dragestimationfluid dynamicshomework-and-exercisesrocket-science
I'm trying to model a rocket's position with air resistance as a factor. The rocket is fairly low-powered (Estes A8-3, 9.7N of thrust for 0.7 seconds), and I estimate it won't be going faster than 50 metres per second.
Should I be using linear or quadratic air resistance for this?
I found online that linear air resistance is used for relatively low velocities. How low is 'low velocity'?
Specific impulse is usually defined as $I_{sp} = \frac{F_T}{\dot m ~ g_0} $
That's true only if you use standard metric units. With force expressed in pounds-force and mass flow rate expressed in pounds, one simply divides the force (in pounds-force) by the mass flow rate (in pounds/second) and voila! you have specific impulse in lbf·s/lb. For example, the first stage of the Saturn V produced 7,715,150 pounds-force of thrust at launch while consuming fuel at a rate of 29,157.58 pounds/second. Divide 7,715,150 by 29,157.58 and you get 264.6, the specific impulse at launch. Properly, this value of 264.6 is in units of lbf·s/lb. If you do the math, it is also numerically equal to the specific impulse in seconds.
Alternatively, one could convert that force to newtons and the mass to kilograms, yielding 34.3817 meganewtons of thrust and 13.22565 metric tons per second of fuel consumption. Now the division yields 2594.9 m/s. Divide by g0=9.80665 m/s2 and you get 264.6 seconds.
A second alternative is to convert that force in pounds-force to kilogram-force. This conversion yields a force of 3,499,530 kilograms-force. Now we're back to using the trick of simply dividing the force (in kg-f) by the mass flow rate (in kg/s): 3499530/13225.65 = 264.6.
Germany was the leading European developer of rocketry up until 1945. They used kilogram-force to express thrust rather than newtons. The Americans and Russians took over after that point. The Americans stuck to using customary units, pounds-force and pounds-mass. The Russians followed the German tradition of expressing thrust in kilograms-force. In both cases, simply dividing force by mass yields specific impulse in seconds.
In fact, despite being a banned unit, most European aerospace engineers tended to express thrust in kilograms-force rather than newtons up until the 1980s, and some still do. It is a very convenient unit for spacecraft and aircraft that operate near the Earth.
I apologize in advance, but I'm going to make the problem look worse first so that I can explain it:
Imagine that instead of you measuring the velocity from the station where the rocket is launched - let's call it station A - you measure instead from station B, which is moving away from station A at some high speed. And imagine that station A launches two identical rockets, one toward you and one away from you.
Since we're operating under the well-tested current theory of special relativity, we understand that we can either say "B is moving away from A" or with equal correctness "A is moving away from B" since we can declare either one as "stationary" for purposes of measuring velocities with respect to our chosen coordinate system. So let's start over and say that Station A is moving at speed away from Station B and launches two rockets.
Of course, before they launch from A, both rockets will have a certain kinetic energy based on their motion, and that energy will be equal for the two identical rockets. When we launch them, however, that begins to change: one of them is starting to move away from us even faster, while the other begins moving away from us slower. After some time, in fact, one will be moving away from us at twice the speed it originally had - that is, four times the kinetic energy - while at the same time the other is moving away from us at zero relative velocity: no kinetic energy whatsoever! Now we're really mixed up, because both rockets did the exact same thing but one ended up with a very high kinetic energy and one with no energy whatsoever.
The issue I hope to highlight with that example is that kinetic energy is frame-dependent and not an intrinsic energy for an object. That's why you can safely ride along in a car moving at 100km/h along a road but not be safely struck by a car moving at 100km/h along a road; if you travel with the car, it has zero kinetic energy in your frame, but if you're standing on the road it has a very high kinetic energy in your frame. Kinetic energy is relative.
On another note, I said before that the two rockets are doing "the same thing," but in any given single frame of reference that's not quite true. Obviously, in my example above, from the frame of station B one rocket is losing kinetic energy and the other is gaining kinetic energy, for example. But that's only considering the rockets themselves: those are being propelled by launching photons the other direction, and those photons are expelled with a certain energy themselves. If you were to measure the energy of the photons themselves, (attaching a small mirror to the back of the rocket headed toward you so that you can see the "exhaust") you would notice that over time the exhaust of the rocket moving away from you seems less energetic and the exhaust of the rocket moving toward you seems more energetic. That is to say, photons launched toward you are less energetic than photons launched away from you... which was the same conclusion we came to with the rockets themselves launching from Station A, and again is a consequence of the relativity of kinetic energy. But this time we can more directly say that one is "red-shifted" and the other "blue-shifted."
EDIT: It may also be simpler to imagine the solution if you use a normal reaction drive with massive exhaust. The exhaust is low mass but leaves at high speed; the rocket is higher mass and thus gains less velocity from the reaction. The kinetic energy increases by the same amount in opposite directions for each body of mass (taking all the exhaust as a "body"), otherwise we're getting energy for free from somewhere. Also, since massive exhaust has subliminal expulsion speed, you can have a case where the rocket is accelerating away from you but the exhaust is also moving away from you, which isn't possible with photon drives. But in all cases, photons or not, the total energy of the system is conserved: measured from an inertial coordinate system, the rocket "gains" the same amount of energy that the exhaust "loses," and the measured exchange rate does increase quadratically.
Best Answer
The quantity to check is the Reynolds number $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density of air, $v$ is the relative velocity, $L$ the typical dimension of the rocket and $\mu$ the (dynamic) viscosity.
For small Reynolds numbers, i.e. $Re\ll 1$, viscous forces dominate and you have laminar flow. In this regime, the drag force is proportional to the velocity. For large Reynolds numbers, drag is dominated by turbulences and is proportional to the velocity squared.
Plugging in some numbers: $\rho \approx 1~\textrm{kg}~\textrm{m}^{-3}$, $\mu\approx 2~ 10^{-5}~\textrm{kg}~\textrm{m}^{-1}\textrm{s}^{-1}$, $v\approx 20~\textrm{m}~\textrm{s}^{-1}$, $L\approx 1~\textrm{m}$, you get a Reynolds number on the order of a million, which is much larger than 1. This suggests (as expected) that you are in the turbulent regime and should use the force proportional to the velocity squared.