Geosynchronous Planetary Ring Challenge
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
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.
orbital-mechanics
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
asked Aug 26 at 19:56
Taalkeus Blank
212
add a comment |Â
up vote
3
down vote
favorite
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.
orbital-mechanics
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
asked Aug 26 at 19:56
Taalkeus Blank
212
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
1
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.).
– uhoh
Aug 27 at 0:57
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.
– uhoh
Aug 27 at 0:58
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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.
orbital-mechanics
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
asked Aug 26 at 19:56
Taalkeus Blank
212
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.
orbital-mechanics
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
asked Aug 26 at 19:56
Taalkeus Blank
212
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
edited Aug 26 at 20:40
Russell Borogove
70.4k2219300
70.4k2219300
asked Aug 26 at 19:56
Taalkeus Blank
212
asked Aug 26 at 19:56
Taalkeus Blank
212
asked Aug 26 at 19:56
Taalkeus Blank
212
212
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
1
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.).
– uhoh
Aug 27 at 0:57
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.
– uhoh
Aug 27 at 0:58
add a comment |Â
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
1
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.).
– uhoh
Aug 27 at 0:57
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.
– uhoh
Aug 27 at 0:58
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
1
1
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.).
– uhoh
Aug 27 at 0:57
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.).
– uhoh
Aug 27 at 0:57
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.
– uhoh
Aug 27 at 0:58
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.
– uhoh
Aug 27 at 0:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
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.
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
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.
– Dragongeek
Aug 26 at 20:46
3
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
@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.
– Loren Pechtel
Aug 26 at 21:34
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
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.
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
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.
– Dragongeek
Aug 26 at 20:46
3
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
@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.
– Loren Pechtel
Aug 26 at 21:34
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
 |Â
show 7 more comments
up vote
7
down vote
accepted
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.
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
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.
– Dragongeek
Aug 26 at 20:46
3
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
@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.
– Loren Pechtel
Aug 26 at 21:34
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
 |Â
show 7 more comments
up vote
7
down vote
accepted
up vote
7
down vote
accepted
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.
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
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.
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
answered Aug 26 at 20:39
Russell Borogove
70.4k2219300
70.4k2219300
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.
– Dragongeek
Aug 26 at 20:46
3
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
@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.
– Loren Pechtel
Aug 26 at 21:34
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
 |Â
show 7 more comments
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.
– Dragongeek
Aug 26 at 20:46
3
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
@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.
– Loren Pechtel
Aug 26 at 21:34
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
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.
– Dragongeek
Aug 26 at 20:46
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.
– Dragongeek
Aug 26 at 20:46
3
3
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.)
– Russell Borogove
Aug 26 at 20:53
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.)
– Russell Borogove
Aug 26 at 20:53
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
Would it work if we spin the ring above the orbital velocity at it's current altitude?
– Dragongeek
Aug 26 at 21:05
2
2
@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.
– Loren Pechtel
Aug 26 at 21:34
@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.
– Loren Pechtel
Aug 26 at 21:34
1
1
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
I dispute your definition, as does Wikipedia ("an orbit is the gravitationally curved trajectory of an object"), but you do you.
– Russell Borogove
Aug 28 at 2:27
 |Â
show 7 more comments
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f30306%2fgeosynchronous-planetary-ring-challenge%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Just read Larry Niven's Ringworld series
– DJohnM
Aug 26 at 20:00
LOL. I already have, but if you remember; that was a stellar ring, not a planetary ring.
– Taalkeus Blank
Aug 26 at 20:00
@TaalkeusBlank Story begins with rich people preparing to leave their home planet and go to a ring that orbits it (which had lasers)?
– uhoh
Aug 27 at 0:51
1
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.).
– uhoh
Aug 27 at 0:57
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.
– uhoh
Aug 27 at 0:58