Or, just how far out is that Highport? Does anyone know a formula that could be used to determine the radius of a Geosynchronous orbit for a given planetary size?
Geosynchronous orbits
- Thread starter Spartan159
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To find geosynch, you also need to know sol length.
You need whatever distance produces an equatorial orbit the same length as the day.
There is no convenient way to figure the length of day... because it's a widely variable result. Venus, for example, is retrograde and several hundred days, but is, for traveller purposes, the same size as earth... Mars is much smaller than earth, and has a 17,000 km "areostationary" orbit. the Sol is 24:39:35.244...
wikipedia has a formula: https://en.wikipedia.org/wiki/Areostationary_orbit
Thanks Aramis. World Builders Handbook has a formula for randomly determining rotation period that I can use. It occurs to me that a tidally locked planet could not have a stationary orbit as such although I suppose it could use a Lagrange Point, L1 and L2 would be relatively close but I'm not sure how stable they are. Then again nothing says a Highport does not have station keeping thrusters either.
On a side note it would be interesting to see a Highport so massive that it rings the planet, I would think quite possible on smaller planets, I have seen the concept done in a few places. Pretty sure the first Starship Troopers movie had one about earth. Dirty Pair: Project Eden had one that was even connected to the planet with space elevators. Before the Dirty Pair blew it up of course :devil:
On a side note it would be interesting to see a Highport so massive that it rings the planet, I would think quite possible on smaller planets, I have seen the concept done in a few places. Pretty sure the first Starship Troopers movie had one about earth. Dirty Pair: Project Eden had one that was even connected to the planet with space elevators. Before the Dirty Pair blew it up of course :devil:
shaunhilburn
SOC-12
Given:
Rotation (in seconds)
Mass (in earth masses)
Radius (equatorial, in kilometers)
Assuming a perfectly spherical planet, geostationary altitude (km) = cuberoot( (Rotation/5068)² × Mass ) × 6378 - Radius
I have a more complex version that accounts for oblateness.
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