Delta-v to move from GEO to GEO
To move a satellite in geostationary orbit, 166°55′E, to the antipode, 13° 4' 3.2" W, what delta-V would be needed for this to be accomplished?
As for time constraints, I do not know what are the possibilities with contemporary technology. What would be the fastest? What would be the delta-V if the time required was four times slower than the fastest option already discussed?
orbital-mechanics orbital-maneuver delta-v geostationary
asked 7 hours ago
Bob516
1,3651316
|
show 3 more comments
To move a satellite in geostationary orbit, 166°55′E, to the antipode, 13° 4' 3.2" W, what delta-V would be needed for this to be accomplished?
As for time constraints, I do not know what are the possibilities with contemporary technology. What would be the fastest? What would be the delta-V if the time required was four times slower than the fastest option already discussed?
orbital-mechanics orbital-maneuver delta-v geostationary
asked 7 hours ago
Bob516
1,3651316
6
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
2
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
1
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
1
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
2
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago
|
show 3 more comments
To move a satellite in geostationary orbit, 166°55′E, to the antipode, 13° 4' 3.2" W, what delta-V would be needed for this to be accomplished?
As for time constraints, I do not know what are the possibilities with contemporary technology. What would be the fastest? What would be the delta-V if the time required was four times slower than the fastest option already discussed?
orbital-mechanics orbital-maneuver delta-v geostationary
asked 7 hours ago
Bob516
1,3651316
To move a satellite in geostationary orbit, 166°55′E, to the antipode, 13° 4' 3.2" W, what delta-V would be needed for this to be accomplished?
As for time constraints, I do not know what are the possibilities with contemporary technology. What would be the fastest? What would be the delta-V if the time required was four times slower than the fastest option already discussed?
orbital-mechanics orbital-maneuver delta-v geostationary
orbital-mechanics orbital-maneuver delta-v geostationary
asked 7 hours ago
Bob516
1,3651316
asked 7 hours ago
Bob516
1,3651316
edited 5 hours ago
asked 7 hours ago
Bob516
1,3651316
asked 7 hours ago
Bob516
1,3651316
asked 7 hours ago
Bob516
1,3651316
1,3651316
6
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
2
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
1
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
1
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
2
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago
|
show 3 more comments
6
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
2
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
1
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
1
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
2
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago
6
6
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
2
2
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
1
1
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
1
1
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
2
2
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago
|
show 3 more comments
1 Answer
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Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.
In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.
So, the decisive figure is how long you want to take to move your satellite.
If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.
The semi-major axis of an orbit with period $t$ is:
$$a = sqrt[3]{frac{mu t^2}{4pi^2}}$$
Where $mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:
$$Delta v_1
= sqrt{frac{mu}{r_1}}
left( sqrt{frac{2 r_2}{r_1+r_2}} - 1 right)$$
Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.
To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.
If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.
The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.
answered 4 hours ago
Russell Borogove
82.2k2273356
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
add a comment |
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Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.
In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.
So, the decisive figure is how long you want to take to move your satellite.
If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.
The semi-major axis of an orbit with period $t$ is:
$$a = sqrt[3]{frac{mu t^2}{4pi^2}}$$
Where $mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:
$$Delta v_1
= sqrt{frac{mu}{r_1}}
left( sqrt{frac{2 r_2}{r_1+r_2}} - 1 right)$$
Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.
To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.
If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.
The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.
answered 4 hours ago
Russell Borogove
82.2k2273356
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
add a comment |
Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.
In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.
So, the decisive figure is how long you want to take to move your satellite.
If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.
The semi-major axis of an orbit with period $t$ is:
$$a = sqrt[3]{frac{mu t^2}{4pi^2}}$$
Where $mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:
$$Delta v_1
= sqrt{frac{mu}{r_1}}
left( sqrt{frac{2 r_2}{r_1+r_2}} - 1 right)$$
Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.
To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.
If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.
The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.
answered 4 hours ago
Russell Borogove
82.2k2273356
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
add a comment |
Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.
In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.
So, the decisive figure is how long you want to take to move your satellite.
If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.
The semi-major axis of an orbit with period $t$ is:
$$a = sqrt[3]{frac{mu t^2}{4pi^2}}$$
Where $mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:
$$Delta v_1
= sqrt{frac{mu}{r_1}}
left( sqrt{frac{2 r_2}{r_1+r_2}} - 1 right)$$
Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.
To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.
If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.
The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.
answered 4 hours ago
Russell Borogove
82.2k2273356
Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.
In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.
So, the decisive figure is how long you want to take to move your satellite.
If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.
The semi-major axis of an orbit with period $t$ is:
$$a = sqrt[3]{frac{mu t^2}{4pi^2}}$$
Where $mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:
$$Delta v_1
= sqrt{frac{mu}{r_1}}
left( sqrt{frac{2 r_2}{r_1+r_2}} - 1 right)$$
Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.
To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.
If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.
The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.
answered 4 hours ago
Russell Borogove
82.2k2273356
edited 1 hour ago
answered 4 hours ago
Russell Borogove
82.2k2273356
answered 4 hours ago
Russell Borogove
82.2k2273356
answered 4 hours ago
Russell Borogove
82.2k2273356
82.2k2273356
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
add a comment |
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
@RussellBorgove For the really fast option, how long would that take?
– Bob516
3 hours ago
1
1
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
It takes half a day, one shortcut orbit -- rewrote to clarify.
– Russell Borogove
3 hours ago
add a comment |
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6
What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations)
– Jack
6 hours ago
2
This is mathematically interesting as it depends on the initial and final longitudes.
– uhoh
6 hours ago
1
@Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are.
– uhoh
6 hours ago
1
I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated.
– uhoh
6 hours ago
2
@Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter.
– Russell Borogove
5 hours ago