Hello Folks,
It has always been a sort of holy grail for me to try and find ways to
represent ship movement within a star system, along with planetary movements
concurrent with ship movement. One of those pesky things I've not been able to
pin down was planetary movement along an ellipse. I'm hoping that perhaps even
THAT issue will be overcome as of tonight.
See, here's the problem. The velocity of a planet within its orbit along an
ellipse, is variable. The further away from the sun the planet is, the slower
its velocity becomes. The closer to its sun the planet it is, the faster its
velocity becomes. Kepler's second law of planetary motion states that "A line
joining a planet and the Sun sweeps out equal areas during equal intervals of
time". This in turn, gives me an idea for how to determine where a planet is
supposed to be at any given time...
Suppose for example, you have a planet whose period is 365.24 days. You know
that the planet is 1 AU's distance (on average) from the Sun, and that its
orbital eccentricity is say, .02, then you know for instance, that at its
shorest axis, the planet's distance is 147,093,602 km, while its furthest
Distance is 152,097,053. Area of that particular Elipse is equal to:
Pi * A * B where A is the distance from the center of the Ellipse along its
shortest axis, and B is the Distance from the Center of the Ellipse along its
longest Axis.
Or
Pi * 147,093,602 * 152,097,053 km or 70,285,292,258,994,195.79 square
kilometers.
Now, if there are 365.24 days to that planet's orbit, then there are 86400
seconds to the year in question. The area swept out by the planet's revolution
around its star would therefore result in an area being swept out equal to its
total area divided by 86400 seconds or 813,487,178,923.54 square kilometers per
second.
All that remains then, is to determine what triangle will produce an area equal
to the value of 813,487,178,923.54 square kilometers where you know two of the
three sides of the triangle's length if you know the included angle between
those two sides. Once you know the included angle between two sides, you can
determine the last side of the triangle, and compute the area of the triangle
where it is equal to 1/2 Base times Height.
Does this make sense? Have I missed anything? If this works, then I may
finally have a means for plotting the location of any given space craft for use
in a GURPS campaign (be it GURPS TRAVELLER, GURPS TRANSHUMAN SPACE or even
GURPS SPACESHIPS.
Thoughts? Comments?
Hal
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