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.

  • $\begingroup$ Just read Larry Niven's Ringworld series $\endgroup$
    – DJohnM
    Commented Aug 26, 2018 at 20:00
  • $\begingroup$ LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring. $\endgroup$ Commented Aug 26, 2018 at 20:00
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    $\begingroup$ 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
    Commented Aug 27, 2018 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
    Commented Aug 27, 2018 at 0:58

2 Answers 2


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.

  • $\begingroup$ 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
    Commented Aug 26, 2018 at 20:46
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    $\begingroup$ 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.) $\endgroup$ Commented Aug 26, 2018 at 20:53
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    $\begingroup$ @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. $\endgroup$ Commented Aug 26, 2018 at 21:34
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    $\begingroup$ @ChristopherJamesHuff That's not an orbit, though -- Earth's surface is moving about 400 m/s space-fixed at the equator, versus circular orbit velocity of about 7900 m/s. Very stable indeed, though, and has many other convenient properties -- easy to reach, easy to maintain a shirt-sleeve-comfortable environment, etc. $\endgroup$ Commented Aug 27, 2018 at 22:13
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    $\begingroup$ I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you. $\endgroup$ Commented Aug 28, 2018 at 2:27

This question has already had a good answer which is that such a thing is not in orbit. That being said I thought it would be entertaining to compute the radius of such a thing. Before I do that I'll add the obvious three notes:

  1. this is a variant of Ringworld;
  2. such Ringworlds are famously unstable as Larry Niven discovered after the initial book in the series was published;
  3. such structures require implausibly strong materials, of which Niven was aware, hence scrith, shadow-square wire and so on.

That being said, here's the maths.

Let's assume that the structure is in a circular, equatorial, 'orbit' around a planet of mass $M$ and angular velocity $\omega$, which is therefore the same as the angular velocity of the ring. The radius of the ring will be $r$. $g$ is the acceleration due to gravity on the surface of the planet: $g \approx G M/R^2$ where $R$ is the radius of the planet.

The radial acceleration of the ring is then

$$a_c = r\omega^2$$

and the gravitational acceleration of the ring is

$$a_G = \frac{G M}{r^2}$$

For for a geostationary orbit then we can just equate these two things and get

$$r^3 = \frac{G M}{\omega^2}$$

And we can plug numbers into this to get the height of geostationary satellites.

But this isn't what we want: we want the effective inward acceleration to be $g$ on the ring. In other words we want

$$r\omega^2 - g = \frac{G M}{r^2}$$


$$r^3 - \frac{g}{\omega^2}r^2 = \frac{G M}{\omega^2} = 0$$

This is a cubic in $r$. And solving cubics is one of the reasons algebra systems were invented as far as I'm concerned: even if I knew the cubic formula there is essentially zero chance I could write down the following expression without error. So, resorting to an algebra system we get the following real solution for $r$:

\begin{equation*} \begin{split} r = \frac{1}{3} &\left(\frac{\sqrt[3]{\frac{3 \sqrt{3} \sqrt{4 g^3 G M+27 G^2 M^2 \omega ^4}}{\omega ^4}+\frac{2 g^3}{\omega ^6}+\frac{27 G M}{\omega ^2}}}{\sqrt[3]{2}}\right.\\ &\quad +\left.\frac{\sqrt[3]{2} g^2}{\omega ^4 \sqrt[3]{\frac{3 \sqrt{3} \sqrt{4 g^3 G M+27 G^2 M^2 \omega ^4}}{\omega ^4}+\frac{2 g^3}{\omega ^6}+\frac{27 G M}{\omega ^2}}}+\frac{g}{\omega ^2}\right) \end{split} \end{equation*}

And we can plug the values of $M$ and $\omega$ for Earth into this, and we get

$$r \approx 1.85\times 10^9\,\mathrm{m}$$

So that's where you would need to build such a thing. Note that this is further than a geostationary orbit by a factor of about $44$ and just under 5 times the semimajor axis of the Moon. Even assuming active stabilisation you would almost certainly need to demolish or move the Moon to avoid the stability problems it would cause for such a structure.


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