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#140 by ChrisC on 16 Nov, 2011 02:32
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The orbital plane precesses about 6 degrees per day in geocentric reference because of Earth's equatorial oblateness. So the geocentric plane of the departure asymptote is sweeping across the celestial sphere at some combination of this rate, earth's axial tilt, and Earth's movement around the sun.
I was disoriented by this response earlier today because at first I thought it had been posed by Jim, not JimO, and it was so elegantly written ...
And now JimO comes along and blows his previous explanation out of the water.
I asked about this topic here a year or so ago and got the brief oblateness explanation. Now THIS is a masterpiece. Thanks JimO for generously taking the time to dig up the explanation from your book!
And thanks MichaelJ for that piece you did on the legal ramifications. Very interesting.
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#141 by Lee Jay on 16 Nov, 2011 03:12
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Oh no. I'm frightened to say Jim O's explanation is just about exactly what I had envisioned, but it seemed so simple I didn't say it. Must be reading too many of Jorge's posts. On the other hand, I'm proud to say I still don't understand what this has to do with an asymptote to a Mars transfer orbit since it's so highly inclined.
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#142 by just-nick on 16 Nov, 2011 03:40
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#143 by Comga on 16 Nov, 2011 04:39
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Oh no. I'm frightened to say Jim O's explanation is just about exactly what I had envisioned, but it seemed so simple I didn't say it. Must be reading too many of Jorge's posts. On the other hand, I'm proud to say I still don't understand what this has to do with an asymptote to a Mars transfer orbit since it's so highly inclined.
Try this:
The trajectory to Mars leaves Earth in a particular direction.
If the axis of rotation of that orbit were pointed in that direction, the spacecraft would have to make a right angle turn, after killing off all of the orbital velocity. Obviously that won't work.
The idea is to add to the orbital momentum. This works when the trajectory to Mars is in the plane of the orbit. When the plane of the orbit rotates too far from the trajectory, the correction becomes unaffordable from an energy and momentum standpoint.
Or this:
Thing of the Mars departure in reverse. The line to Mars runs back towards the Earth. Wherever it would pass the Earth, the line swings around Earth in a hyperbola. If the collection of these lines is limited to those where the perigee of the hyperbola matches some altitude, you get this hyperbolic cylinder, with the tangent points forming the circle.
The center of the circle is the line along the trajectory through the center of the Earth.
If a satellite passes through that point, it then hits that circle at normal incidence. If it gets a quick boost to escape velocity, plus a specific velocity, it leaves along one of the hyperbolas and is off to Mars.
That circle is fixed in space, or rather, moving very slowly over a period of days. If the orbit normal is regressing around the Earth, it may not intersect the circle, never mind go through the center of it. The energy to force it to the center is the energy for a big plane change, which we know is beyond the S/C capability.
Is that any better? -
#144 by iamlucky13 on 16 Nov, 2011 04:47
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Oh no. I'm frightened to say Jim O's explanation is just about exactly what I had envisioned, but it seemed so simple I didn't say it. Must be reading too many of Jorge's posts. On the other hand, I'm proud to say I still don't understand what this has to do with an asymptote to a Mars transfer orbit since it's so highly inclined.
Inclination is the problem. If everything is in plane, it's merely a matter of timing the burn to give you a proper asymptote.
But there's multiple inclinations involved - Fobos-Grunt to earth, earth to the sun, Mars to the sun, and Phobos to Mars. If the Fobos-Grunt around earth plane doesn't line up close enough with the desired heliocentric plane, your existing orbital momentum is largely wasted and you need a lot more delta-V to make up for it.
However, if I'm following the geometry properly here, the precession at this altitude is a 60 day cycle, so it flips 180 degrees after 30 days giving you proper alignment on the other half of your orbit.
A related thought I had - the launch window extended through November 21, I believe. However, the extremes of the acceptable launch window still provides some propellant margin because obviously, problems can come up after launch. So...if sometime around 30 days after launch the precession of the orbital plane has flipped it 180 degrees, might there still be a chance of reaching Mars if a recovery occurs by that time? -
#145 by sdsds on 16 Nov, 2011 07:06
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Naively making the simplifying assumption that Earth, Mars and spacecraft heliocentric orbits are coplanar, a propulsively efficient elliptical trajectory from Earth to Mars needs to depart in the direction of the Earth's motion around the Sun. I can't visualize how the F-G geocentric orbit at its high inclination can ever have a velocity vector pointing in that direction.
Thus I'm assuming the intended heliocentric transfer ellipse was not in (e.g. was "above") the ecliptic plane. Where am I going with this? It seems like after 180 degrees of precession, the F-G geocentric orbit would facilitate an equivalent transfer ellipse that was "below" the ecliptic plane. My question for those who understand this stuff is whether that analysis is simply too naive to be useful? -
#146 by Proponent on 16 Nov, 2011 07:16
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Regardless of inclination with respect to the earth's equator, the velocity vector can align with the earth's orbital motion about the sun at some point in the orbit. The craft may will be above or below the ecliptic plane at the time, but that hardly matters, since the out-of-plane distance is just a few thousand kilometers, which is a small fraction of the distance to the sun.
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#147 by ugordan on 16 Nov, 2011 08:13
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Oh no. I'm frightened to say Jim O's explanation is just about exactly what I had envisioned, but it seemed so simple I didn't say it. Must be reading too many of Jorge's posts. On the other hand, I'm proud to say I still don't understand what this has to do with an asymptote to a Mars transfer orbit since it's so highly inclined.
Allow me to point you to a nice primer on interplanetary orbital mechanics posted on this very same forum several years ago:
http://forum.nasaspaceflight.com/index.php?topic=1337.msg84606#msg84606
Has some nice illustrations which should make understanding why F-G would need to be in a specific plane (since its inclination is fixed) to hit a specific TMI trajectory.
In particular, this post is the kicker for F-G's situation: http://forum.nasaspaceflight.com/index.php?topic=1337.msg84837#msg84837 -
#148 by Michael J on 16 Nov, 2011 12:32
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The latest from Ria Novosti. Attempts to communicate with the spacecraft continue to fail, and the apogee of the spacecraft's orbit has decreased.
http://tinyurl.com/cnylco3 -
#149 by Lee Jay on 16 Nov, 2011 13:09
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Regardless of inclination with respect to the earth's equator, the velocity vector can align with the earth's orbital motion about the sun at some point in the orbit. The craft may will be above or below the ecliptic plane at the time, but that hardly matters, since the out-of-plane distance is just a few thousand kilometers, which is a small fraction of the distance to the sun.
This is exactly where I went wrong. I was thinking you'd have to be coplaner and in the ecliptic at the same time. Duh...Mars is far away, that's irrelevant! -
#150 by pm1823 on 16 Nov, 2011 13:22
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On the other hand, I'm proud to say I still don't understand what this has to do with an asymptote to a Mars transfer orbit since it's so highly inclined.
FG and Mars.
(Don't embed images - it breaks the internet
- Chris).
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#151 by ChileVerde on 16 Nov, 2011 13:50
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Any guesses as to why P-G is slowly ascending? Off the top of my head, it's either propellent outgassing or possibly excess RCS activity caused by the IAU trying to square the circle of LEO as opposed to a deep space cruise.
It's just the perigee that seems to be ascending. The apogee is coming down even faster and the mean altitude with it. Just what is causing the perigee behavior isn't clear, though outgassing or attitude thruster firings have been suggested. There's also some slight possibility that the orbital elements coming out of Space Command are in error -- I think that's unlikely, though it's happened before.
Uh-oh. I just got the latest orbital elements and they show the mean altitude increasing, then decreasing, then increasing again. I'm really not sure what's going on here. -
#152 by simonbp on 16 Nov, 2011 14:57
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Chile, do you have a table of the mean altitude/semimajor axis values you could post?
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#153 by simplex on 16 Nov, 2011 15:42
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Fobos - Grunt and the Chinese satellite both have, amongst other things, UHF antennas and communication systems.
see: http://nssdc.gsfc.nasa.gov/nmc/spacecraftDisplay.do?id=YINGHUO-1
UHF transceivers are used by Ham radio operators to communicate with cubesats and other small satellites placed in low orbits.
So the stories about being unable to communicate with Phobos - Grunt, because the earth transmitters are too strong and their beams too narrow, are not credible. This can be the case for the X band but not for UHF. -
#154 by just-nick on 16 Nov, 2011 15:48
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Fobos - Grunt and the Chinese satellite both have, amongst other things, UHF antennas and communication systems.
But this is useless unless those UHF systems are configured to allow commanding of the spacecraft. And, for that matter, unless they are turned on.
--N -
#155 by ChileVerde on 16 Nov, 2011 16:06
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Chile, do you have a table of the mean altitude/semimajor axis values you could post?
Here you go:
UTC Epoch Mean Altitude
YYDDD.dddddddd kilometers
11312.89839094 274.8521
11313.14788776 281.6713
11313.39819696 281.2676
11313.58551366 281.0564
11313.89772518 280.9048
11313.89772540 280.8227
11314.14749491 280.6525
11314.33481518 280.4674
11314.77184893 280.0384
11314.95914282 280.1417
11315.20886687 280.1761
11315.83319279 280.0165
11316.08288834 279.8168
11316.39501126 279.8157
11316.83198086 279.4874
11317.01923030 279.1055
11317.08163703 279.0343
11317.51854193 278.5565
11317.83055630 278.2568
11318.01776824 278.1408
11318.26738376 278.0658
11318.89142147 277.5679
11319.07858799 277.4156
11319.45294171 277.1089
11319.70248387 276.7967
11320.13914578 277.0736
11320.26397368 276.7125
11320.45111420 276.7691
I'd suggest not trusting the altitudes much past the second place to the right of the decimal.
Edit: To convert day of year to the civil calendar, see http://disc.gsfc.nasa.gov/julian_calendar.shtml -
#156 by Nittany Lion on 16 Nov, 2011 16:09
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However, the original swerve to the left (westwards) is NOT
counterbalanced, so the satellite’s orbital plane has been effectively
shifted a small amount.
Something is missing; either from this explanation or from my brain. Help me decide which.
For JimO’s equator crossing from the southwest, the satellite is initially pulled to the west and, after the equator crossing, pulled back to the east. So, I’m thinking, there is no change in the plane of the orbit.
If the shifts are asymmetric, i.e., the westward shift is larger than the eastward shift, then the relative magnitude of the shifts is reversed one-half orbit later when the equator is crossed from the north. Again, I’m thinking, there is no change in the plane of the orbit.
If the rotation of the earth during the equator crossing is a factor (a possible reason why the westward shift is larger than the eastward), then why do retrograde orbits shift to the east?
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#157 by Jim on 16 Nov, 2011 16:12
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So the stories about being unable to communicate with Phobos - Grunt, because the earth transmitters are too strong and their beams too narrow, are not credible. This can be the case for the X band but not for UHF.
Yes, they are. UHF is not for commanding the spacecraft and is not viable for deep space. UHF is only for a science experiment at Mars.
So what is your point?
Why do you keep thinking there is a conspiracy. -
#158 by simplex on 16 Nov, 2011 16:18
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But this is useless unless those UHF systems are configured to allow commanding of the spacecraft. And, for that matter, unless they are turned on.
No spacecraft, so big, relies on a single communication unit. All of them have, beside high frequency high gain antenna, UHF or VHF systems all the time.
It simply does not make sense to send the spaceship just with an active X band antenna and later using it to activate the UHF module, the backup, which is slow but much more reliable than a dish antenna, because it can communicate in any condition and at any orientation.
Mars Rovers have also UHF modules and this is the emergency antenna. In the first moments on the surface it is this antenna that send and receives data not the high gain one.
"each rover carries 3 much smaller antennas — one that can move in different directions called the high gain antenna, and two others that are fixed — the low gain and Ultra High Frequency (UHF) antennas. When a rover uses the high gain antenna to speak directly with Earth, it is important that Earth be in view of the rover. That's because this type of communication is a direct shot to Earth, so Earth must be above the martian horizon. The high gain antenna must also be in the correct position. It can spin around so that the entire rover doesn't have to waste energy twisting and turning to get in the right orientation to face Earth. However, in the first few hours of the mission, before the high gain antenna is up and running, the low gain (low-bandwidth) antenna is put into action. Again, Earth must be above the martian horizon as the low gain is also a direct-to-Earth communicator. But the low gain antenna does not have to be pointed at the Earth. Its signal is sent out in all directions at once. We say it’s “omnidirectional”.
see: http://athena.cornell.edu/kids/bn_special_report.html
The fact that Russians do not have UHF like mars rovers and many other spacecrafts is simply not credible. Just to rely on the self capabilities of a dish antenna to orient itself to the earth is simply irresponsible. Only and idiat could have designed such a spacecraft. -
#159 by Jim on 16 Nov, 2011 16:23
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No spacecraft, so big, relies on a single communication unit. All of them have, beside high frequency high gain antenna, UHF or VHF systems all the time.
It simply does not make sense to send the spaceship just with an active X band antenna and later using it to activate the UHF module, the backup, which is slow but much more reliable than a dish antenna, because it can communicate in any condition and at any orientation.
Wrong. They do only rely on frequency system. MSL, Juno, etc use only X-band for comm with earth. UHF is not used for deep space. they have multiple transmitters and receivers. But if the spacecraft software is messed up, then it does matter how many there are.
Again, UHF is only for a science experiment at Mars.
UHF on MSL is only used to communicate with other spacecraft in MSL orbit and not with earth.
You are showing the lack of understanding spacecraft design.
- Chris).