Author Topic: Orbits Q&A  (Read 96298 times)

Offline Danderman

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Re: Orbits Q&A
« Reply #40 on: 03/31/2011 04:25 PM »
And a related question:

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

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

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

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


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"

Offline jabe

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Re: Orbits Q&A
« Reply #41 on: 04/28/2011 10:55 PM »
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

Offline glanmor05

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Re: Orbits Q&A
« Reply #42 on: 06/06/2011 10:02 PM »
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. 
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Offline Robotbeat

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Re: Orbits Q&A
« Reply #43 on: 06/06/2011 11:57 PM »
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.
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Offline IsaacKuo

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Re: Orbits Q&A
« Reply #44 on: 06/07/2011 02:11 AM »
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...

Quote
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.

Offline Robotbeat

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Re: Orbits Q&A
« Reply #45 on: 06/07/2011 06:00 AM »
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...

Quote
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.
« Last Edit: 06/07/2011 06:06 AM by Robotbeat »
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Offline kevin-rf

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Re: Orbits Q&A
« Reply #46 on: 06/07/2011 12:19 PM »

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.
« Last Edit: 06/09/2011 12:37 AM by kevin-rf »
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Offline Robotbeat

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Re: Orbits Q&A
« Reply #47 on: 06/07/2011 09:40 PM »
(you edited the quotes wrong... I didn't say that, I was quoting someone else. :) )
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Offline kevin-rf

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Re: Orbits Q&A
« Reply #48 on: 06/09/2011 12:39 AM »
(you edited the quotes wrong... I didn't say that, I was quoting someone else. :) )

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Offline deltaV

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Re: Orbits Q&A
« Reply #49 on: 07/31/2011 07:08 PM »
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.

Online mmeijeri

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Re: Orbits Q&A
« Reply #50 on: 07/31/2011 08:07 PM »
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.
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Offline deltaV

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Re: Orbits Q&A
« Reply #51 on: 07/31/2011 08:39 PM »
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.
« Last Edit: 07/31/2011 08:54 PM by deltaV »

Offline deltaV

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Re: Orbits Q&A
« Reply #52 on: 07/31/2011 09:07 PM »
I just found a thread which may answer my questions:

Offline IsaacKuo

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Re: Orbits Q&A
« Reply #53 on: 07/31/2011 09:52 PM »
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.

Offline sdsds

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Re: Orbits Q&A
« Reply #54 on: 08/01/2011 03:39 AM »
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)?
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Offline IsaacKuo

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Re: Orbits Q&A
« Reply #55 on: 08/01/2011 10:53 AM »
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.

Offline baldusi

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Re: Orbits Q&A
« Reply #56 on: 09/30/2011 06:12 PM »
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?

Offline IsaacKuo

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Re: Orbits Q&A
« Reply #57 on: 10/01/2011 01:01 AM »
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.

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Re: Orbits Q&A
« Reply #58 on: 11/30/2011 06:14 AM »
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?
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Offline IsaacKuo

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Re: Orbits Q&A
« Reply #59 on: 11/30/2011 02:49 PM »
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.

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