Why does Webb orbit L2, is it because of the Moon?

In summary, the JWST was not designed with the intention of lasting 10 years. It was designed to be cheaper and simpler than the other space telescopes currently in development.
  • #1
bland
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I was already puzzled by the concept of orbiting a Lagrange point and then I find out it's about the same size orbit as the Moon. I am thinking that if there was no Moon that the Earth and the Sun are far enough away to be treated as points and so that there would be an exact distance further out that the Webb could maintain the same velocity, or would it still need to orbit?

So I'm now wondering if it's our Moon that causes Webb to need to orbit L2 in a large orbit in order to be stable enough?
 
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  • #2
bland said:
I was already puzzled by the concept of orbiting a Lagrange point...
Lagrange points are not stable in terms of the combined potential. Objects "orbit" around them due to Coriolis force.

220px-Lagrangian_points_equipotential.png
 
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  • #3
A.T. said:
Objects "orbit" around them due to Coriolis force.

OK, I would never have suspected that the Coriolis force would have anything to do with it.
 
  • #4
Of the Lagrange points L1, L2, and L3 are unstable and L4 and L5 are stable. Seen from the rotating frame of reference, where the Lagrange point under consideration is fixed you have of course an interplay of the gravitational force and the inertial forces (i.e., both the Coriolis and centrifugal forces). In the Lagrange point itself a body being precisely there at rest the body stays at rest, because there the gravitation and centrifugal force (seen in the frame corotating with Sun and Earth) cancel each other (which is the definition of the Lagrange points in the first place).

The JWST is positioned at L2 for its advantageous properties concerning astronomical observations, and that's why there are and have been also several other space probes positioned there (like WMAP and Planck, measuring the cosmic microwave background). For details, see

https://solarsystem.nasa.gov/resources/754/what-is-a-lagrange-point/

PS: The pdf linked at the end of this website really explains the mechanics nicely.
 
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  • #5
vanhees71 said:
Of the Lagrange points L1, L2, and L3 are unstable and L4 and L5 are stable. Seen from the rotating frame of reference,
None of them are stable based on the potential in the rotating frame alone. L4 and L5 are maxima of the potential, not minima. But they are plateau like maxima and the Coriolis force let's stuff circle around them. You could call this semi stable. L1, L2, and L3 are saddle points of the potential and thus maybe even less stable.
 
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  • #6
I think the original question asks why the JW telescope will be in orbit around L2, as opposed to simply being stationed at, or near, L2. My understanding is this...

L2 is a point of unstable equilibrium, A craft stationed there would need many regular course corrections to maintain its position.

However, there are orbits around L2 that are stable (or very nearly so). A craft in such an orbit requires far less fuel to maintain its orbit compared to a craft stationary relative to L2. That’s why the JW telescope will (hopefully) be positioned in such an orbit.

For more details try these:
https://en.wikipedia.org/wiki/Lissajous_orbit
https://en.wikipedia.org/wiki/Halo_orbit
 
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  • #8
vanhees71 said:
Indeed, every 23 days you have to correct the telescope's orbit, and that's one of the limiting factors of its lifetime. The fuel is calculated to last for about 10 years.

https://solarsystem.nasa.gov/resources/754/what-is-a-lagrange-point/
Ten years isn't long for such an expensive item!

Surely it (the JWST) has been designed with facilities to enable refuelling (and other forms of simple servicing) by robotic systems. Not to do so would seem a huge omission.
 
  • #10
Steve4Physics said:
Surely it (the JWST) has been designed with facilities to enable refuelling (and other forms of simple servicing) by robotic systems. Not to do so would seem a huge omission.
From another thread. It sounds like these later projects will surpass the benefits of JWST, thus deprecating the need for a longer life for the JWST.
anorlunda said:
JWST is not the only one. There are four other space telescope projects in the works. All four of them are designed to orbit the L2 point. According to the video, all four are intended to be simpler and cheaper than the JWST; not surprising because technology marches on.

2:54 Habitable Exoplanet observatory (HabEx)
8:53 Lynx X-Ray observatory
12:00 Origins Space Telescope (OST)
17:17 Large Ultraviolet, Optical, Infrared Surveyor (LUVOIR)

 
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  • #11
Steve4Physics said:
Surely it (the JWST) has been designed with facilities to enable refuelling (and other forms of simple servicing) by robotic systems. Not to do so would seem a huge omission.
Because it wasn't expensive enough?
 
  • #12
Vanadium 50 said:
Because it wasn't expensive enough?
A removeable fuel-cap would have just pushed it over the $10bn mark.
 
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  • #13
Steve4Physics said:
A removeable fuel-cap would have just pushed it over the $10bn mark.
Doesn't cost a lot in virtual money like bitcoin... :oldbiggrin:
 
  • #14
Why not place it on the moon?
 
  • #15
MarkL said:
Why not place it on the moon?
Not cool.
 
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  • #16
A.T. said:
Not cool.
So harsh.

:-p
 
  • #17
A couple of points but first.. hi all it has been ages since I posted here! Any boy remember me?
Now... refueling JWST. The attachment ring used to connect to the launch vehicle is still in place and usable however there are no ports for fuel so you cannot onload fuel. Now what they could do it fly up a thruster unit with fuel, connect to the ring and use the new thruster. How It would effect dynamics of the telescope? I have no idea.

Next. I have been been looking at the derivations of the L2 point and simply cannot understand them. My understanding that Lagrange and Euler were working on maxima minima problems and one of the them was
Grad(##/phi##}=0 . Solutions to this ought to yield areas of low or no gravitational attraction.

All of the on line derivations start with a balance of centripetal forces. These plateaus on potential exist whether or not a body is there. Am I wrong?

As to orbiting L2 I think this would be much easier to understand if someone were to post a potential map through L2 and perpendicular to the Earth sun axis. In this axis we should see a set of closed equipotential lines around L2. Hop on one of these lines and you are in orbit.
 
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  • #18
Integral said:
hi all it has been ages since I posted here! Any boy remember me?

I have been been looking at the derivations of the L2 point and simply cannot understand them.
Hi!

The L2 point is outside the Earth's orbit on a line extended from the Sun through the Earth. Object X, initially positioned at L2, experiences real forces of gravity from the Sun and the Earth that add to give a centripetal force on X. L2 is located such that this real net centripetal force causes an angular velocity for X that is the same as the angular velocity of the Earth, 360 degrees per year. Since X and the Earth have the same angular velocity, if X is given the correct initial linear velocity, the the Sun, Earth, and X remain colinear as time evolves.

Without the the Earth's gravity, an object in circular orbit around the Sun at distance L2 would have an angular velocity that is less than 360 degrees per year. For example, after one Earth year, Mars has not completed one orbit, i.e., less that 360 degrees.
 
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  • #19
Thank you George. That seems a better explanation then I typically see. However I am trying to align that with my understanding. That the L points are solutions to Grad(phi)=0, Where phi is the gravitation potential due to sun and earth. These areas having a constant potential would experience little or no attraction to either Earth or sun. So they are gravitationally stable, with the instability being due to the empty space of the L region tracks with earth. However any massive body must adhere to orbital mechanics, so it would have the wrong period for this orbit so will tend to seek its correct orbit, that of earth. Where is my error?
 
  • #20
Integral said:
A couple of points but first.. hi all it has been ages since I posted here! Any boy remember me?
What a long hiatus?
What happened that you returned posting here?
 
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  • #21
Integral said:
Thank you George. That seems a better explanation then I typically see. However I am trying to align that with my understanding. That the L points are solutions to Grad(phi)=0, Where phi is the gravitation potential due to sun and earth. These areas having a constant potential would experience little or no attraction to either Earth or sun. So they are gravitationally stable, with the instability being due to the empty space of the L region tracks with earth. However any massive body must adhere to orbital mechanics, so it would have the wrong period for this orbit so will tend to seek its correct orbit, that of earth. Where is my error?
After playing around off-and-on with this, I think that I can get all five Lagrange points from a potential formulation. I made several sign mistakes along the way, and I might still have some glitches, but it is starting to look okay. I am not sure if I am going to type this into the PF interface, or whether I will attach a pdf.

The key is moving to a rotating frame that has the centre of mass of the Earth-Sun system on the rotation axis, and that has a period of one year. Then, a test mass will have three "forces" acting on it, the gravitational attraction of the Earth, the gravitational attraction of the Sun, and the centrifugal pseudoforce associated with the rotating frame. There is a potential associated with each of these three "forces". There is no Coriolis force, as we want the test mass to be stationary in the rotating frame.
 
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  • #22
Well, "stationary" is an ideal case which cannot be achieved accurately in practice. That's why there must be regular corrections of the orbit and that's limiting the lifetime of the telescope, because there's only a finite amount of fuel. As was discussed in another thread fortunately the launch of the satellite was so accurate that one needed much less of the fuel for the initial orbit corrections, and thus the limiting time from the fuel is enhanced to 20 years rather than the planned 10 years. Let's keep the fingers crossed that not something else goes wrong till the telescope gets operational!
 
  • #23
MathematicalPhysicist said:
What a long hiatus?
What happened that you returned posting here?
Been a while since I encountered a problem where I needed reliable help. No other place on the web like this.
Being retired now, my biggest problem has been when to crawl out of bed and make a pot of coffee.

George Jones said:
After playing around off-and-on with this, I think that I can get all five Lagrange points from a potential formulation. I made several sign mistakes along the way, and I might still have some glitches, but it is starting to look okay. I am not sure if I am going to type this into the PF interface, or whether I will attach a pdf.

The key is moving to a rotating frame that has the centre of mass of the Earth-Sun system on the rotation axis, and that has a period of one year. Then, a test mass will have three "forces" acting on it, the gravitational attraction of the Earth, the gravitational attraction of the Sun, and the centrifugal pseudoforce associated with the rotating frame. There is a potential associated with each of these three "forces". There is no Coriolis force, as we want the test mass to be stationary in the rotating frame.
Glad to hear you are making some headway. While sure that this is closer to the problem that Lagrange solved I am no Lagrange. My first attempt has been to search for some topography along the Earth sun axis I am afraid my scaling to a undiminsioned variable has erased all signs of small changes. I am still working.
 
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  • #24
 
  • #25
I am sure glad that some one posted that image. It is just wrong. Consider that the L4 and L5 points are 97e6 mi away from Earth and the sun, while L2 is 1e6 mi from Earth and the 98e6 mi from the sun. how in any way can L4 and L5 higher potential then L2. I simply refuse to to believe that l4 and l5 are huge ridges dominating Earth's orbit. IF you look at the caption for that it claims that it is a plot of potential and centrifugal force. How do you do that? I am not sure what this is a plot of but it is not a potential plot of the L points.
 
  • #26
Integral said:
. how in any way can L4 and L5 higher potential then L2. I simply refuse to to believe that l4 and l5 are huge ridges dominating Earth's orbit
Perhaps it is I misinterpreting the graph but...

L2 is deeper in Earth's g-well than L5, yes?

Which scenario makes more sense:
1. A satellite at L5 getting destabilized and "falling" to L2, or
2. A satellite at L2 getting destabilized and "rising" to L5?
(hopefully we agree #1 makes sense)

It's not that the ridges are "huge", its that they're very flat - meaning very little change in potential over a very large distance. L4 and L5 are relatively flat space, and far from Earths steep sided well.
 
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  • #27
Integral said:
I simply refuse to to believe that l4 and l5 are huge ridges dominating Earth's orbit.
I think you need to look again. If the Earth gravity is not included those points L3,L4, and L5 are in the bottom of the effective (psuedo) potential for objects with fixed angular momentum orbiting the sun. They are stable wrt radial fluctuations. Troughs not ridges.
 
  • #28
Did this oblique view a few years ago.
pic_lagrange.gif
 
  • #29
I am sorry that does not make any sense to me. The all of the Earth sun L point are determined by the Earth sun potential field. As worked out by Lagrange and Euler in the late 1770s.
 
  • #30
  • #31
Integral said:
What is that a plot of.
It's just an oblique angle of the same diagram, showing potential a little better. Doesn't really address your concerns

Please see my comments and questions in post 26.
 
  • #32
DaveC426913 said:
Perhaps it is I misinterpreting the graph but...

L2 is deeper in Earth's g-well than L5, yes?

Which scenario makes more sense:
1. A satellite at L5 getting destabilized and "falling" to L2, or
2. A satellite at L2 getting destabilized and "rising" to L5?

It's not that the ridges are "huge", its that they're very flat - meaning very little change in potential over a very large distance. L4 and L5 are relatively flat space, and far from Earths steep sided well.
neither of those make any sense. L2 is the throne and crown of the L points. The Earth sun L4 and L5 are tiny dimples at 97 million miles from earth. L2 is a mountain in the back yard.
 
  • #33
Integral said:
neither of those make any sense. L2 is the throne and crown of the L points. The Earth sun L4 and L5 are tiny dimples at 97 million miles from earth. L2 is a mountain in the back yard.
AFAIK, nothing in the diagram contradicts what you say.

It sounds like you are mixing extent with gradient of potential

L5 is almost flat - very little change in potential over distance - which necessarily means its very large in extent.

Or look at it another way, L5 is so weak, you'd have to travel millions of miles from it to even notice a change.

L2 OTOH is so strong but also so compact that a small deviation will be a huge climb uphill. Thats what makes it so stable. A satellite is confined to a very small space that also has a strong restorative force. Ideal conditions for an L point
 
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  • #34
DaveC426913 said:
It's just an oblique angle of the same diagram, showing potential a little better. Doesn't really address your concerns

Please see my comments and questions in post 26.
Ok, just a rendition.
It would be nice if one could just add the potential of Earth to the potential of sun. Unfortunately the common expression of potential is the result of the integral of force, a constant of integration is dropped. I see the Earth and sun forces acting in parallel along the Earth sun axis. Thus there has to be a potential cone emanating from Earth to L2 and beyond.
 
  • #35
DaveC426913 said:
AFAIK, nothing in the diagram contradicts what you say.

It sounds like you are mixing extent with gradient of potential

L5 is almost flat - very little change in potential over distance - which necessarily means its very large in extent.

Or look at it another way, L5 is so weak, you'd have to travel millions of miles from it to even notice a change.

L2 OTOH is so strong bit also so compact that a small deviation will be a huge climb uphill.
excep that way I see it L2 is the lowest potential on the map are these not potential maps?
 

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