Orbital atmosphere (Karman Line -1000km)

In summary: ISS has to fight against the Earth's gravity every day and it's still up there!In summary, there is a small amount of drag that satellites have to deal with in space. The direction of the drag is determined by the orbital direction of the satellites. The wind speeds vary depending on the direction of the drag.
  • #1
darkdave3000
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TL;DR Summary
How do air particles in orbital space travel in relation to the rotating Earth below? Are they dragged along or do they lag behind?
The Earth Rotates at complete revolution at 24 hours, the troposphere and stratosphere are dragged along with it at the equator, but what about 100km-1000km above the equator? Is there a formula or chart that astronomers can use to predict the tiny amount of drag that satelites have to encounter in space and does it matter which direction their orbit is? That is will there be stronger winds from west to east vs east to west? Is there a chart or formula predicting the velocity of the air relative to ground at those altitudes?

David
 
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  • #2
Can you think of any force that could cause them to 'drag' behind the orbit of say, a satellite at the same height and velocity? It would have to come from somewhere other than (the spinning) Earth.
 
  • #3
So is the air 100km-1000km above the equator rotating in sync with the ground under it? I need to know to simulate this in my specialized software simulator.
 
  • #4
darkdave3000 said:
So is the air 100km-1000km above the equator rotating in sync with the ground under it?
How much air is there above 100 km ?
It will be totally ionised to a plasma. The temperature and velocity of the ions will be very high. What is there will be rotating with the air below. But there is almost nothing there. Solar storms and auroral activity will significantly modify the average movement of ions.
 
  • #5
Baluncore said:
What is there will be rotating with the air below.
Are you sure about this? How is this possible? The radius of the Earth is 6500km and the air at 1000km you say will be in sync with the troposphere which is in sync with the rotating ground? Wouldnt centrepetal/centrefugal forces be too high for the air to be rotating with the air below as you say?

Please ignore the incredibly low air density of the air at these levels. The point of my simulation is to simulate everything not everything significant.David
 
  • #6
darkdave3000 said:
Please ignore the incredibly low air density of the air at these levels.
You cannot ignore the zero air pressure, or the satelites orbiting below you, and space junk.
Do you mean air, or plasma ?
What is the mean free path for ions at that altitude ?
How can you call that air ?

To reach that altitude, ions must thermally escape from the atmosphere below. The average thermal velocity will be that of the atmosphere they rise from, but they will then be swept by the solar wind.

Find the temperature of ions at that altitude. Work out the velocity of a nitrogen ion at that altitude.
 
  • #7
I understand this is a complex topic with many variables that's why I want to deal with the neutral air first then the ions. Can we restrict our discussions to just the neutral charged air particles for now? Let's pretend I am writing a simulator to model in real time the neutrally charged air particles between-100km-1000km. Can I assume they will rotate at the exact same rotation of the troposphere? Or that they slightly lag? Are you sure they are not slightly slower? The reason I ask this is because my simulator has to model the drag that is experienced by satellites so that we can study and simulate the way satellites continue to stay in orbit with station keeping. Once again please for now let's restrict our discussions to neutral charged air and then later we can continue the discussions for ions.
 
  • #8
darkdave3000 said:
Wouldnt centrepetal/centrefugal forces be too high for the air to be rotating with the air below as you say?
This is a good point and the effect must kick in at some stage. It seems a hard problem to solve - good luck with any simulation.

At altitudes where the density of the atmosphere is high enough, the molecules will be swept along with a 24 hour period due to the averaging effect of many collisions. They are only up there because of thermal energy. Way above this, individual molecules will be following orbits according to their velocity (speed and direction), which depends on the thermal motion they started off with. When they are moving away, their orbits will follow elliptical paths in all directions and you can ignore the speeds of the atmosphere (if they don't hit anything) which will bring them back down into more dense gas and they will lose / share their energy with other molecules.
The distribution of molecular velocities could include some with escape velocity into attraction from the Sun or Moon but only if the atmosphere were hot enough. I think this would be what happened to Mercury's early atmosphere and to the Moon's atmosphere which would have been pulled off by Earth. But molecules with the sort of speeds that the ISS has (2.66km/s) would be very rare (temperature around 700K(?)) This is obvious when you think how low the drag is on craft in LEO; not many of them get up that high.

As far as the simulation goes, my reaction is basically the same as @Baluncore 's; you can only simulate so far and simulating everything is pie in the sky. Go back and decide what the actual problem is that you want to solve and the numbers of significant figures you expect to be working with. If it's for a game or a bit of fun then just remember that simulations are not real life so you can't rely on taking the results outside the simulation. But who's going to argue with you about accuracy?
 
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  • #9
sophiecentaur said:
But molecules with the sort of speeds that the ISS has (2.66km/s)
You're kidding right?
 
  • #12
darkdave3000 said:
neutral charged air particles

darkdave3000 said:
neutral charged air
Which is it? "Neutral," or "Charged?" Can't be both simultaneously.
 
  • #13
Bystander said:
Which is it? "Neutral," or "Charged?" Can't be both simultaneously.
Neutral charge = no charge
 
  • #14
You must work out what the thermal velocity and mean free path are at that altitude.
https://en.wikipedia.org/wiki/Mean_free_path#Kinetic_theory_of_gases

The speed of sound in gases is related to the average speed of particles in the gas, Vrms;
Vrms = √( 3 * k * T / m )
where; k is the Boltzmann constant = 1.38×10-23 J/K
m is the mass of each (identical) particle in the gas, in kg.
and T is the absolute temperature, in K.
 
  • #15
darkdave3000 said:
Please ignore the incredibly low air density of the air at these levels. The point of my simulation is to simulate everything not everything significant.
RIght, so are you already taking into account the pull of the the Moon? The Sun? Jupiter? Other planets? A million asteroids? Other artificial satellites? Solar wind? Using GR instead of Newtonian mechanics?

your profile said:
Favorite Area of Science: Improved Euler method in computational maths
Unless your profile is out of date you will improve the accuracy of your simulation a great deal by using a more suitable integration method.

If you want to learn how to write a good orbital simulator a good place to start is Howard Curtis's Orbital Mechanics for Engineering Students. A bad place to start is asking specific questions about things you think are relevant.
 
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  • #16
darkdave3000 said:
Once again please for now let's restrict our discussions to neutral charged air and then later we can continue the discussions for ions.
Start by putting numbers on things.
The temperature of particles in Low Earth Orbit is about 1300 K.
The molecular weight of an N2 molecule is 14g * 2 = 0.028 kg.
1 mole is 6.022e+23 particles.
The mass of an N2 molecule is; 0.028 / 6.022e+23 = 4.65e-26 kg.
Vrms = Sqrt( 3 * 1.38e-23 * 1300 / 4.65e-26 kg ) = 1075 m/s.
1 km/sec is the speed of a rifle bullet, insufficient to enter or remain in orbit.

Assuming it did not react or hit anything, if an average N2 molecule was traveling upwards, how high would it go before stopping and falling back due to gravity ?
Solve for; PE = KE; then; m·g·h = ½·m·v2; So; h = ½ · v2 / g.
10752 / ( 2 * 9.8 ) = 58.96 km.

So the very few air molecules up there are NOT in orbit. They are thermally quite well connected to the atmosphere below.
 
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  • #17
pbuk said:
RIght, so are you already taking into account the pull of the the Moon? The Sun? Jupiter? Other planets? A million asteroids? Other artificial satellites? Solar wind? Using GR instead of Newtonian mechanics?Unless your profile is out of date you will improve the accuracy of your simulation a great deal by using a more suitable integration method.

If you want to learn how to write a good orbital simulator a good place to start is Howard Curtis's Orbital Mechanics for Engineering Students. A bad place to start is asking specific questions about things you think are relevant.
The moon for now , I'll worry about the sun and other above mention celestial objects later once I understand your answer to my question a out how the ground rotates in relation to those neutral air particles.

Yes my profile is outdated, I now use Runge Kutta,
 
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  • #18
Baluncore said:
You must work out what the thermal velocity and mean free path are at that altitude.
https://en.wikipedia.org/wiki/Mean_free_path#Kinetic_theory_of_gases

The speed of sound in gases is related to the average speed of particles in the gas, Vrms;
Vrms = √( 3 * k * T / m )
where; k is the Boltzmann constant = 1.38×10-23 J/K
m is the mass of each (identical) particle in the gas, in kg.
and T is the absolute temperature, in K.
But how is this helpful? I assume these velocities cancel out so they don't indicate an overall general net direction relative to the Earth or it's rotating ground? How do I work out the total momentum relative to Earth or it's rotating ground?
 
  • #20
darkdave3000 said:
But how is this helpful? I assume these velocities cancel out so they don't indicate an overall general net direction relative to the Earth or it's rotating ground?
It shows that, (what little atmosphere is up there), is coupled to the Earth via exchange with the average atmosphere below.
The neutral molecules are nowhere near being in orbit, and will typically fall back into the atmosphere within a few seconds, if they are not ionised first.
 
  • #21
darkdave3000 said:
The moon for now , I'll worry about the sun and other above mention celestial objects later once I understand your answer to my question a out how the ground rotates in relation to those neutral air particles.
I didn't answer your question, but I will now: forget about modelling the atmosphere. The fact is that there are a whole load of perturbations of much greater magnitude than atmospheric drag (except in Very Low Earth Orbits (VLEOs)) affecting the path of an artificial satellite, some of which are chaotic, and others even if theoretically deterministic have unknown initial conditions.
darkdave3000 said:
Yes my profile is outdated, I now use Runge Kutta, check out my website photonbytes.com
Improved Euler is a Runge Kutta method so it is not clear what method you are referring to, however all single-step explicit methods suffer from similar problems due to truncation errors, generally resulting in loss of energy and therefore spiralling inwards. Learn about symplectic methods.

For fully developed tools you should look at NASA's Orbit Determination Toolbox and General Mission Analysis Tool (GMAT), both of which are open source projects, to see the end result of thousands of man years of effort.
 
  • #22
Simplistically, I would expect that most of the particles in the OP's middle zone will be on suborbital trajectories. They will rise out of the atmosphere when they are given enough velocity by impacts to do so (tantamount to heating), but, being short of orbital velocity, they will fall back again.

I would further expect that - on average - compared to the velocity of the upper atmo due to Earth's rotation, they will fall behind.

The net effect, from the FoR of Earth's surface, is a rarefied envelope of molecules mostly rising and falling, with the lesser effect of moving slowly counter to Earth's-spin (though from a remote FoR, they are actually moving pro-Earth-spin, just slightly slower than the atmo).
 
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  • #23
DaveC426913 said:
I would further expect that - on average - compared to the velocity of the upper atmo due to Earth's rotation, they will fall behind.
What is "on average". How does the geographical latitude and hence variation in radius of rotation influence the velocity?
 
  • #24
darkdave3000 said:
Yes my profile is outdated, I now use Runge Kutta, check out my website photonbytes.com
I did and I have a few comments:
  • Your videos are unwatchable, you should use something like OBS Studio.
  • You have a model of a supersonic ballistic trajectory; how are you modelling the complex relationship between velocity, density and drag?
  • You talk about building a wind tunnel to investigate the drag coefficient of e.g. the Saturn V rocket. I think you are underestimating the
  • resources that would be required to do this. Besides there is a great deal of data available on the Saturn and other missions both from NASA and from other enthusiasts for example https://web.archive.org/web/20170313142729/http://www.braeunig.us/apollo/saturnV.htm.
  • The site linked above reminded me of the MSISE-90 Model of the Upper Atmosphere which both answers some of your questions above and indicates variability according to solar radiation conditions which you do not seem to be taking into account.
 
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  • #25
Baluncore said:
What is "on average". How does the geographical latitude and hence variation in radius of rotation influence the velocity?
Particles all have their intrinsic speeds and directions. Some will be ejected forward i.e. faster than the atmo is spinning with the Earth.

But on average (taken as a whole mass), it seems they will fall behind the spinning atmo.

As to latitude, I had not considered it.
 
  • #27
  • #28
DaveC426913 said:
Those winds south of Africa's point exceed 430km/h!
Yes, but it is pretty thin at only 1% of the sea level density. It is also seasonal.
Looking at different altitudes ...
1000 hPa = surface
850 hPa = 4,425 ft
700 hPa = 9,500 ft
500 hPa = 18,000 ft
250 hPa = 34,000 ft = Jet streams
70 hPa = 58,000 ft
10 hPa = 85,000 ft
 

1. What is the Karman Line?

The Karman Line is an imaginary boundary located at an altitude of 100 kilometers above the Earth's surface. It marks the boundary between the Earth's atmosphere and outer space.

2. How was the Karman Line determined?

The Karman Line was determined by Hungarian-American engineer and physicist, Theodore von Karman, in the 1950s. He calculated that at an altitude of 100 kilometers, the Earth's atmosphere becomes too thin for aeronautical purposes.

3. What is the significance of the Karman Line?

The Karman Line is significant because it is used as a boundary to define where space begins. It is also used as a reference point for determining the altitude of spacecraft and satellites.

4. Is the Karman Line a universally recognized boundary?

No, the Karman Line is not a universally recognized boundary. Different organizations and countries may have their own definitions of where space begins. For example, the US Federal Aviation Administration considers the boundary to be at an altitude of 80 kilometers, while the International Astronomical Union defines it at 50 kilometers.

5. Can the Karman Line change over time?

Yes, the Karman Line can change over time due to various factors such as changes in Earth's atmosphere, solar activity, and human activities. It is also a subject of ongoing scientific debate and may be redefined in the future as our understanding of space and the Earth's atmosphere improves.

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