Posted on 03/28/2006 12:09:01 PM PST by orionblamblam
Again, *relative* rotation. You keep assuming that there is some essential difference and there is not. Ernst Mach proved that back in the 1800's. Your assumption that you could launch stright north and be in a polar orbit is simply not true.
The issue is *relative* rotation. I can understand why you refuse to admit that, however. You have no poit if you do.
It seems intuitive that if it launches due east from 26 deg latitude, the 'natural' inclination will be ~ 26 degrees.
You can show differently?
> Your assumption that you could launch stright north and be in a polar orbit is simply not true.
That's true on Real Earth, sure. But on your mythical non-rotating Earth, it would be true for any location on Earth.
> The issue is *relative* rotation. I can understand why you refuse to admit that, however.
No idea what you're talking about.
>Ok, so what is the big, big, BIG, Huge! difference?
About 1040 mph.
At least, that’s the difference for Alcantara launch facility in Brazil, and it qualifies as a big, big, BIG, huge difference.
Of course, I’ll have to show you how to derive this, so here goes. . .
Case 1: Stationary, non-rotating reference frame fixed to the Earth (your geocentric model).
There’s no rotation of the Earth in this model, so a rocket on a launch pad anywhere in the world is completely stationary when at rest. This means that it can launch in any direction for the same fuel cost, and there would be no need for an analogy of “taking off into the wind.” (If you actually meant that literally, then you were way off the mark).
Case 2: Stationary, non-rotating reference frame centered on Earth but with Earth rotating.
Allowing Earth to rotate counterclockwise when viewed from the north pole means that the launch pad would be traveling eastward, and we can calculate the initial velocity. One sidereal day is the time required for the Earth to spin 360 degrees (one full rotation). Given that:
Sidereal day = 23 hr. 56 min. 4.09053 sec (23.934 hr.)
Earth equatorial mean radius = 6378 km [3963 mi.]
Latitude of Alcantara launch facility, Brazil = 2° 17’ South (2.283°)
Then:
v = ù•r•cos(ã)
= {(2ð radians)/(23.934 hr)}•(6378 km)•cos(2.283°)
= 1673 km/h [1040 mph]
So in Case 1, the rocket is assumed to be stationary before launch, but in Case 2, it starts with an eastward velocity of 1040 mph. That means that for eastward prograde launches, I can use this initial velocity to my benefit and either carry a heavier payload or use less fuel. Your launch would still work, but you carried more fuel than needed or less payload than you could have.
But for westward retrograde launches, I’m starting 2080 mph in the hole, so to speak, and I have to account for that with extra fuel. You, on the other hand, would assume that there’s no difference, and you’d be left scratching your head when your rocket came crashing down after running out of fuel. Or maybe you’ll achieve orbit, but it would be a lot lower than you were planning (and quite likely useless).
Big, big, BIG, huge difference!
Now do you see why most launches are prograde? Why take a 2080 mph handicap unless you absolutely have to? Yes, at higher latitudes, this handicap is reduced, but you still need to account for it.
You tried to use the following quote from the website “Orbital Dynamics for the Compleat[sic] Idiot” (http://www.nwc.navy.mil/press/Review/1999/winter/s&d1-w99.htm) as a prooftext for your position:
“The coordinate system used in this tutorial has its origin at the center of Earth. In this coordinate system the Sun goes around the earth, contrary to what you may have heard. There is actually no harm in this viewpoint, since the origin of coordinates can always be whatever is convenient. It is true that Earth would be an awkward origin if we were interested in the motion of other planets (since they have complicated orbits in Earth-centered coordinates), but the planets do not sensibly affect Earth satellites, the subject of interest here, so the center at Earth will do fine.”
That was either dishonest or ignorant of you because the next paragraph, which you didn’t quote, is as follows:
“The problem is how to orient the x, y, and z axes. The z axis is easy; it will point to the North Pole along Earth's axis of rotation. We want to have a direction for the x axis that does not rotate with Earth, so that an observer could cling to the axis frame and see all the stars other than the Sun as stationary—an inertial frame of reference. There are many such directions. The convention is to orient the x axis in the direction of the vernal equinox (), which is by definition the direction to the Sun on the day in spring when day and night have the same length. Any time you can see the constellation Pleiades, you are looking approximately in the direction . Since is in Earth's equatorial plane, there is a right angle between the x and z axes. Now orient the y axis to be perpendicular to them both, and you have our coordinate system. You might wish to imagine a transparent sphere centered on these coordinates, with every star other than the Sun fixed to it. This is the celestial sphere. If you sit on top of this sphere and look down at Earth, you will see it rotating counterclockwise on its z axis.”
What you omitted is that for this tutorial, they are allowing the Earth to rotate within the stationary reference frame, and they are telling that this is so that they can have an inertial reference frame. That’s a very important distinction that obviously escaped you, and when Orion pointed it out, you ignored it.
So for your education, I’ll finish my post with a somewhat lengthy discussion of inertial reference frames quoted directly from a first-year engineering textbook (JL Meriam & LG Kraige, “Engineering Mechanics, Volume 2: Dynamics, 2nd Edition”, John Wiley and Sons, 1986, pages 109-111):
~~~~~~~~~~
The basic relationship between force and acceleration is found in Newton’s second law of motion, [F=ma], the verification of which is entirely experimental. We will describe the fundamental meaning of this law by an ideal experiment in which force and acceleration are assumed to be measured without error. A mass particle is isolated in the primary inertial system (an imaginary set of reference axes which are assumed to have no translation or rotation in space) and is subjected to the action of a single force F1. The acceleration a1 of the particle is measured, and the ratio F1/a1 of the magnitudes of the force and acceleration will be some number C1. . .The experiment is now repeated. . .Again the ratio of the magnitudes F2/a2 will produce a number C2. The experiment is repeated as many times as desired. We draw two important conclusions from the results. First, the ratios of applied force to corresponding acceleration all equal the same number. . .a constant
We conclude that the constant. . .is a measure of some property of the particle which does not change. This property is the inertia of the particle which is its resistance to rate of change of velocity. . .
. . .The second conclusion we draw from the ideal experiment is that the acceleration is always in the direction of the applied force. . .
. . .Whereas the results of the ideal experiment are obtained for measurements made relative to the “fixed” primary inertial system, they are equally valid for measurements made with respect to any nonrotating reference system which translates with a constant velocity with respect to the primary system. . . the acceleration measured in a system translating with no acceleration is the same as that measured in the primary system. Thus Newton’s second law holds equally well in a nonaccelerating system, so that we may define an inertial system as any system in which [F=ma, in vector form] is valid.
If the ideal experiment described were performed on the surface of the earth and all measurements were made relative to a reference system attached to the earth, the measured results would show a slight discrepancy upon substitution into [F=ma]. This discrepancy would be due to the fact that the measured acceleration would not be the correct absolute acceleration. The discrepancy would disappear when we introduced the corrections due to the acceleration components of the earth. These corrections are negligible for most engineering problems which involve the motions of structures and machines on the surface of the earth. In such case the accelerations measured with respect to reference axes attached to the surface of the earth may be treated as “absolute,” and [F=ma] may be applied with negligible error to experimental measurements made on the surface of the earth*.
There is an increasing number of problems, particularly in the fields of rocket and spacecraft design, where the acceleration components of the earth are of primary concern. For this work it is essential that the fundamental basis of Newton’s law be thoroughly understood and that the appropriate absolute acceleration components be employed.
Before 1905 the laws of Newtonian mechanics had been verified by innumerable physical experiments and were considered the final description of the motion of bodies. The concept of time, considered an absolute quantity in the Newtonian theory, received a basically different interpretation in the theory of relativity announced by Einstein in 1905. The new concept called for a complete reformulation of the accepted laws of mechanics. The theory of relativity was subjected to early ridicule but has had experimental check and is now universally accepted by scientists the world over. . .(The theory of relativity demonstrates that there is no such thing as a preferred primary inertial system and that measurements of time which are made in two coordinate systems which have a velocity relative to one another are different. . .) Important problems dealing with atomic and nuclear particles, for example, involve calculations based on the theory of relativity and are of basic concern to both scientists and engineers.
* An example of the magnitude of the error introduced by neglect of the motion of the earth may be cited for the case of a particle which is allowed to fall from rest (relative to the earth) at a height h above the ground. We can show that the rotation of the earth gives rise to an eastward acceleration (Coriolis acceleration) relative to the earth and, neglecting air resistance, that the particle falls to the ground a distance
x = (2/3)•ù•SQRT((2•h^3)/g)•cos(ã)
east of the point on the ground directly under that from which it was dropped. The angular velocity of the earth is ù = 0.729E-4 rad/s, and the latitude, north or south, is ã. At a latitude of 45° and from a height of 200 m, this eastward deflection would be x = 43.9 mm.
~~~~~~~~~~
Sorry, the reference frame rotates instead of the earth.
>Sorry, the reference frame rotates instead of the earth.
Sorry, a rotating reference frame is NOT an inertial reference frame. I take it you didn't bother reading my post, which is unsurprising as you didn't bother reading the sources you quoted either.
So does the Earth rotate along with your rotating reference frame? In other words, is the Earth fixed to your rotating reference frame? Or is it fixed to a non-rotating reference frame?
Careful how you answer. Think about the consequences to your argument. . .
I didn't say that a rotating reference frame was intertial. All I said was that the reference frame rotates instead of the earth. It does this to maintain it's position with the stars that are fixed wrt that rotating reference frame.
The earth is not fixed to either a rotating reference frame nor is it fixed to a non-rotating reference frame.
The reference frame is rotating relative to a fixed-earth. As such, the calculation are the same as for an assumed rotating earth within a fixed reference frame.
Orion's contention was that there was some essential difference due to the heliocentric model and there is not. The only contention is whether the reference frame rotates or the earth.
There is no way to know.
Ash
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