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Old October 16 2011, 10:50 PM   #57
CorporalCaptain
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Re: TOS Enterprise Question...

Using the parameters suggested by blssdwlf, I have plotted a theoretically real orbit for the camera, according to Newtonian mechanics. This solution is offered "as is", but I'll explain my work, so that it can be checked. Note that other alternatives are possible; I am describing only one proposal.



The displayed units are SI, so distance is measured in meters. The planet's surface is red, the camera's orbit is green, and the Enterprise's orbit is blue. The center of the planet is market by a cross, and the point of closest approach between the camera and the Enterprise is marked by a circle on the left. Both the Enterprise and the camera orbit counterclockwise in the diagram. This diagram is to scale.

The planet is assumed to be a sphere that can function as a point mass. The radius of the planet is 6371000 meters and its mass is 5.9736*10^24 kilograms, both Earth values. The value of Netwon's gravitational constant G that I am using it is 6.67384*10^(-11) in SI units. The mass of the Enterprise is assumed to be zero (which is a reasonable assumption for otherwise if its mass were not vanishingly small relatively speaking, the Enterprise would disrupt every solar system it visited).

The Enterprise orbits at an altitude of 6600 miles in a circular orbit, as suggested by blssdwlf. There, it will orbit the planet in 6.122994278617534 hours. It travels at a constant speed of 4843.676699517414 meters per second in the Newtonian inertial frame.

The orbit of the camera that I am plotting is in the same plane as the orbit of the Enterprise. For simplicity, I have chosen the point of closest approach to be the apogee of the camera. At this point of closest approach, the camera, the Enterprise, and the center of the planet lie in the same line, and the camera is 417 feet further from the center of the planet than the Enterprise (which is one ship width, or about that depending on your source; this is done just so that the ship and camera do not collide). To completely determine the rest of the camera's orbit, we need only know its speed at the instant when it reaches apogee. For this value I choose 357 feet per second slower than the Enterprise, in accordance with what blssdwlf has suggested. Using the vis viva equation and slugging things out using the properties of the ellipse and elliptical orbits, we determine that the eccentricity of the camera's orbit is 0.04441833885412092, its perigee has an altitude of 5701.960117206670 miles, and its period is 5.736632021918684 hours. So, although the camera is moving slower than the Enterprise during the hypothetical stock footage in question, it will fall in towards the planet soon and outrun the Enterprise on the next lap around the planet. The camera will even survive that lap easily, in that it will come nowhere near burning up in the atmosphere.

Now for the part which may come as some surprise to some. Over the course of about 4 seconds, the trajectory of both the camera and the Enterprise may be very accurately approximated by constant speed straight line motion. I will leave it as an exercise for the reader to run calculations that will convince him- or herself that this is so in the case of the Enterprise; these calculations are straightforward since the Enterprise orbits the planet in a circular orbit at constant speed. Convincing oneself that this is so in the case of the camera is somewhat more involved which I will now partially describe, hopefully adequately enough to convince the reader.

The speed of the camera is controlled by Kepler's second law: it must sweep out equal areas of the ellipse in equal amounts of time. Although a closed form solution for the motion of the camera does exist, and is within the ability of a very advanced student of first semester integral calculus to grasp, presenting that solution and mathematically proving the point will not really be helpful in this forum. Therefore, I will simply present the following observation: if the motion of the camera is approximately at a constant speed and in a straight line over the four seconds of the shot, then the area swept out by the triangle accumulated along that line of motion between the center of the planet will be equal to the fraction of the area of the whole ellipse that is four seconds divided by the period of the orbit. This is so to 16 digits of accuracy, if one takes the line of motion and speed to be the values the camera has at apogee.

Furthermore, over this four second interval, if the camera is pointed in a fixed direction and not rotating according to the Newtonian inertial frame, then the celestial sphere (on which the stars lie) will appear to turn at most .06532751490506261 degrees, which is hardly at all. However, by tweaking the camera zoom setting, I suppose some apparent stellar motion could perhaps be seen. I say these things only to indicate how rapidly the stars might appear to move; naturally, the camera is free to pivot and point in any direction during the shot, in order to frame the ship in a cinematic manner.

So, therefore and in other words, straight line motion, against a nearly fixed stellar background, is what the camera will appear to see, over the four seconds of the shot.


ETA: The lower bound for how much the celestial sphere will appear to turn in four seconds is .06385944919511107 degrees. The actual amount is in between this value, and the upper bound value given above, but it is almost exactly equal to this lower bound given here.

The lower bound is the value assuming that the camera orbits (hypothetically and non-gravitationally) at its tangent velocity at its apogee. The upper bound is taken according to the orbital velocity of the Enterprise itself, which at all times during the four seconds of the shot is moving faster than the camera. As you can see, these values are very close to each other; the lower bound is .0224723948566697 percent less than the upper bound.
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Last edited by CorporalCaptain; October 17 2011 at 02:05 AM. Reason: clarified turn rate of celestial sphere
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