Author Topic: Orbits Q&A  (Read 88621 times)

Offline baldusi

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Orbits Q&A
« on: 01/27/2011 01:38 PM »
I have a couple of questions about orbits. In particular, GSO.
1) I understand that there are "slots" given to each country. Are this governed by ITU? How are they given?
2) How wide are this "slots"? I've seen quoted integer degrees, but just Argentina has two or three slots, so they should be more.
3) What's the optimum maneuver to "move" between two slots? The initial and final angular momentum and the relationship of potential and kinetic energy are the same. So I would assume that it can be made very cheaply if not very fast.
4) Regarding Polar Orbits. It seems that USA launches southward, and I'm assuming Plesetsk, launches Northward. The Indians, launch also southward, but on the opposite side of USA. I don't know how the Chinese launch their polar orbit satellites. But the main question is: don't you end up with polar orbits going in opposite directions? I guess not opposite since that would leave the earth's orbit, but they should cross at a very steep angle. Of course since a lot of polar orbit satellites "don't exists" it's not like there's a lot of coordination. But isn't it dangerous?
5) If it is dangerous, couldn't some country that's worried about another's spy satellite launch a "scientific" instrument in polar orbit that "just" happens to accidentally crash on one of those "satellites that don't exist"?

Offline Jim

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Re: Orbits Q&A
« Reply #1 on: 01/27/2011 02:45 PM »
I have a couple of questions about orbits. In particular, GSO.
1) I understand that there are "slots" given to each country. Are this governed by ITU? How are they given?
2) How wide are this "slots"? I've seen quoted integer degrees, but just Argentina has two or three slots, so they should be more.
3) What's the optimum maneuver to "move" between two slots? The initial and final angular momentum and the relationship of potential and kinetic energy are the same. So I would assume that it can be made very cheaply if not very fast.
4) Regarding Polar Orbits. It seems that USA launches southward, and I'm assuming Plesetsk, launches Northward. The Indians, launch also southward, but on the opposite side of USA. I don't know how the Chinese launch their polar orbit satellites. But the main question is: don't you end up with polar orbits going in opposite directions? I guess not opposite since that would leave the earth's orbit, but they should cross at a very steep angle. Of course since a lot of polar orbit satellites "don't exists" it's not like there's a lot of coordination. But isn't it dangerous?
5) If it is dangerous, couldn't some country that's worried about another's spy satellite launch a "scientific" instrument in polar orbit that "just" happens to accidentally crash on one of those "satellites that don't exist"?


1.  Don't know

2.  A few degrees, it depends on the frequency band that the spacecraft is going to broadcast on.  Spacecraft can occupy the same slots.

3.  They drift into position

4.  No and no. It is all about timing.  The orbit is fixed in space, relationship to the earth is not fixed by the launch site.  Also think about this.  On one hemisphere, the spacecraft is going north and on the other side it is going south.

5.  They use radars to make sure there are no conjunctions.  That is what COLA's are for.

Offline baldusi

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Re: Orbits Q&A
« Reply #2 on: 01/27/2011 03:42 PM »
Regarding Polar Orbits. It seems that USA launches southward, and I'm assuming Plesetsk, launches Northward. The Indians, launch also southward, but on the opposite side of USA. I don't know how the Chinese launch their polar orbit satellites. But the main question is: don't you end up with polar orbits going in opposite directions? I guess not opposite since that would leave the earth's orbit, but they should cross at a very steep angle. Of course since a lot of polar orbit satellites "don't exists" it's not like there's a lot of coordination. But isn't it dangerous?
1 No and no. It is all about timing.  The orbit is fixed in space, relationship to the earth is not fixed by the launch site.  Also think about this.
2 On one hemisphere, the spacecraft is going north and on the other side it is going south.
1 So you basically have two polar orbits, one that goes North to South and another that goes South to North if looking from the Sun to Earth, right? And all polar orbit satellites are on those two orbits at different phase, right? They should cross at two points.
2 I said that India launches southward and Russia and China northward.

Offline Jim

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Re: Orbits Q&A
« Reply #3 on: 01/27/2011 04:16 PM »

1 So you basically have two polar orbits, one that goes North to South and another that goes South to North if looking from the Sun to Earth, right? And all polar orbit satellites are on those two orbits at different phase, right? They should cross at two points.
2 I said that India launches southward and Russia and China northward.

no, there are infinite number of polar orbits.  Spin a coin on a table, that represents all the orbits.  They all cross.

there could be a polar orbit 90 degrees from the line from the  Sun to Earth

Offline Targeteer

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Re: Orbits Q&A
« Reply #4 on: 01/27/2011 04:30 PM »
I have a couple of questions about orbits. In particular, GSO.
1) I understand that there are "slots" given to each country. Are this governed by ITU? How are they given?
2) How wide are this "slots"? I've seen quoted integer degrees, but just Argentina has two or three slots, so they should be more.
3) What's the optimum maneuver to "move" between two slots? The initial and final angular momentum and the relationship of potential and kinetic energy are the same. So I would assume that it can be made very cheaply if not very fast.
4) Regarding Polar Orbits. It seems that USA launches southward, and I'm assuming Plesetsk, launches Northward. The Indians, launch also southward, but on the opposite side of USA. I don't know how the Chinese launch their polar orbit satellites. But the main question is: don't you end up with polar orbits going in opposite directions? I guess not opposite since that would leave the earth's orbit, but they should cross at a very steep angle. Of course since a lot of polar orbit satellites "don't exists" it's not like there's a lot of coordination. But isn't it dangerous?
5) If it is dangerous, couldn't some country that's worried about another's spy satellite launch a "scientific" instrument in polar orbit that "just" happens to accidentally crash on one of those "satellites that don't exist"?

1.  They are assigned by either a treaty or agreement.  Nations/companies apply for slots and retain granted slots for a fixed period of time.  If they don't fill them they are lost.

2.  I think they are in 0.5 degree increments.  One of the ways I believe they try to avoid interference between adjacent satellites is by alternating polarization of frequencies.  Several companies cluster satellites very closely at an assigned location as well...
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Offline Danderman

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Re: Orbits Q&A
« Reply #5 on: 03/04/2011 03:37 PM »
This is a question about gravitational swingby maneuvers - it is well known that swingbys of Jupiter or other planets can provide significant changes in delta-V for spacecraft, and lunar swingbys less so. I would imagine that one variable in calculating the benefit of a swingby is distance from the c/g of the planet or moon, so that a lunar swingby maneuver should pass very closely to the Moon.

What is the lower limit to the mass of the object providing the delta-v,  ie could some useful amount of delta-V be gained from passing within 100 meters of a small asteroid?

Offline IsaacKuo

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Re: Orbits Q&A
« Reply #6 on: 03/04/2011 06:10 PM »
What is the lower limit to the mass of the object providing the delta-v,  ie could some useful amount of delta-V be gained from passing within 100 meters of a small asteroid?

No.  Roughly, the amount of delta-v benefit you could gain with a gravitational swingby is on the order of the escape velocity (at your minimum altitude).  Escape velocity is proportional to the square root of mass divided by radius.  So a big asteroid like Ceres has an escape velocity of 500m/s (significant), but an asteroid with a tenth the radius would have an escape velocity of only 50m/s (not significant).

Offline Danderman

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Re: Orbits Q&A
« Reply #7 on: 03/04/2011 06:21 PM »
What is the lower limit to the mass of the object providing the delta-v,  ie could some useful amount of delta-V be gained from passing within 100 meters of a small asteroid?

No.  Roughly, the amount of delta-v benefit you could gain with a gravitational swingby is on the order of the escape velocity (at your minimum altitude).  Escape velocity is proportional to the square root of mass divided by radius.  So a big asteroid like Ceres has an escape velocity of 500m/s (significant), but an asteroid with a tenth the radius would have an escape velocity of only 50m/s (not significant).

How does proximity to the object figure into this? I can't seem to get any delta-V from Jupiter from my living room, so proximity must be a factor.
« Last Edit: 03/04/2011 06:46 PM by Danderman »

Offline IsaacKuo

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Re: Orbits Q&A
« Reply #8 on: 03/04/2011 06:30 PM »
How does proximity to the object figure into this. I can't seem to get any delta-V from Jupiter from my living room, so proximity must be a factor.

Proximity determines escape velocity.  At your current distance from Jupiter, escape velocity is hardly anything.  Escape velocity is inversely proportional to the square root of your distance to the center of the object.  Escape velocity is maximized by standing on the object's surface.

Offline Danderman

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Re: Orbits Q&A
« Reply #9 on: 03/04/2011 06:47 PM »
How does proximity to the object figure into this. I can't seem to get any delta-V from Jupiter from my living room, so proximity must be a factor.

Proximity determines escape velocity.  At your current distance from Jupiter, escape velocity is hardly anything.  Escape velocity is inversely proportional to the square root of your distance to the center of the object.  Escape velocity is maximized by standing on the object's surface.

A wonderful explanation, thank you!

Offline Danderman

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Re: Orbits Q&A
« Reply #10 on: 03/05/2011 04:44 PM »
I was thinking about this further, and there must be a velocity component to this, as well. For example, Earth swingbys are usually preceded by very long trajectories out past Mars or Venus, followed by the return to Earth, with the subsequent  swingby maneuver to Jupiter or some comet.  There has to be some reason related to desired velocity why these trajectories travel so far from Earth, unless its to allow the target to line up with the Earth swingby trajectory.

Could a useful Earth swingby be achieved simply by traveling out a few hundred thousand miles, or the swingby significantly enhanced by flight very far from Earth?
« Last Edit: 03/05/2011 04:50 PM by Danderman »

Offline Danderman

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Re: Orbits Q&A
« Reply #11 on: 03/06/2011 05:16 PM »
And a related question:

Let's say that you need to launch a spacecraft to Mars on a given day, and you have no tolerance for a delay (for whatever reason). So, you cannot afford to hope that it doesn't rain at the launch site.

Is it possible to launch, say, 100 days before the actual planned launch date, fly away from the Earth for 50 days, return, and then use a swingby to go to Mars on the desired date? Yes, I know this means exposing the spacecraft to space for an extra 100 days, but assuming that the 50 day out and 50 days back trajectory is feasible, what is the downside to this approach?   Why hasn't it been tried as a way to ensure that a payload makes it to Mars during any particular launch window?

Oh, in the event that the launch at T - 100 days is itself delayed, I assume that we have the computers today that can quickly generate a 98 day trajectory that accomplishes the same purpose and would allow a launch at T - 98 days.

Also, any chances of obtaining a velocity boost during the Earth swingby?

Offline mmeijeri

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Re: Orbits Q&A
« Reply #12 on: 03/06/2011 05:17 PM »
You can do this from a Lagrange point.
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Offline Danderman

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Re: Orbits Q&A
« Reply #13 on: 03/06/2011 05:44 PM »
You can do this from a Lagrange point.

I would imagine that the delta-V required to go to a Lagrange point would be more than anything gained from the resulting Earth swingby, so in this case, the cost of having an absolutely positive departure date for Mars from Earth would be that extra Delta-V to hang around the Lagrange point.

Is there a "Free" trajectory where a spacecraft could just depart Earth, and return without a requirement for some additional maneuvering?

Offline mmeijeri

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Re: Orbits Q&A
« Reply #14 on: 03/06/2011 06:14 PM »
I would imagine that the delta-V required to go to a Lagrange point would be more than anything gained from the resulting Earth swingby, so in this case, the cost of having an absolutely positive departure date for Mars from Earth would be that extra Delta-V to hang around the Lagrange point.

That depends on how you get to the Lagrange point. If you use a slow but relatively efficient three body trajectory which doesn't need a second burn to insert you into a Lagrange point orbit then the perigee lowering burn almost cancels out against the saved insertion burn. This would work for unmanned flights like probes or MTV components / consumables. Having the MTV cycle between high Earth and high Mars orbit instead of descending deep into each gravity well and then climbing out of it again would be a good idea anyway.

Quote
Is there a "Free" trajectory where a spacecraft could just depart Earth, and return without a requirement for some additional maneuvering?

If it's really free then you cannot influence the launch date, you would merely delay it by a fixed amount of time, which would be no improvement. But doing it with only a neglible amount of maneuvering might be possible. It's subtle effects that make the three body trajectories work, so maybe there's a way to make it work.
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Offline rklaehn

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Re: Orbits Q&A
« Reply #15 on: 03/06/2011 06:32 PM »
You can do this from a Lagrange point.
I would imagine that the delta-V required to go to a Lagrange point would be more than anything gained from the resulting Earth swingby, so in this case, the cost of having an absolutely positive departure date for Mars from Earth would be that extra Delta-V to hang around the Lagrange point.

Unless you are planning to use a HLV to launch your spacecraft in a single launch, you will need a staging point to assemble your spacecraft prior to departure from earth. And you need some way to swing by earth to use the oberth effect. Unless you have some miracle propulsion system, this is not optional. The delta-v for a mars hohmann transfer from earth C3=0 using the oberth effect is 0.6 km/s, without it is 3.7km/s.

Quote
Is there a "Free" trajectory where a spacecraft could just depart Earth, and return without a requirement for some additional maneuvering?

Earth/Moon L1/L2 has various advantages as a staging point, such as being outside the van allen belts and close to earth C3=0. But in principle something like a highly elliptical earth orbit (a more extreme form of GTO) would do as well. But then you have the issue that your spacecraft has to pass through the van allen belts every few days, and has to cope with very different thermal environments.

So many people have independently come to the conclusion that the small delta-v penalty of EML1/2 is worth it.
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Offline Danderman

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Re: Orbits Q&A
« Reply #16 on: 03/06/2011 06:41 PM »
Earth/Moon L1/L2 has various advantages as a staging point, such as being outside the van allen belts and close to earth C3=0. But in principle something like a highly elliptical earth orbit (a more extreme form of GTO) would do as well. But then you have the issue that your spacecraft has to pass through the van allen belts every few days, and has to cope with very different thermal environments.

You're almost on point there.

Let's look at a mission like Rosetta or Cassini, where they flew a trajectory that returned to Earth after *years*, and used an Earth swingby to travel to their final destination.

I am asking if its possible to duplicate this effort with shorter travel times, maybe not all the way out to Mars, but possible 50 or 100 days out, and then use an Earth swingby to travel to Mars. It would be nice to pick up some extra velocity from the maneuver, but the prime requirement is to be 100% sure of the Earth departure date - by launching ahead of time. Of course, another way of doing this is to launch into LEO ahead of time, but I am not thrilled about cold soaking the escape stage, CONTOUR tried this with disasterous results. Even if the escape stage works, there are potential underburns and the like which would ruin the mission. I am more interested in seeing if there is a way to send a probe beyond LEO and use a swingby to generate the appropriate Earth departure date.

Concerning the Van Allen belts, yeah, that's a problem, that's why I am looking at a 50 or 100 day trajectory, so there are minimum transits through the Van Allen belts, and they are quick transits.
« Last Edit: 03/06/2011 06:42 PM by Danderman »

Offline IsaacKuo

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Re: Orbits Q&A
« Reply #17 on: 03/07/2011 01:36 PM »
Let's look at a mission like Rosetta or Cassini, where they flew a trajectory that returned to Earth after *years*, and used an Earth swingby to travel to their final destination.

I am asking if its possible to duplicate this effort with shorter travel times, maybe not all the way out to Mars, but possible 50 or 100 days out, and then use an Earth swingby to travel to Mars.

No, it is not.  You can use the Moon for various things, which complicates things, but that's not what you're asking about.  For your purposes, we'll pretend the Moon doesn't exist.

It boils down to relative velocities.  With an unpowered gravity assist, the speed going in equals the speed going out--relative to the body in question.  It's similar to bouncing off of a wall.  Changing the angle of the wall can change the direction you bounce to, but it won't change the speed.

The orbital mechanics of leaving the Earth and coming back are such that you'll return with the same speed you left, relative to Earth, unless you use some thrust or you do a swing by some other body.  (Actually, you can return with a slightly different speed, due to the fact that Earth has a slightly non-circular orbit and your elliptical solar orbit can cross Earth's orbit in two different places.)

That said, it is possible to use the Oberth effect with multiple passes (also known as powered gravity assists).  If you are using a thruster with low acceleration, you can save propellant by using elliptical orbits where the perigee is close to Earth.  You thrust near perigee, which has the effect of incrementally raising your apogee.  For practical purposes, this only works until you reach Earth escape velocity.  In theory, you could continue to use this effect by waiting until you pass near Earth again, but this could take decades.

Offline Danderman

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Re: Orbits Q&A
« Reply #18 on: 03/07/2011 04:28 PM »
Although the last few responses are interested, and I will comment on them, let me try to restate the problem:

Are there heliocentric orbits departing from Earth that can later intersect the Earth's orbit while the Earth is at the intersect point that would allow gravitational swingby maneuvers leading to other destinations, such as Mars or Jupiter?

The answer clearly is: yes, because its been done. In the case of Rosetta, the spacecraft entered a heliocentric orbit with an apohelion beyond Mars, returned to the vicinity of Earth (at perihelion), and then used a swingby manuever to gain velocity and change direction to intercept a comet.

This begs the question as to what is required to enter a heliocentric orbit that would re-intercept the Earth at perihelion, is the required duration some number of years or are there shorter trajectories? Are there trajectories between Earth departure and Earth intercept with a duration of days or weeks or months?

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Offline IsaacKuo

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Re: Orbits Q&A
« Reply #19 on: 03/07/2011 05:04 PM »
Are there heliocentric orbits departing from Earth that can later intersect the Earth's orbit while the Earth is at the intersect point that would allow gravitational swingby maneuvers leading to other destinations, such as Mars or Jupiter?

Leading to other destinations, sure.  You can change direction with a graviational swingby.  You can't use an unpowered gravitational assist to gain more orbital energy than you could have started with.

Quote
The answer clearly is: yes, because its been done. In the case of Rosetta, the spacecraft entered a heliocentric orbit with an apohelion beyond Mars, returned to the vicinity of Earth (at perihelion), and then used a swingby manuever to gain velocity and change direction to intercept a comet.

Rosetta started off with an aphelion beyond Mars.  That's plenty of orbital energy to get to another destination within that region, but not enough to get to Jupiter.

Quote
This begs the question as to what is required to enter a heliocentric orbit that would re-intercept the Earth at perihelion, is the required duration some number of years or are there shorter trajectories? Are there trajectories between Earth departure and Earth intercept with a duration of days or weeks or months?

Unless you go out of your way to aim at Venus or Mars or some other planet, EVERY boost to Earth escape will eventually return to Earth.  You will be in an elliptical orbit of some sort, which intersects with Earth's orbit in at least one place (probably two places).  The only question is to ensure your orbital period is a suitable fractional multiple of 1 year, so that Earth arrives at that point at the same time you do.

If you want a duration of days, then you're actually still in Earth orbit.  You're just on an elliptical Earth orbit which returns from apogee soon enough.

If you want a duration of weeks, then you're in a fuzzy grey area which is sort of still in Earth's domain of influence, but where the Sun's influence is also significant.

If you want a duration of months, then you're in an elliptical solar orbit which takes a "hop" away from Earth before returning on the second intersection point.  Roughly speaking, you start off accelerating toward or away from the Sun.

In all three of these cases, you don't get any sort of helpful boost from an unpowered gravitational swingby of Earth--but you could use it to change your direction.  This might be useful for an extended mission to another target.

It seems that you are not very familiar with orbital mechanics, and could gain a lot of insight from reading up on the Oberth effect, gravity assists, and orbital equations.  Start with basic two body orbital mechanics (like Kepler's laws).  You need to first comprehend how a planet or drifting spacecraft moves before moving on to actual maneuvers.

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