the rocket proper corresponding to the part of the rod inside the boundary, and the exhaust cloud corresponding to the part of the rod outside of the boundary. What's the difference between this rocket+gas cloud and the rod?

1Aside: There are situations where Newton's laws fail but the conservation laws still hold. The conservation laws are in a very real sense much deeper than Newton's three laws. Noether's Theorem in turn is even deeper than the conservation laws.

5. Mar 11, 2012

### joe_cool2

Well, I haven't looked at conservation of momentum stuff since high school, and my current class hasn't gotten there yet, so I'm not exactly sure how the laws can be used to explain why the tip isn't accelerating.

But, if that last question you posed isn't meant to be rhetorical, I would answer that unlike the rod, the rocket's center of mass would not vary sinusoidally, and therefore the center of mass would not cycle between one part and the other, but it would simply move towards the gas cloud.

And since you're making an analogy, I tend to conclude that the rocket-gas system is analogous to that of the rod, that the tip of this changing system boundary is not accelerating.

But I don't know for sure. Does the way in which the center of mass changes change the behavior of that end point? I really don't have a way to analyze this with my present knowledge.

I'll probably learn more in my class, and if the explanation is too much trouble, it is likely that I'll be able to understand more later. I apologize for any lack of understanding on my part, but it is purely a matter of not having the time to re-learn this, as I have midterms coming up. That is to say, my original question has been answered sufficiently, and you need not give an in depth explanation if you feel it is better for me to learn this on my own later on.

Last edited: Mar 11, 2012