Orbits Q&A

[baldusi]

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"?

[Jim]

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.

[baldusi]

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.

[Jim]

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

[Targeteer]

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...

[Danderman]

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?

[IsaacKuo]

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).

[Danderman]

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.

[IsaacKuo]

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.

[Danderman]

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!

[Danderman]

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?

[Danderman]

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?

[mmeijeri]

You can do this from a Lagrange point.

[Danderman]

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?

[mmeijeri]

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.

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.

[rklaehn]

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.

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.

[Danderman]

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.

[IsaacKuo]

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.

[Danderman]

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?

   

[IsaacKuo]

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.

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.

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.

[Danderman]

Re-stating my question would be helpful.

Let's say you have a spacecraft (MAVEN) and for whatever reason you want it to leave the vicinity of Earth on a specific date: December 7, 2013, no matter what. Therefore, a backup probe and rocket are irrelevant unless magic intervenes and they can be launched on December 7, as well if the prime mission fails (so let's ignore the possibility of a backup).

The other requirements are that the trans-Mars burn (or at least the major part of the burn) occur on the day of launch, and that transits through the Van Allen belt should be minimize. This means that long duration parking orbits in LEO or HEO really aren't useful.

What I am looking for is to see if the spacecraft could be launched weeks or months ahead of time into a heliocentric orbit that would later intercept the Earth's orbit while the Earth is there, and use the gravity of the Earth for a swingby on December 7 2013 to go to Mars.  If this is possible, what are the drivers for the duration of the loop away from the Earth? Its mentioned above that some fraction of a year is required, but what determines the size of the fraction? Is that fraction actually a full year?

 

[mmeijeri]

What's wrong with using a Lagrange point? The penalty is no more than tens of m/s, and that doesn't count any reduction in gravity losses through lower thrust requirements.

[Danderman]

What's wrong with using a Lagrange point? The penalty is no more than tens of m/s, and that doesn't count any reduction in gravity losses through lower thrust requirements.

Maybe that's a good choice if the actual delta v requirements to enter and depart the LaGrange point is that low. And assuming that a return to the vicinity to Earth for a gravitation swingby could be achieved from that location.

[IsaacKuo]

Let's say you have a spacecraft (MAVEN) and for whatever reason you want it to leave the vicinity of Earth on a specific date: December 7, 2013, no matter what.
[...]
The other requirements are that the trans-Mars burn (or at least the major part of the burn) occur on the day of launch, and that transits through the Van Allen belt should be minimize. This means that long duration parking orbits in LEO or HEO really aren't useful.

Why not?  Neither would imply spending more time in the Van Allen belts.

What I am looking for is to see if the spacecraft could be launched weeks or months ahead of time into a heliocentric orbit that would later intercept the Earth's orbit while the Earth is there, and use the gravity of the Earth for a swingby on December 7 2013 to go to Mars.

No.  It is not possible.  In order to intercept the Earth again in such a short time, the aphelion must be small.  This lacks the orbital energy required to reach Mars.  An unpowered gravity assist will not help.

You could use a powered gravity assist, but this will not save you any fuel compared to simply going straight for Mars on the desired date (either from LEO or from Earth's surface).

But my big question for you is--what is the point?  What are you trying to accomplish with this idea?  Why not simply launch during the normal launch window?

[sdsds]

Tha ballistic transfers described by Parker (http://ccar.colorado.edu/nag/papers/AAS%2006-132.pdf) are your friends.  They just barely enter the realm of solar influence before returning to e.g. the L1 attractor.

[Danderman]

What I am looking for is to see if the spacecraft could be launched weeks or months ahead of time into a heliocentric orbit that would later intercept the Earth's orbit while the Earth is there, and use the gravity of the Earth for a swingby on December 7 2013 to go to Mars.

No.  It is not possible.  In order to intercept the Earth again in such a short time, the aphelion must be small.  This lacks the orbital energy required to reach Mars.  An unpowered gravity assist will not help.

You could use a powered gravity assist, but this will not save you any fuel compared to simply going straight for Mars on the desired date (either from LEO or from Earth's surface).

I think you are getting close to nailing the problem - since the requirement is for most of the trans-Mars burn to occur on the day of launch (for various reasons), then the duration of the loop away from the Earth will be a function of the magnitude of that burn. If the burn is Earth C3 + a small delta-v, yep, the aphelion will indeed be small.

On the other hand, if the Earth escape burn is, say, 100 m/s under what is required to reach Mars, then the loop away from the Earth will take years.

So, if the initial burn is small enough that the loop away from the Earth is short, the resulting Earth swingby on December 7 2013 must incorporate a fairly significant burn to enable the spacecraft to actually get to Mars.  This is not an "extra" burn in the sense of requiring more propellant than a direct injection to Mars on December 7 2012, but it does entail the risk of splitting the trans-Mars injection into two distinct burns, precisely what killed the Nozomi mission.

[mmeijeri]

Maybe that's a good choice if the actual delta v requirements to enter and depart the LaGrange point is that low. And assuming that a return to the vicinity to Earth for a gravitation swingby could be achieved from that location.

Based on the delta-v's for ballistic transfers to L1/L2 and the cost of an insertion burn into an L1/L2 orbit on a fast transfer trajectory (which should be equal to the cost of a fast Earth swingby) it would be that cheap. Huntress' exploration architecture was based around use of Lagrange points. You could do a swingby from EML1/2 or, as Farquhar figured out, even more efficiently (but less flexibly as Kirk Sorensen pointed out) both an Earth and a moon swingby from SEL1/2.

[IsaacKuo]

I think you are getting close to nailing the problem

What is the problem?

- since the requirement is for most of the trans-Mars burn to occur on the day of launch (for various reasons), then the duration of the loop away from the Earth will be a function of the magnitude of that burn. If the burn is Earth C3 + a small delta-v, yep, the aphelion will indeed be small.

Huh?  You can make the aphelion pretty much whatever you want.  The reason I said you would need a small aphelion is because you want to return to Earth for some bizarre reason within a few weeks or months.  You can't go far away from Earth and expect to return so soon.

So, the Earth swingby must incorporate a fairly significant burn to enable the spacecraft to actually get to Mars.  This is not an "extra" burn in the sense of requiring more propellant than a direct injection to Mars on December 7 2012, but it does entail the risk of splitting the trans-Mars injection into two distinct burns, precisely what killed the Nozomi mission.

Actually, this second burn would be relatively small.  You already have greater than escape velocity--meaning you come in with a positive C3 already.  You only need to boost your C3 by a small amount to get to the desired transfer orbit to Mars.  Maybe 300m/s delta-v or less, depending on the specifics.  It's small change compared to the 11+km/s you needed to escape Earth in the first place.

Anyway, the Nozomi mission involved using the Moon for gravity assist maneuvers.  Like I said, the existence of the Moon allows for various possibilities.

So again--what is the point of all of this?  You keep saying there is some "problem", but what is the "problem" you are concerned with?

[Danderman]

You could do a swingby from EML1/2 or, as Farquhar figured out, even more efficiently (but less flexibly as Kirk Sorensen pointed out) both an Earth and a moon swingby from SEL1/2.

What would you gain from the lunar swingby?

[mmeijeri]

What would you gain from the lunar swingby?

I got the impression the idea was to gain efficiency through use of two powered swingbys.

[Danderman]

What would you gain from the lunar swingby?

I got the impression the idea was to gain efficiency through use of two powered swingbys.

You are correct, but I am always on the lookout for a free lunch!

If you go back to the premise of this thread, the requirement for all this is to absolutely positively ensure an Earth departure date for Mars on December 7 2013. Gaining efficiencies as part of the architecture is a desirement, not a requirement.

[Danderman]

You could do a swingby from EML1/2 or, as Farquhar figured out, even more efficiently (but less flexibly as Kirk Sorensen pointed out) both an Earth and a moon swingby from SEL1/2.

It appears that the Delta-V required to achieve EML1 or EML2 is fairly high for a direct injection (3.8 meters/second) - I am not sure how much of this could be recaptured after an Earth swingby for a Mars trajectory.

[mmeijeri]

It appears that the Delta-V required to achieve EML1 or EML2 is fairly high for a direct injection (3.8 meters/second) - I am not sure how much of this could be recaptured after an Earth swingby for a Mars trajectory.

The injection into a fast transfer trajectory is a little under 3.2km/s while the injection into a slow one is a little over 3.2km/s. The fast trajectory requires a ~0.6km/s insertion burn, while the slow one doesn't require one at all, instead relying on perturbation of the transfer orbit by the Sun. After a 0.6km/s perigee lowering burn for the swingby you should recover the 3.2km/s over LEO orbital velocity at perigee. This suggests the penalty is ~0.6km/s and not the mere tens of m/s I mentioned above. Of course in the case of an MTV that still disregards the large savings of cycling between the edges of gravity wells instead ascending / descending them completely.

EDIT: I suspect you could use the ballistic trajectory in reverse too, in which case the penalty should really be mere tens of m/s be zero(?), but at the cost of another 100 days travel time, which may or may not be an advantage for a manned mission. This would be similar to Farquhar's SEL1/2 scheme.

[Danderman]

It appears that the Delta-V required to achieve EML1 or EML2 is fairly high for a direct injection (3.8 meters/second) - I am not sure how much of this could be recaptured after an Earth swingby for a Mars trajectory.

The injection into a fast transfer trajectory is a little under 3.2km/s while the injection into a slow one is a little over 3.2km/s. The fast trajectory requires a ~0.6km/s insertion burn, while the slow one doesn't require one at all, instead relying on perturbation of the transfer orbit by the Sun. After a 0.6km/s perigee lowering burn for the swingby you should recover the 3.2km/s over LEO orbital velocity at perigee. This suggests the penalty is ~0.6km/s and not the mere tens of m/s I mentioned above. Of course in the case of an MTV that still disregards the large savings of cycling between the edges of gravity wells instead ascending / descending them completely.

EDIT: I suspect you could use the ballistic trajectory in reverse too, in which case the penalty should really be mere tens of m/s be zero(?), but at the cost of another 100 days travel time, which may or may not be an advantage for a manned mission. This would be similar to Farquhar's SEL1/2 scheme.

Yeah, I was looking for a free lunch. I guess the point solution for my particular question is a heliocentric orbit that returns to the vicinity of the Earth when the Earth is there, that's the quick and easy approach.

[Proponent]

You can't use an unpowered gravitational assist to gain more orbital energy than you could have started with.

Actually, it is possible and is done.  Here's Wikipedia's explanation.  Basically, viewed in the frame of the planet, the spacecraft's energy is conserved, and it arrives and departs at the same speed though different directions.  In the sun's frame, however, the spacecraft can gain or lose energy.  Energy is still conserved, because in the sun's frame the planet's speed changes a (very tiny) bit too.

[IsaacKuo]

You can't use an unpowered gravitational assist to gain more orbital energy than you could have started with.

Actually, it is possible and is done.  Here's Wikipedia's explanation.  Basically, viewed in the frame of the planet, the spacecraft's energy is conserved, and it arrives and departs at the same speed though different directions.  In the sun's frame, however, the spacecraft can gain or lose energy.  Energy is still conserved, because in the sun's frame the planet's speed changes a (very tiny) bit too.

That's true in general, but this was in reference to a specific situation--using Earth only (and pretending that the Moon doesn't exist).  You could only use it to gain energy up until what you could have started with--by thrusting directly along the path of Earth's orbit.  You can change your direction relative to Earth, but not your speed.  So your maximum speed relative to the Sun is equal to your speed relative to Earth plus Earth's orbital speed.

[baldusi]

If you were in the equator, in the middle of the Atlantic. Would it be easier, worse or the same than being in Antartica for launching polar orbits? Or SSO?

[kevin-rf]

Polar orbits are unable to use that extra bump from the earths rotation, and actually have to counter it. So yes you take a payload hit. But the name of the game is logistics, not the optimal launch profile. Antarctica would be darn expensive to operate out of.

[baldusi]

Polar orbits are unable to use that extra bump from the earths rotation, and actually have to counter it. So yes you take a payload hit. But the name of the game is logistics, not the optimal launch profile. Antarctica would be darn expensive to operate out of.

what about Ushuaia? It's got an airport, a deep water port and it's home to electronics manufacturing. They are also developing a deep space radar station around there.

[Jim]

Polar orbits are unable to use that extra bump from the earths rotation, and actually have to counter it. So yes you take a payload hit. But the name of the game is logistics, not the optimal launch profile. Antarctica would be darn expensive to operate out of.

what about Ushuaia? It's got an airport, a deep water port and it's home to electronics manufacturing. They are also developing a deep space radar station around there.

Plesetsk is closer to a pole

[Danderman]

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?

I answered my own question, as the chart below shows.  All that is necessary is to use phasing loops to put the payload into an elliptical orbit several weeks/months prior to a final lunar swingby that injects the payload into a heliocentric orbit. 

The answer is basically not only "yes", but its been done several times. The effect of using this maneuver is to widen the launch window to a specific planet considerably.  I was re-inventing the wheel on this one.

This chart is from a paper called "DESIGN OF LUNAR GRAVITY ASSIST FOR THE
BEPICOLOMBO MISSION TO MERCURY"

[jabe]

on a slightly off topic question but it will help me sort out an orbit question, I think I am getting the terminology right, If you were in a space craft and needed to know my velocity Vector relative to a planet as I was entering that planets Sphere of influence, what is the best way to figure that out?  My orbit would obviously hyperbolic around the planet, put how can you ensure you are aimed along the proper asymptote  to get the required periapsis?
jb

[glanmor05]

Apologies if this isn't in the correct thread, but it seemed the best fit and I didn't want to start a new one for what I'm sure will end up being a stupid question.

When a vehicle (or for that matter anything) enters the earths atmoshere and there's "loads" of heat energy generated by friction, does the Earth's rotation slow by a very, very small amount or just the rotation of the atmoshere (or are they the same thing or linked)?

If so, over a very, very long time (if we continue to re-enter bigger and more numerous things (e.g. ISS)) might we eventually cause something "bad" to happen?

It should be clear to anyone reading this that I know naught about physics, so I'd be grateful if anyone who takes the time to answer would be kind...

Thanks. 

[Robotbeat]

Apologies if this isn't in the correct thread, but it seemed the best fit and I didn't want to start a new one for what I'm sure will end up being a stupid question.

When a vehicle (or for that matter anything) enters the earths atmoshere and there's "loads" of heat energy generated by friction, does the Earth's rotation slow by a very, very small amount or just the rotation of the atmoshere (or are they the same thing or linked)?

If so, over a very, very long time (if we continue to re-enter bigger and more numerous things (e.g. ISS)) might we eventually cause something "bad" to happen?

It should be clear to anyone reading this that I know naught about physics, so I'd be grateful if anyone who takes the time to answer would be kind...

Thanks. 

Don't worry about it. Angular momentum is conserved, so the change in rotation is basically the same (but of opposite sign) as the amount the rotation was changed when it was launched. Besides, the Moon changes the Earth's rotation by A LOT more than some puny station.... imagine the force of all the world's tides... I suppose if the Moon didn't retreat from orbit, eventually we would be tidally locked with the Moon as the Moon is with the Earth.

[IsaacKuo]

Don't worry about it. Angular momentum is conserved, so the change in rotation is basically the same (but of opposite sign) as the amount the rotation was changed when it was launched.

There is a very slight amount of mass that reaches Earth escape velocity, though.  Spacecraft which de-orbit fire thrusters in the direction of their motion briefly, which can give the thruster exhaust Earth escape velocity.  Since all of these spacecraft have been in prograde orbits, the overall effect is to extremely slightly reduce Earth's rotation...

Besides, the Moon changes the Earth's rotation by A LOT more than some puny station....

...but of course, this effect is utterly insignificant compared to the Moon and other factors.

[Robotbeat]

Don't worry about it. Angular momentum is conserved, so the change in rotation is basically the same (but of opposite sign) as the amount the rotation was changed when it was launched.

There is a very slight amount of mass that reaches Earth escape velocity, though.  Spacecraft which de-orbit fire thrusters in the direction of their motion briefly, which can give the thruster exhaust Earth escape velocity.  Since all of these spacecraft have been in prograde orbits, the overall effect is to extremely slightly reduce Earth's rotation...

Besides, the Moon changes the Earth's rotation by A LOT more than some puny station....

...but of course, this effect is utterly insignificant compared to the Moon and other factors.

Not all spacecraft deorbit, though. But basically all have an insertion burn, which should have exact same effect (small amount of propellant at escape velocity), but in opposite direction. Never mind. The insertion burn exhaust would have less specific energy than the satellite, not more as in the case of a deorbit burn. You're right.

[kevin-rf]

When a vehicle (or for that matter anything) enters the earths atmoshere and there's "loads" of heat energy generated by friction, does the Earth's rotation slow by a very, very small amount or just the rotation of the atmoshere (or are they the same thing or linked)?

In most cases it would be the opposite, most non polar orbits (with the exception of Israeli, who's satellites are in retrograde orbits) orbit in the direction of earths rotation.

So when you de-orbit one, you go into a transfer orbit that reenters in the same direct that earth rotates, hence the capture gives the atmosphere around it a shove in the direction that the earth rotates.

But as others have mentioned, this is a minor effect compared to all the other effects on earths rotation. There are several large factors that have measurable effects on the earths rotation, including the moon, weather, earthquakes, season accumulation and melting of snow, erosion, ect. It is not a constant.

[Robotbeat]

(you edited the quotes wrong... I didn't say that, I was quoting someone else. )

[kevin-rf]

(you edited the quotes wrong... I didn't say that, I was quoting someone else. )

Some days you win, some days the text editor bites you in the incorrect... Fixed now

[deltaV]

The nodes of the orbit of a LEO propellant depot at a roughly Cape Canaveral inclination would precess with a period of a couple of months (claim based on equations at http://en.wikipedia.org/wiki/Sun_synchronous). Such a depot's orbital plane would therefore be poorly aligned for departures much of the time. My question is how exploration architectures with propellant depots or other long-term structures in LEO typically handle this issue. Perhaps they just accept that this effect will limit available earth departure time windows? Or do they do a maneuver such as raising apogee to 100,000 km or so, do the plane change at apogee, and then do the rest of the earth departure burn at the next perigee (for Oberth effect)? The latter maneuver would be pretty similar to the proposal I've heard of an EML-1 rendezvous and then doing the trans-mars injection during an earth flyby.

[mmeijeri]

Using L1/L2 as a staging point would be the natural solution. That's a good idea for a long list of other reasons too. So much so that not using L1/L2 would be a bad idea.

[deltaV]

Using L1/L2 as a staging point would be the natural solution. That's a good idea for a long list of other reasons too. So much so that not using L1/L2 would be a bad idea.

Can anyone recommend a relatively unbiased summary of the pros and cons of rendezvous in LEO, Earth-Moon Lagrange points, low lunar orbit, and lunar surface? For NEO missions, substitute "during the month-long coast to the NEO" and "at the NEO" for "low lunar orbit" and "lunar surface" respectively. I've found plenty of proposals advocating various architectures, but no clear unbiased summaries. The ESAS gives some discussion of LEO vs. LLO, but next to nothing about lunar surface or Lagrange points.

Edit: even a biased summary of the advantages of Lagrange point rendezvous would be nice.

[deltaV]

I just found a thread which may answer my questions:

[IsaacKuo]

The biggest advantage of LEO staging is that it's useful for satellites to GEO.  In other words, LEO staging is economically useful.

As for staging for interplanetary missions:

Recently, I contemplated a new option--staging a tanker at in a solar co-orbit perhaps 2+ million kilometers from Earth.  This co-orbit is a quasi-orbit around a point roughly 2 million kilometers ahead of or behind Earth, such that it's at an angle around 30 degrees off.  (This is roughly equal to the hyperbolic escape half angle.)

Unlike EML2, this co-orbit is always in the correct right ascension.  There's no waiting for the right time of the month.  The delta-v requirements to get to a near parabolic orbit and back are minimal (i.e. less than EML2), so the tanker itself can make the round trip to meet up with the client vehicle before returning outside Earth's Hill sphere.  The client vehicle still has to get itself to the appropriate elliptical orbit before rendezvous.

As an extra bonus, the starting velocity is a bit above escape velocity rather than below escape velocity.

[sdsds]

Recently, I contemplated a new option--staging a tanker at in a solar co-orbit perhaps 2+ million kilometers from Earth.  This co-orbit is a quasi-orbit around a point roughly 2 million kilometers ahead of or behind Earth, such that it's at an angle around 30 degrees off.  (This is roughly equal to the hyperbolic escape half angle.)

Do I understand correctly these co-orbits are roughly half way to the SEL4 and SEL5 points (which are roughly 60 degrees off)?

[IsaacKuo]

Recently, I contemplated a new option--staging a tanker at in a solar co-orbit perhaps 2+ million kilometers from Earth.  This co-orbit is a quasi-orbit around a point roughly 2 million kilometers ahead of or behind Earth, such that it's at an angle around 30 degrees off.  (This is roughly equal to the hyperbolic escape half angle.)

Do I understand correctly these co-orbits are roughly half way to the SEL4 and SEL5 points (which are roughly 60 degrees off)?

No these co-orbits are near Earth--maybe 2 or so million kilometers away.  SEL4 and L5 are about 1AU away (150 million kilometers away).

By 30 degrees off, I mean how it looks from Earth.  If you were on Earth and you looked in the direction of Earth's orbital motion, you would see the point 2 million kilometers ahead of Earth's orbit directly ahead.  The spacecraft would appear to revolve around this point, staying roughly 30 degrees away from this point.

Incidentally, SEL4 would also appear to be 30 degrees away from this point, but it's much further away.

[baldusi]

Let's say that time is of not great importance. How much delta-v would be required to got to a different place withing Earth's orbit? For example a troyan? Significantly more than C3=0? Because, intuitively, once escaping along the correct vector, you might be pretty much along the orbit, right?

[IsaacKuo]

Let's say that time is of not great importance. How much delta-v would be required to got to a different place withing Earth's orbit? For example a troyan? Significantly more than C3=0? Because, intuitively, once escaping along the correct vector, you might be pretty much along the orbit, right?

You don't need significantly more than C3=0.  You just need to escape on a vector that puts you on a solar orbit with a slightly lower or higher orbital period.  This more or less puts you directly into a horseshoe orbit.  You will orbit around the Sun slightly faster or slower than Earth.  After some decades or centuries, you will end up in the desired place, and a small thrust can adjust your orbital period to exactly 1 year.

[Galactic Penguin SST]

When one describes the characteristics of a sun-synchronous orbit (SSO), it is convenient to describe the mean local time at the ascending/descending node of the satellite. What is the time system used for the MLT?

[IsaacKuo]

When one describes the characteristics of a sun-synchronous orbit (SSO), it is convenient to describe the mean local time at the ascending/descending node of the satellite. What is the time system used for the MLT?

I am not an expert or an authority, but there's only one answer that makes any sense to me--it's simply a conversion from degrees to 24 hour clock.

For example, if the ascending node is exactly 0 degrees from the Sun's position, then that translates to 12:00 Noon.

[Jim]

When one describes the characteristics of a sun-synchronous orbit (SSO), it is convenient to describe the mean local time at the ascending/descending node of the satellite. What is the time system used for the MLT?

Time system, meaning 12 or 24 hr?

[Galactic Penguin SST]

When one describes the characteristics of a sun-synchronous orbit (SSO), it is convenient to describe the mean local time at the ascending/descending node of the satellite. What is the time system used for the MLT?

Time system, meaning 12 or 24 hr?

Nah. What I am meaning is that whether that time corresponds to the local apparent solar time, the mean solar time, or the sidereal time, or something else...

[Jim]

it doesn't really matter, the differences are small.

[deltaV]

I'm wondering what ranges of declination of the launch asymptote (DLA) are encountered in beyond LEO mission planning. (The reason I'm wondering is this affects the inclination of a LEO depot.)

The declination of the moon never exceeds 28.6 degrees. If I understand lunar trajectories correctly DLA for a moon mission won't exceed this value by much. Cape Canaveral has about that inclination so I'm guessing excessive DLA is not a significant issue for lunar missions.

Apparently Mars Odyssey had a DLA of 51.7 degrees: page 14 of http://www.boeing.com/defense-space/space/delta/kits/d284_mars_odyssey.pdf. (Thanks to Jim for pointing out Mars Odyssey's declination in another thread.) Presumably this large declination is due to the Earth's 30 km/s orbital speed making even a few degrees of plane change a big deal. Does anyone know what the maximum DLA from Earth one can encounter when planning conjunction class (i.e. Hoffman transfer both ways) Mars missions?

For NEOs someone (Antares I think jongoff) pointed out in another thread that many NEOs have large declinations. Can anyone quantify this? For example a list of top NEO opportunities and their DLAs?

Edit: interesting article on Apollo launch windows and trajectory planning: http://history.nasa.gov/afj/launchwindow/lw1.html .

[deltaV]

Does anyone know what the maximum DLA from Earth one can encounter when planning conjunction class (i.e. Hoffman transfer both ways) Mars missions?

Answering part of my own question, it looks like Mars Odyssey was near the upper end of the range of Mars transfer DLAs. See figure 5 of  http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/19062/1/98-0259.pdf and many figures of http://www.scribd.com/doc/33801471/NASA-MSFC-Mission-Design-Handbook .

Edit: the closest I've found for an answer to the part of my question about NEOs is in a Lockheed Martin Plymouth Rock paper: http://www.lockheedmartin.com/data/assets/ssc/Orion/Toolkit/OrionAsteroidMissionWhitePaperAug2010.pdf . On pages 20-21 they mention that some NEO missions require a declination of more than 50 degrees and that KSC is limited to inclinations of at most 55 degrees due to range safety constraints.

Edit: here are some more NEO mission planning sources:
http://www.kiss.caltech.edu/workshops/space-challenge2011/references/human-exploration-on-neo/Asteroids%20Destinations%20Accessible%20for%20Human%20Exploration_A%20Preliminary%20Survey_Adamo%20et%20al.pdf
http://www.kiss.caltech.edu/workshops/space-challenge2011/references/asteroids-classification-and-study/A%20Comprehensive%20Ongoing%20Survey%20of%20the%20Near-Earth%20Asteroid%20Population%20for%20the%20Human%20Mission%20Accessibility_Barbee%20et%20al.pdf
http://www.nasa.gov/pdf/474223main_Johnson_ExploreNOW.pdf
It seems that there are quite a few NEO opportunities with asymptote declinations less than 40 degrees.

[baldusi]

Silly question, but is there an easy formula to get the period for a revisit to a point by each orbit (Assuming circular), given the inclination, altitude and also assuming that the orbit starts above the point?
I'm wondering how to chose an orbit to maximize launch windows. If I was on the Equator that would be easy :-p, but for the rest of the cases I'm wondering if there's a "simple" formula, or some open source program to calculate it.
I'm not asking about maximizing over multiple launch site, non circular orbits, etc. The simplest possible case.

[IsaacKuo]

Roughly, you can design the orbit to have a period which is a fraction of about one sidereal day.  This lets you reach the orbit once per day.  It's a little bit off, due to orbital precession, but the essential idea is to have one window per day.

Depending on the specific acceptable launch angles from the launch site, it may be possible to design an orbit to provide two windows per day.  Generally, though, one per day is the most you could get.

[kevin-rf]

After unexpectedly spotting the NOSS 3-5 pair last night, and then noticing how unnatural it looks for two satellites to be flying in such close formation, a question formed in my mind  (Watching the two cross the sky I really thought I was witnessing debris from a breakup event).

Anyway the question is how did the earlier NOSS first and second generation satellites maintain the triangle formation? Was the triangle constant through out the orbit, or was the geometry constantly evolving as the three satellites orbited the earth. I just can not see how one satellite would always be to one side of the orbital plane.

Anyone know how this formation flying was done from an orbital mechanics standpoint? Was there always a triangle, or only over certain parts of the orbit?

[IsaacKuo]

Anyway the question is how did the earlier NOSS first and second generation satellites maintain the triangle formation? Was the triangle constant through out the orbit, or was the geometry constantly evolving as the three satellites orbited the earth. I just can not see how one satellite would always be to one side of the orbital plane.

Anyone know how this formation flying was done from an orbital mechanics standpoint? Was there always a triangle, or only over certain parts of the orbit?

I don't know anything about what was done with NOSS, but in theory it's possible to set up a triangular formation.

First off, imagine the center point is a circular orbit.  Then, each of the satellites maintains an elliptical orbit with the same semi-major axis, but is inclined enough so that it appears to make a tight circle around that centerpoint.

If we take the imaginary central orbit to be around the equator, then the inclination provides near sinusoidal north-south motion.

The elliptical eccentricity provides cyclic east-west motion (it goes faster or slower based on altitude).

The proper combination of this north-south and east-west motion results in a circular motion around that imaginary central point.

Each of the three satellites can be in such orbits, but with orbital parameters 120 degrees apart from each other, so they always form a roughly equilateral triangle.

[C5C6]

Is there any easy-to-understand ISS rendezvous procedure overview? I still don't understand what type of burns are necessary, how the spacecraft orbit changes on each burn and every time I see an orbit graphic I don't understand what I see...

[Danderman]

I'm wondering what ranges of declination of the launch asymptote (DLA) are encountered in beyond LEO mission planning.

I am guessing that by "declination" you mean the plane of the target body's orbit in relation to the plane of the Earth's orbit around the Sun. Or, in the case of the Moon, the inclination of the Moon's orbit around the Earth in relation to the Equator.

[mmeijeri]

I am guessing that by "declination" you mean the plane of the target body's orbit in relation to the plane of the Earth's orbit around the Sun.

No, it's the angle between the launch asymptote and the equatorial plane, as someone explained to me a while back. For escape missions you just care about the asymptote (and your velocity), not the (geocentric) orbital plane. Any plane containing the asymptote will do, though not all of them are equally easy to reach from the ground of course.

[jabe]

not exactly sure where to ask it so i will ask it here.
I am going to be talking to my physics students about using EML-1 as a hypothetical spot to build a spacestation and then launch a probe to mars.  To get the probe to mars do they do a small burn to get it into a large elliptical orbit around earth (apogee of EML1 distance and perigee altitude of about 200km(?) ) and then do a burn at perigee to get the probe into the earth mars transfer orbit?
is 1) this right way to do it..and
   2) any major advantage to use EML2 instead of EML1?
jb

[sdsds]

A few comments about the approximations involved. A) Most analysis uses a circular-restricted three body problem model, which assumes the Earth and Moon are in circular orbits around their combined barycentre, but in reality they're not. B) Even if reality met the assumptions of the CR3BP model it would be essentially impossible to balance exactly at a lagrange point without some station-keeping, as even the tiniest deviation leads onto trajectories that head away from the lagrange point. (Thus the expectation is that a spacestation would instead be in a stable "orbit" around a lagrange point.) C) The departure burn you describe would not lead to an exactly elliptical trajectory because the Moon would continue to exert some gravitational influence even after the burn.

That said these approximations are reasonable and should lead to solutions that are close to reality.

As for your question 2, some seem to prefer EML2 because it allows a powered swing-by of the Moon, thus decreasing the total delta-v needed to get onto a trajectory headed for your target perigee. IMO those who push this may not have adequately considered the counter-intuitive possibility of a lunar swing-by after an EML1 departure. The trajectories involved are strange, for sure. Note that a lunar swing-by to save delta-v almost certainly incurs an increase in delta-t.

Caveats: I am not an expert. Don't plan a real mission based on these observations. Your delta-v's may vary from the EPA estimates depending on your driving style.

[jabe]

A few comments about the approximations involved. A) Most analysis uses a circular-restricted three body problem model, which assumes the Earth and Moon are in circular orbits around their combined barycentre, but in reality they're not.

thnx for that clarification.

B) Even if reality met the assumptions of the CR3BP model it would be essentially impossible to balance exactly at a lagrange point without some station-keeping, as even the tiniest deviation leads onto trajectories that head away from the lagrange point. (Thus the expectation is that a spacestation would instead be in a stable "orbit" around a lagrange point.)

did not realize that it could not actually stay there.  I realized they could orbit though.  I'm assuming the"field lines" for the diagram attached are gravitational potential field lines?  Or am I off base there as well.

C) The departure burn you describe would not lead to an exactly elliptical trajectory because the Moon would continue to exert some gravitational influence even after the burn.

That said these approximations are reasonable and should lead to solutions that are close to reality.

I realized that they are approximations..good to know they are a reasonably accurate ones.

As for your question 2, some seem to prefer EML2 because it allows a powered swing-by of the Moon, thus decreasing the total delta-v needed to get onto a trajectory headed for your target perigee. IMO those who push this may not have adequately considered the counter-intuitive possibility of a lunar swing-by after an EML1 departure. The trajectories involved are strange, for sure.

Never thought of swing by.  Is EML1 a good point to place a station for ease of use when leaving the lunar surface or is a spacestation at EML4/5 a better choice?

Note that a lunar swing-by to save delta-v almost certainly incurs an increase in delta-t.

for increase delta-t I guess only matters if cargo is humans

Caveats: I am not an expert. Don't plan a real mission based on these observations. Your delta-v's may vary from the EPA estimates depending on your driving style.

dam..I have to now rebuild my spaceship that I was doing..now I know my EPA estimates were way off.

so one more question...for now
EML3 any use at all?
jb

[baldusi]

I can't really understand the Lissajous orbit, so I'm going to ask. Can you bias a Lissajous orbit around EML2, so that you spend more time looking to the moon poles than the equator? In other words, can you make it have an "apogee" perpendicular to the moon orbit plane and the "perigee" when it intersect it?
I'm thinking of having three satellites on such an orbit for permanent LOS to the poles, sort of like the Russians do with their Molnyia orbits.

[MP99]

not exactly sure where to ask it so i will ask it here.
I am going to be talking to my physics students about using EML-1 as a hypothetical spot to build a spacestation and then launch a probe to mars.  To get the probe to mars do they do a small burn to get it into a large elliptical orbit around earth (apogee of EML1 distance and perigee altitude of about 200km(?) ) and then do a burn at perigee to get the probe into the earth mars transfer orbit?
is 1) this right way to do it..

The answer here is "it depends".

If your probe has a very weak but high-Isp engine (IE Solar Electric Propulsion, like Dawn), then your probe doesn't want to struggle out of Earth's gravity well. Dawn only fired it's SEP once Delta II had injected it past escape velocity.

This mode would also work spiralling away from EML, after you'd perhaps used a chemical stage to deliver your probe to the Moon, though the Dawn mode (chemical injection past escape, then SEP in interplanetary space) is probably better.



If your probe has a powerful rocket engine, then you get an advantage from the Oberth effect if it perform it's TMI burn as close as possible to the Earth. If it starts in the vicinity of the Moon, a relatively small burn will leave the Moon and put the probe into an elliptical orbit that has it's perigee at LEO. The Oberth benefit at perigee when it does the TMI is bigger than the cost of the departure burn from the Moon.

But, it's worth noting that EML is not a free lunch. Imagine your spacecraft was launched through TLI, but the Moon happened to be in the wrong part of the sky. The s/c would end up in an elliptical orbit around the Earth with it's apogee near the Moon's orbit and it's perigee at LEO. The s/c could perform it's TMI burn at any subsequent perigee and gain the Oberth benefit mentioned above.

In order to go to EML you need to add some velocity to circularise the orbit. Then, you need another burn to reverse that and leave EML again, and further dV to reduce the orbit's perigee down to LEO, and that's just to put you back into essentially the same orbit you were in just after you'd finished TLI. That's quite a lot of extra dV, and is badly counter-productive if all the manoeuvres are performed by a single stage. Far more efficient if your stage just performs a simple TMI directly from LEO immediately after it's launched, exactly like all current Mars missions have done.



To get any benefit from EML you need a more complicated mission plan - some form of assembly of components:-

A chemical stage that boosts your stage through TLI, docks and gets refueled at EML, before leaving, reducing perigee and performing TMI (via the Lunar flyby).

Or, the probe and it's TMI stage delivered separately to EML, docking together and then performing TMI. This mode would work for either a chemical or SEP TMI stage.



As far as getting to EML, if the element can survive a long transit (months) then it can can get to EML quite efficiently using a ballistic capture trajectory. For quick transits, EML2 needs less total dV. This may allow crew to quickly transit to a hab or TMI stage that is bigger because it was able to get there via a slow-boat trajectory.

For chemical propulsion, I believe you can get from EML2 into that elliptical orbit that takes advantage of the Oberth effect with less dV than from EML1.

cheers, Martin

[MP99]

B) Even if reality met the assumptions of the CR3BP model it would be essentially impossible to balance exactly at a lagrange point without some station-keeping, as even the tiniest deviation leads onto trajectories that head away from the lagrange point. (Thus the expectation is that a spacestation would instead be in a stable "orbit" around a lagrange point.)

did not realize that it could not actually stay there.  I realized they could orbit though.  I'm assuming the"field lines" for the diagram attached are gravitational potential field lines?  Or am I off base there as well.

See http://en.wikipedia.org/wiki/Halo_orbit.

For EML2 this is actually an advantage, because the probe is not eclipsed by the Moon, so can communicate with the Earth.

cheers, Martin

[jabe]

As far as getting to EML, if the element can survive a long transit (months) then it can can get to EML quite efficiently using a ballistic capture trajectory. For quick transits, EML2 needs less total dV. This may allow crew to quickly transit to a hab or TMI stage that is bigger because it was able to get there via a slow-boat trajectory.

For chemical propulsion, I believe you can get from EML2 into that elliptical orbit that takes advantage of the Oberth effect with less dV than from EML1.

cheers, Martin

some great points i hadn't thought of, Ion vs chemical.
I have both of Bruno's books. his Fly me to the moon  is excellent.  A MUST read for anyone who has interest in non hohmann orbits..VERY easy read. His other one is just a tad over my head but interesting to sorta follow.
My premise to the discussion is that  station is there and we have a probe ready to fly..so what best to do..  For high school can't get too much into math..just the general train of thought.

I appreciate everyones input..

jb

[aero]

What is BEO?

[mmeijeri]

What is BEO?

Beyond Earth Orbit, often confused with beyond Low Earth Orbit. The moon is beyond LEO, but not BEO.

[truth is life]

Stimulated by some comments made [topic=29081.150]here[/topic] simonbp and jcm, I was wondering some things about the orbits into which space observatories are put:

1: Clearly, a lot of observatories benefit from being outside of LEO, in highly elliptical orbits or even at SEL1/2. Just as clearly, a lot of observatories don't, or at least don't benefit enough for it to be worthwhile paying for the bigger rocket to escape LEO. What sorts of factors lead to the selection of the observatory's orbit? Clearly cost and launch vehicle capability is one of them, but are there any frequency bands which benefit particularly from certain locations, for example?

2: At what point were the possible advantages of these alternative orbits, especially SEL-2, identified? The 1960s, the 1970s, later? Because of the Shuttle and the possibility of repair and refurbishment, in the 1970s and 1980s many observatories were going to be placed into LEO, correct? (I know at least three, Hubble, Compton, and SIRTIF were). Do you think that without Shuttle there might have been more consideration of alternate orbits for these vehicles?

(This seemed to belong here, since it *is* about orbits...)

[Jim]

2: At what point were the possible advantages of these alternative orbits, especially SEL-2, identified? The 1960s, the 1970s, later? Because of the Shuttle and the possibility of repair and refurbishment, in the 1970s and 1980s many observatories were going to be placed into LEO, correct? (I know at least three, Hubble, Compton, and SIRTIF were).

Where did SIRTIF eventually go?

GRO was gamma rays which are not affected by anything.

[mmeijeri]

At what point were the possible advantages of these alternative orbits, especially SEL-2, identified?

Interesting question. The Lagrange points have been known for a very long time, having been discovered by Euler (hence the name ), but to the degree early exploration plans used high energy staging orbits, they focused on highly elliptical ones, or perhaps general high altitude circular orbits, not Lagrange points specifically.

[truth is life]

Where did SIRTIF eventually go?

SIRTIF eventually became Spitzer and ended up in a heliocentric orbit. But it was originally supposed to be a Spacelab pallet and get carried up by Shuttle for week-long missions. Obviously the Challenger disaster killed any idea of Shuttle rapidly flying astronomical payloads, and then Dan Goldin's tenure led to a lot of changes which delayed the first flight. But way back in the early '80s, which is what I was asking about, it was supposed to be in LEO, as a Shuttle payload.

GRO was gamma rays which are not affected by anything.

Well, the Earth is certainly opaque to them...

I guess I'm not quite understanding what you're trying to say here? There isn't a lot in space (on Solar System levels, ignoring interstellar extinction and the like) which affects EM radiation. There's zodiacal light, especially for infrared telescopes; there's not burning out your delicate scope by pointing it at an excessively intense source, like the lit Earth or the Sun; and there's not exposing your telescope to too much radiation and screwing with the electronics. There's nothing exceptionally special about gamma rays in that regard, except possibly for the electronics being less sensitive.

The off-hand comment was just pointing out that Hubble and Compton were put in LEO, and could not have been put into a higher orbit, and that SIRTIF was originally just a Spacelab pallet, and not capable of independent flight (although they had some notion of possibly adapting it to free-flight later).

@mmeijeri: Lagrange, not Euler Sorry for being a tad pedantic, but...eh, I'm a physicist, so...

[mmeijeri]

@mmeijeri: Lagrange, not Euler Sorry for being a tad pedantic, but...eh, I'm a physicist, so...

The collinear ones had already been discovered by Euler. Apparently Euler already had so many things named after him that they decided to give the honour to Lagrange... Very often these things aren't named after their original discoverer.

[Jim]

Where did SIRTIF eventually go?

SIRTIF eventually became Spitzer and ended up in a heliocentric orbit. But it was originally supposed to be a Spacelab pallet and get carried up by Shuttle for week-long missions. Obviously the Challenger disaster killed any idea of Shuttle rapidly flying astronomical payloads, and then Dan Goldin's tenure led to a lot of changes which delayed the first flight. But way back in the early '80s, which is what I was asking about, it was supposed to be in LEO, as a Shuttle payload.

GRO was gamma rays which are not affected by anything.

Well, the Earth is certainly opaque to them...

I guess I'm not quite understanding what you're trying to say here? There isn't a lot in space (on Solar System levels, ignoring interstellar extinction and the like) which affects EM radiation. There's zodiacal light, especially for infrared telescopes; there's not burning out your delicate scope by pointing it at an excessively intense source, like the lit Earth or the Sun; and there's not exposing your telescope to too much radiation and screwing with the electronics. There's nothing exceptionally special about gamma rays in that regard, except possibly for the electronics being less sensitive.

The off-hand comment was just pointing out that Hubble and Compton were put in LEO, and could not have been put into a higher orbit, and that SIRTIF was originally just a Spacelab pallet, and not capable of independent flight (although they had some notion of possibly adapting it to free-flight later).

What was saying is that GRO went into LEO because a higher orbit would not have brought any real benefits, where SIRTF did.  There were many other things than Golden that led to its many delays.  One was affordability as a Titan IV Centaur payload

The shuttle paradigm thinking also limited what was to be done.

[Robotbeat]

@mmeijeri: Lagrange, not Euler Sorry for being a tad pedantic, but...eh, I'm a physicist, so...

The collinear ones had already been discovered by Euler. Apparently Euler already had so many things named after him that they decided to give the honour to Lagrange... Very often these things aren't named after their original discoverer.

Freaking Euler. Is there anything he didn't do?

[truth is life]

On a completely different note, I was looking at the proposed Exploration Platform at EML-2 the other day, and got to thinking about the powered swing-by maneuver that really helps L-2's economics (compared to L-1). It occurred to me that during the powered swing-by, you have to pass quite close to the Moon, so that you could take the lander, undock, perform an LOI (to a phasing orbit; I assume that the swingby is not necessarily positioned perfectly for the desired landing site) and land, all the while Orion goes on and does the swing-by, then rendezvous and docks with the platform. It seems that this could provide a somewhat better performance than taking the lander all the way to L-2 then returning it to the Moon, albeit at a (small, I suspect, due to the presence of the platform) safety risk. It would also reduce the time spent in deep space, since the crew would be on the lunar surface while Orion was climbing towards L-2 (a saving of about three days), and would not need to undergo a transit from the platform to the Moon's surface (another three days less).

Assuming, however, that global access is a requirement, would the LOI burn require a large plane change? Or could the departure trajectory from Earth be selected to allow coplaner injection into whatever phasing orbit you want? Put another way, to what extent are the large plane changes demanded by ESAS-type all-LOR architectures determined by global access requirements versus a desire for "anytime return" that isn't relevant in an L-2 based architecture?

[baldusi]

Do Mars missions that land do a direct insertion? In other words, they come at planetary speeds and brake directly to the atmosphere? Thus, do the Mars orbiters need to spend fuel to break into orbit?
If that is the case, a Mars Synchronous Orbit would require more delta-v than a lander, but less than Low Mars Orbit orbiter?
Any idea of how much extra delta-v after a TMI for a MSO orbit?

[MP99]

On a completely different note, I was looking at the proposed Exploration Platform at EML-2 the other day, and got to thinking about the powered swing-by maneuver that really helps L-2's economics (compared to L-1). It occurred to me that during the powered swing-by, you have to pass quite close to the Moon, so that you could take the lander, undock, perform an LOI (to a phasing orbit; I assume that the swingby is not necessarily positioned perfectly for the desired landing site) and land, all the while Orion goes on and does the swing-by, then rendezvous and docks with the platform.

If that works, I suspect it would only allow for an equatorial landing site.

cheers, Martin

[truth is life]

Do Mars missions that land do a direct insertion?

Usually, yes. The Vikings and possibly some of the Russian probes were exceptions out of necessity, but nowadays direct insertion to the Martian surface is the rule.

Thus, do the Mars orbiters need to spend fuel to break into orbit?

Yes and no. They need to spend propellant to insert into initial Martian orbit; no one has ever aerocaptured (without subsequently and immediately landing), to my knowledge. However, there have been several probes, most prominently Mars Global Surveyor, which used atmospheric drag (and thus much less propellant) to circularize their orbit.

If that is the case, a Mars Synchronous Orbit would require more delta-v than a lander, but less than Low Mars Orbit orbiter?

It would depend on the details. In general I would expect that inserting into MSO would require a pretty substantial amount of delta-V, though. There would be no real scientific benefit from it, so I'm not sure why you would want to, at least in the near future.

Any idea of how much extra delta-v after a TMI for a MSO orbit?

Unfortunately, no.

@MP99: Well, if you would read the rest of the post that is exactly what I am asking, isn't it? To state it more technically, the question is whether the departure vector from LEO can be selected so that coplanar (ie., minimum delta-V) insertion into an arbitrarily inclined (ie., allowing you to land wherever you please) low lunar orbit is possible while still allowing for the powered L-2 swing-by maneuver.

[kevin-rf]

Doesn't Deimos' orbit pass near or through MSO? I though it did, but I could be wrong.

[truth is life]

Doesn't Deimos' orbit pass near or through MSO? I though it did, but I could be wrong.

Almost, but not quite. It's got an orbital period of 30.3 hours, about 6 hours too long to be in MSO. Semi-major axis is apparently about 3000 km too large as well.

[baldusi]

So, my next question is: doing the standard Hoffman transfer, would Mar's gravity capture an Orbiter on some "free" orbit, or would it simply deviate it? In other words, does the transfer speed relative to Mars is more or less than Mars escape velocity?

[ugordan]

It's more than Mars escape velocity. Without some fancy 3-body action (and nothing of significant mass is available at Mars) you cannot get a "free" capture by Mars and need to shed some energy.

You can minimize the incoming hyperbolic excess velocity if you can match Mars' heliocentric orbit parameters as closely as possible with your incoming spacecraft, but as you can imagine this is not possible with a Hohmann transfer from Earth. Such matching can be done by solar powered ion craft like Dawn, though.

[sdsds]

The intuition behind the "patched conics approximation" technique is that as you approach Mars its gravitational influence begins to dominate the trajectory and the influence of the Sun can be largely ignored. The spacecraft is effectively approaching Mars "from infinity" and its Mars-relative trajectory will thus be parabolic at a minimum, and most likely hyperbolic.

It's possible that the gravitational influences of the moons of Mars could be used in some tricky way, but that would depend on them being in exactly the right positions for it, which might constrain your launch window a bit!

[Hop_David]

So, my next question is: doing the standard Hoffman transfer, would Mar's gravity capture an Orbiter on some "free" orbit, or would it simply deviate it? In other words, does the transfer speed relative to Mars is more or less than Mars escape velocity?

With regard to Mars, the Earth to Mars Hohmann path is approximately a hyperbola.

Speed of a hyperbola is sqrt(Vesc^2 + Vinf^2). This can be easily remembered with this device:


In this case Vinf is about 2.7 km/s.

Vesc gets smaller and smaller as you get further from Mars

To get the burn to circularize an orbit, you would subtract circular orbit velocity from hyperbola velocity.

Since escape velocity is sqrt(2) circular orbit velocity, escape velocity can be visualized as the hypotenuse of an isosceles right triangle with circle orbit speed as the the legs.



The red portion is circle orbit speed subtracted from hyperbola speed. This is the burn needed.

Assuming a Vinf of about 2.65 km/s, it'd take a 1.9 km/s burn to go from a hyperbolic orbit to a circular orbit at Mars synchronous altitude.

I used Hohmann.xls to get that figure:

[baldusi]

Wow! That's the best answer ever!!! Thank you so much!!! I'll print it for posterity. BTW, those graph are golden!
So, forgetting the issue of plane change for a while, it's 100m/s more than a GTO transfer with a 1800m/s deficit. Doesn't seems like that much outside of what a stock commercial satellite does, right? Of course the plane change could be a nasty issue, and the navigational issues are way beyond what a stock communications satellites does. But at least from a delta-v perspective doesn't seems that difficult. An Atlas V 431 can do 3.87tonnes to TMI. Thus that's a normal commsat with a 1.800m/s deficit.
Now that I come to think of it, is such a problem the plane change? Since you are coming from outside the gravitational sphere of influence, you could chose any plane you want by selecting where you do your burn. I mean, assuming no rotational tilt with respect to your transfer plane, if you approach from the north you could get to polar, if you approach from the east you could get to equatorial, right?

[Jason1701]

Would the delta-v to get from the lunar surface to a Lagrange point be equal to that required to make the return trip?
Thanks.

[Hop_David]

Now that I come to think of it, is such a problem the plane change? Since you are coming from outside the gravitational sphere of influence, you could chose any plane you want by selecting where you do your burn. I mean, assuming no rotational tilt with respect to your transfer plane, if you approach from the north you could get to polar, if you approach from the east you could get to equatorial, right?

My models assume coplanar orbits so plane change expense is ignored. If an equatorial orbit is your goal, plane change expense would be hefty since Mars tilts 25 degrees to the ecliptic plane.

A way to mitigate would be doing multiple burns to park at Mars synchronous orbit. The 1.9 figure assumes a single burn to enter Mars synchronous orbit.

1st burn: .7 km/s, just enough to exit Hohmann to a Mars capture orbit with a 300 km periapsis and 570,000 km apoapsis (apoapsis right at the edge of Mars' sphere of influence). At a 570,000 km apoapsis, the ship is only moving .031 km/s

2nd burn: .05 km/s Apoapsis plane change/raising periapsis. At this high altitude a .05 km/s burn suffices for a 25 degree plane change  and raising periapsis to Mars synchronous altitude. The apoapsis should be in the equatorial plane during plane change burn, this would be a matter of timing.

3rd burn: .6 km/s to circularize at Mars synchronous altitude.

These 3 burns total less than 1.4 km/s. You may notice the total is less than the 1.9 km/s single burn figure I gave earlier.

Although this triple burn saves delta V, it'd be more time consuming.

[baldusi]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!
But the thing I'm not quite getting about the orbital mechanics is this: I understand that I'm coming from outside the sphere of influence, and let's say that the minimum approach to the planet takes me over a line that passes through the center of the planet and is normal to my transfer plane. At that time the gravity pull of the planet would be normal to my plane, and thous I would get no "help" of the planet's gravity to counter some of my velocity, but would result (at that instant) in a small plane change, right?
So, any burn that I do at that point would be spent all on my current speed. In other words, I would have to burn in my transfer plane, to get captured by the planet.
I don't know if I made myself clear. My first intuition told me that if you are approaching from the infinity your plane matters not as long as your intended plane ascending/descending node coincides with your approach vector (if you are pointing straight to the center of the planet).

[Hop_David]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!
But the thing I'm not quite getting about the orbital mechanics is this: I understand that I'm coming from outside the sphere of influence, and let's say that the minimum approach to the planet takes me over a line that passes through the center of the planet and is normal to my transfer plane. At that time the gravity pull of the planet would be normal to my plane, and thous I would get no "help" of the planet's gravity to counter some of my velocity, but would result (at that instant) in a small plane change, right?
So, any burn that I do at that point would be spent all on my current speed. In other words, I would have to burn in my transfer plane, to get captured by the planet.
I don't know if I made myself clear. My first intuition told me that if you are approaching from the infinity your plane matters not as long as your intended plane ascending/descending node coincides with your approach vector (if you are pointing straight to the center of the planet).

No matter what direction the ship is coming from, the path will be a hyperbola with the hyperbola's focus at Mars' center. So the hyperbola's plane will be passing through Mars' center.

To get the most bang out of your capture burn, the burn should be anti-parallel to the velocity vector at hyperbola's periapsis. So no plane change during the 1st burn.

This periapsis burn should be at an ascending or descending node of the target orbit, as you say.

I would postpone the plane change until apoapsis. Plane changes are cheaper at a distant apoapsis.

[mgfitter]

So is there an easy way (excel spreadsheet like the one above?) to calculate the approximate dv for just a plane change?

-MG.

[Hop_David]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!

I was also surprised first time I did the numbers. Phobos and Deimos are two of the lowest delta V destinations in the solar system.

[IsaacKuo]

The requirements for the plane change are rather complex, and depend on the exact timing and details of the transfer between Earth and Mars.  In the ideal situation, no plane change is required at all--you simply insert directly into the desired orbit plane.  This desirable situation is possible when the asymptotic approach path of your transfer orbit is parallel to any line which lies on the desired orbit plane.

[Hop_David]

So is there an easy way (excel spreadsheet like the one above?) to calculate the approximate dv for just a plane change?

-MG.

If you're changing direction but not speed, the before and after velocity vectors are sides of an isosceles triangle with the plane change vector as base.



With each isosceles side having the magnitude v, magnitude of plane change vector is 2 * v(sin(a/2)) where a is angle of plane change.

[mgfitter]

Thanks. Can I check my math quick?

a = 15 deg.
v = 7,800 m/s (roughly 200km circular).

So 2 * v(sin(a/2)) = 2,036 m/s, right?

-MG.

[Hop_David]

Thanks. Can I check my math quick?

a = 15 deg.
v = 7,800 m/s (roughly 200km circular).

So 2 * v(sin(a/2)) = 2,036 m/s, right?

-MG.

You got it.

If you wanted a plane change and a speed change, the triangle would no longer be isosceles. Then the Law of Cosines could do the job.

[IsaacKuo]

Sorry I misunderstood the question.  I was assuming the question was about the plane change requirements to get from a Earth-to-Mars transfer orbit to an equatorial synchronous orbit.  In other words, you don't know the angle of the plane change beforehand.

If you do know the angle of the desired plane change, then yes it's a simple calculation.

But the required angle for getting to an equatorial Mars orbit could be anything from 0 degrees to 25 degrees, depending on the details of the transfer orbit (including the date of arrival).

[mgfitter]

You got it.

If you wanted a plane change and a speed change, the triangle would no longer be isosceles. Then the Law of Cosines could do the job.

Thanks. Circular PC's are good enough for me for now, but I now know who to ask if I have more complex questions in the future. Thanks for the education!

-MG.

[baldusi]

Thank you for all this replies, you are being very instructive and exteremely helpful!

[kevin-rf]

Does anyone know the DeltaV and time required to reach the Sun/Earth L3 Lagrangian point?

Yes I know communications would be "non-trivial" (Jim would say non starter), but just curious about it as an alternative for Space Weather Stereo type missions.

[baldusi]

Does anyone know the DeltaV and time required to reach the Sun/Earth L3 Lagrangian point?

Yes I know communications would be "non-trivial" (Jim would say non starter), but just curious about it as an alternative for Space Weather Stereo type missions.

It's a trade off of delta-v and delta-t. Once you reach C3=0, you could have an arbitrary bigger orbit, and drift one way or the other, but it would take longer. Also, you have to consider in which direction you leave on C3=0, because if you leave sunwards or anti sunward, you'd already be on a slightly different orbit. once you reach L3, you have to calculate the delta-v to reach L3. Technically, I think that if you use a very slightly bigger orbit, you'd be pull in. But it would be really slow.

[mgfitter]

Is there actually a stable orbit that would take a craft in a permanent figure-of-8 around both the Earth and the Moon?

If there is, can someone shed some light on what sort of parameters it would have?   Perigee and Perilune, orbital duration etc.

I assume such an orbit wouldn't be completely stable and would require some RCS activity on each orbit, but I'm curious whether some form of "transport hub" in such an orbit might not be a workable idea.

Thanks!

-MG.

[QuantumG]

Is there actually a stable orbit that would take a craft in a permanent figure-of-8 around both the Earth and the Moon?

If there is, can someone shed some light on what sort of parameters it would have?   Perigee and Perilune, orbital duration etc.

I assume such an orbit wouldn't be completely stable and would require some RCS activity on each orbit, but I'm curious whether some form of "transport hub" in such an orbit might not be a workable idea.

See http://cbboff.org/UCBoulderCourse/documents/LunarCyclerPaper.pdf

[sdsds]

Is there actually a stable orbit that would take a craft in a permanent figure-of-8 around both the Earth and the Moon?

See http://cbboff.org/UCBoulderCourse/documents/LunarCyclerPaper.pdf

Wow, from the "AAS 91-105" is it correct to conclude this paper was published in 1991? That's awhile ago! These trajectories would seem to provide strong competition to those that orbit near the Lagrange points. Has there been any comparison of those two classes of trajectories as possible locations for a prototype deep space habitat?

[truth is life]

Short question here: Supposing I have a known C3 and have a known delta-V capability, how can I calculate my final C3 should I choose to use said delta-V capability to the utmost?

[Proponent]

The quantity c₃ is just twice the specific mechanical energy, i.e., twice the sum of the kinetic energy per unit mass and the potential energy per unit mass.  At radius r and speed v the value of c₃ is

    v²/2 - GM/r = c₃/2 .

Solve this for v using the desired c₃, and that tells you the speed needed at radius r.  Subtract the initial speed, and you've got the necessary delta-V, assuming no losses.

[baldusi]

There's something I don't get about GTO launch windows. GSO should be the only orbit that doesn't care about your RAAN, thus, only your geographic latitude should matter. But I've seen lots of times launch windows. Even on the Ariane 5 manual shows the payload penalties around the year. Why is that? Are solar and thermal issues important (like the recent Briz-M failure)? something to do with GTO and the desired orbital slot?

[jongoff]

There's something I don't get about GTO launch windows. GSO should be the only orbit that doesn't care about your RAAN, thus, only your geographic latitude should matter. But I've seen lots of times launch windows. Even on the Ariane 5 manual shows the payload penalties around the year. Why is that? Are solar and thermal issues important (like the recent Briz-M failure)? something to do with GTO and the desired orbital slot?

I think that RAAN still matters because you're trying to get to a specific GEO slot as quickly as possible with the minimal waste of extra propellant.  Just a guess though.

~Jon

[simonbp]

The nodes matter for the transfer orbit because the ascending node of the transfer orbit needs to be 180 degrees from the desired final longitude, so that the apogee of the transfer orbit passes through the final orbit.

But that's in the rotating frame, RAAN is the node in the non-rotating inertial frame. In the rotating frame, the transfer orbit's node is fixed, but in the inertial frame it rotates over the course of a day. So, if the launch window is constrained by RAAN, it's not because of the two-body orbital mechanics, but rather some external constraint. The most likely external constraint would visibility to non-GEO comm sats providing communications to the rocket during ascent.

[kevin-rf]

Dumb question, but I remember a few years back ULA not taking the moons influence into account an having a browns pants day after a GTO launch. The perigee ended up much lower than planned.

Does that have any serious effect on these windows?

[Proponent]

I believe that was because a bi-elliptic transfer rather than a Hohmann transfer was used.  In the bi-elliptic, the satellite is initially injected into an orbit with an apogee considerably higher than GEO.  In that intermediate orbit, the satellite is close to having escape energy and while near apogee it can be pretty strongly influenced by the moon.

[Proponent]

So, if the launch window is constrained by RAAN, it's not because of the two-body orbital mechanics, but rather some external constraint. The most likely external constraint would visibility to non-GEO comm sats providing communications to the rocket during ascent.

Visibility of non-GEO comsats is unlikely to explain a seasonal variation in Ariane 5's GTO capability, though.  Baldusi's guess that it's got to do with solar thermal effects looks plausible to me.  I wonder if there might also be lighting constraints wrt illuminating the payload's solar panels, though presumably these wouldn't show up in the Ariane 5 users' guide, since they would be payload-dependent rather than launch-vehicle-dependent.

[Jim]

The most likely external constraint would visibility to non-GEO comm sats providing communications to the rocket during ascent.

They don't use non-GEO comm sat during ascent

[yinzer]

Dumb question, but I remember a few years back ULA not taking the moons influence into account an having a browns pants day after a GTO launch. The perigee ended up much lower than planned.

Does that have any serious effect on these windows?

I thought that was Sea Launch...

[Jim]

Dumb question, but I remember a few years back ULA not taking the moons influence into account an having a browns pants day after a GTO launch. The perigee ended up much lower than planned.

Does that have any serious effect on these windows?

I thought that was Sea Launch...

It was an Atlas with supersynchronous orbit

[Galactic Penguin SST]

I am trying to determine which of the lunar missions worldwide were intentionally launched on a direct transfer trajectory off-plane from the Moon's orbital plane. Can anyone find which of these transfer orbits had a Right Ascension of the Ascending Node difference from that of the Moon's much greater than 0? Thanks!

(I'm trying to determine the launch time of Chang'e 3 during the early December period, using the methods of http://www.svengrahn.pp.se/histind/N1target/zondlt.htm  )

[baldusi]

To answer my own question about GTO launches from Kourous. The windows are daily, and the reason, as stated by the manual are:

The physical and mathematical definitions of the minimum window are as follows:
• the daily window is 45 minutes long
• the opening of the window corresponds to a solar aspect angle of 65° with respect to the
reference Apogee Motor Firing (AMF) attitude which permits instantaneous transfer from
the reference GTO orbit to geosynchronous orbit at apogee 6 (when the line of apsides is
colinear with the line of nodes).
Reference AMF attitude:
• right ascension: perpendicular to radius vector at apogee 6
• declination: - 7.45 deg with respect to equatorial plane

Can anyone help me understand what it means (btw, 6 degrees is the latitude of Kourou).