3
$\begingroup$

A challenge for those of you with some math skills. we know the elevation and speed required to put a satellite into a geosynchronous orbit(or geostationary if you prefer), and that any further out we go; the object is effected by centripetal and centrifugal forces, causing objects to fly away from the earth.

Here's the challenge. At what elevation and speed, would a planetary ring (around the earth, laughs for other planets) be required to be in both stationary orbit and provide enough rotational forces; to cause a 1G environment, for those that are on/in the ring?

must provide math, and good luck to all of you.

I started thinking about this as a way to provide food for the growing population of earth, while much of the space we still have available is slowly being consumed. with a controlled environment, and sectional production; I believe that we would have food clean or chemicals, insects, and funguses, that the food produced would be healthier then the food we receive now. at the equator, night for the earth would become day for the ring; and night for the ring, would be day on earth.

$\endgroup$
  • $\begingroup$ Just read Larry Niven's Ringworld series $\endgroup$ – DJohnM Aug 26 '18 at 20:00
  • $\begingroup$ LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring. $\endgroup$ – Taalkeus Blank Aug 26 '18 at 20:00
  • 1
    $\begingroup$ also note, "geostationary" is a subset of "geosynchronous", they are not really interchangeable, An inclined or even polar orbit could be "synchronous" if it had a period of 1 sidereal day, even though it wouldn't be stationary. It would also be called a "repeat ground track orbit" if it was a rational fraction of one sidereal day (e.g. 1/2, 2, 3/2, 2/5, etc.). $\endgroup$ – uhoh Aug 27 '18 at 0:57
  • $\begingroup$ Also, If it circled the earth equatorially at a different altitude than GEO but a point on the ring was still stationary with respect to a point on Earth, that's not quite a "geostationary orbit" in my opinion since forces holding it there includes either compression or tension within the ring itself, not just orbital mechanics. $\endgroup$ – uhoh Aug 27 '18 at 0:58
7
$\begingroup$

Here's the challenge. At what elevation and speed, would a planetary ring (around the earth, laughs for other planets) be required to be in both stationary orbit and provide enough rotational forces; to cause a 1G environment, for those that are on/in the ring?

There's no way to achieve that. A body in stable orbit is by definition at zero gee. Any planetary ring that develops significant centrifugal "gravity" must be rotating at substantially more than orbital speed.

| improve this answer | |
$\endgroup$
  • $\begingroup$ In the orbital ring case you don't need to have it spin to stay up. Technically, it could be still and rely on the structure of the ring to keep it aloft. Gravity pulls on all parts of the ring at the same time but the ring is incompressible and doesn't squish. $\endgroup$ – Dragongeek Aug 26 '18 at 20:46
  • 3
    $\begingroup$ In that case, it's not stable -- if any small force moves it off-center, gravity starts working more on the "near" side and it accelerates in the same direction until it crashes into the central body. (Niven didn't realize this when he wrote Ringworld, and had to introduce a system to control the position of the ring in The Ringworld Engineers.) $\endgroup$ – Russell Borogove Aug 26 '18 at 20:53
  • $\begingroup$ Would it work if we spin the ring above the orbital velocity at it's current altitude? $\endgroup$ – Dragongeek Aug 26 '18 at 21:05
  • 2
    $\begingroup$ @Dragongeek No. What happens is the central body pulls more on the bit that's closer, less on the bit that's farther away. The ring is shifted in the direction it's already off-center. Now it's farther off center and shifts even faster. Very soon it hits the central body. $\endgroup$ – Loren Pechtel Aug 26 '18 at 21:34
  • 1
    $\begingroup$ I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you. $\endgroup$ – Russell Borogove Aug 28 '18 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.