38
$\begingroup$

We all know that James Webb Space telescope is planned to be launched in 2018. It has been decided that the orbit of JWST will be elliptical around the Lagrange point L2, which has been declared as one of the unstable points (L1, L2, L3).

So then, why should it orbit the point?

I've got a few points. The telescope's instruments are quite sensitive and should always be maintained away from Sun (Really? Is that so?) and also in a cold environment - both these are satisfied by L2. Wikipedia says this:

The combined gravitational forces of the Sun and the Earth can hold a spacecraft at this point, so that in theory it takes no rocket thrust to keep a spacecraft in orbit around L2. In reality, the stable point is comparable to that of a ball balanced upon a saddle shape. Along one direction any perturbation will drive the ball toward the stable point, while in the crossing direction the ball, if disturbed, will fall away from the stable point. Thus some station-keeping is required, but with little energy expended (only 2–4 m/s per year, from the total budget of 150 m/s)

First, is it really stable at all? If that is so unstable in reality, why should it be positioned there? I mean, L4 & L5 points are fine. Why can't the telescope be positioned in such a way that it's always facing outward from the Sun? (Earth too, if they don't want the reflection from it to crack itself)

$\endgroup$
4
  • $\begingroup$ @geoffc: "... sun gets that way?" That's why I mentioned "positioned facing outward from the sun". IMHO, I think it can be done ;-) $\endgroup$ Jul 17, 2013 at 16:57
  • $\begingroup$ @PearsonArtPhoto Ya, I was thinking L3 more than L4/L5. See en.wikipedia.org/wiki/Co-orbital_configuration Ignore my comment, may delete it. $\endgroup$
    – geoffc
    Jul 17, 2013 at 17:25
  • 1
    $\begingroup$ @CrazyBuddy - regarding the requirement to keep spacecraft away from the Sun. Oh heck yeah!! The telescope will be operating, according to Wikipedia, at "roughly 40 K (−233.2 °C; −387.7 °F)". In space, at the distance from the Sun where JWST will be operating, exposure to direct sunlight would heat it up to something like 200C within a matter of minutes. This would render it useless. $\endgroup$ Jul 22, 2013 at 2:38
  • 2
    $\begingroup$ Totally agreed, We all really do know it's gonna be launched in 2018 $\endgroup$ Dec 25, 2019 at 23:25

3 Answers 3

51
$\begingroup$

There are a couple of reasons.

  1. The distance from the L2 to Earth is only 1.5 million km away. The L4/L5 are 1 AU, or about 150 million km away. That leads to a reduction in link margin of 40 db, or 1/10000. That is quite significant. In order to compensate for that difference, you either need a bigger radio dish, more power, or a loss in data.
  2. As you mentioned, the fuel usage is quite low to maintain that position, only on the order of 150 m/s delta v for the entire mission. That isn't a whole lot, and in fact, is less than what is required to keep a satellite in geostationary orbit.
  3. The satellite is much closer, reducing the time to command an object. Light only will take 5 seconds to reach James Webb, whereas it will take 9 minutes to reach L4/L5. This limits the ability to do real time commands, which occasionally are useful (Think Gamma Ray Bursts, Super Novas, etc).

Bottom line is, the communication problem is simplified with a closer telescope, and that more than makes up for having to take a bit more fuel.

$\endgroup$
1
  • 7
    $\begingroup$ ...and if they ever did have to fix something, hypothetically speaking of course, robotically or manned, the same L2 advantages apply here as well. $\endgroup$
    – uhoh
    Jan 27, 2016 at 13:52
7
$\begingroup$

I believe it is as the Wikipedia sub-article on L2 says:

The Sun–Earth L2 is a good spot for space-based observatories. Because an object around L2 will maintain the same relative position with respect to the Sun and Earth, shielding and calibration are much simpler.

$\endgroup$
5
  • 3
    $\begingroup$ While this is true, the same is true for pretty much every one of the L points. $\endgroup$
    – PearsonArtPhoto
    Jul 23, 2013 at 17:03
  • 4
    $\begingroup$ @PearsonArtPhoto shielding - L2 remains in Earth's shadow, meaning no glare from the Sun. $\endgroup$
    – SF.
    Jan 26, 2016 at 11:03
  • 1
    $\begingroup$ @SF., there is no any shadow at JW orbit - space.stackexchange.com/a/4111/2843. Shielding is easier than in LEO when there is very huge warm IR-sending object is rotating around the telescope at random angles - soo much thermal shocks every hour, and it is impossible to have (actively and fast-rotating) EarthShield to hide from thermal radiation from Earth for cryogenic mirrors. Earth "shadow" in scale - en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra#Penumbra " The full cone extends over 1.32 million km.", and L2 is 1.5 mln. And L2 halo orbit is 100s km away from L2 point. $\endgroup$
    – osgx
    Jan 21, 2018 at 5:40
  • $\begingroup$ @osgx: that only means no full umbra. L2 remains in antumbra which severely reduces the amount of light reaching the point. That is definitely different than "no any shadow" - unlike at every other Earth-Sun Lagrangian point. $\endgroup$
    – SF.
    Jan 21, 2018 at 22:36
  • 3
    $\begingroup$ @SF. and still JWST (and other observatories) are at halo orbit 100s kilometers, trying to keep their path away from antumbra&penumbra as long as possible. Gaia - issfd.org/ISSFD_2014/ISSFD24_Paper_S2-5_Renk.pdf "disadvantage of eclipses ... eclipse avoidance manoeuvre ... partial eclipse .. is undesirable with respect to the thermal balance"; JWST - ntrs.nasa.gov/search.jsp?R=20160001318 "Significant requirements affecting the JWST... the avoidance of any Earth/Moon eclipses... none allowed, Constraint Driver:Power and Thermal" $\endgroup$
    – osgx
    Jan 22, 2018 at 4:14
3
$\begingroup$

About the stability, L2 is unstable in the radial direction: if the probe is a little closer or a little further in the Sun-Earth axis it will be pushed yet further by gravitation.

However L2 is stable in the perpendicular plane, that's why in some animation you see it orbiting L2 in the plane perpendicular to the Sun-Earth axis: gravitation will pull it toward L2.

See this post for a detailed explanation.

$\endgroup$
2
  • $\begingroup$ Do you know about the third direction? I guess I'd phrase it precisely like this. L2 traces an orbit around the sun. Imagine the JWST is at a point on this orbit 1.5 million km of the actual L2 point. Will gravity pull it towards L2? (The fact that it's orbiting around L2 makes me think that surely it must be, but I don't understand how.) $\endgroup$ Jan 4, 2022 at 21:34
  • $\begingroup$ Sorry for not being accurate :) When I say "perpendicular direction" I mean "perpendicular plane". So radial direction = 1 unstable direction, perpendicular plane=2 stable directions. Also when talking about stability, don't forget we are in a rotating referential. Gravity will not pull the spacecraft toward L2 on its own. It's a combination of gravity and Coriolis forces that makes up Lagrangian points. $\endgroup$
    – lolo101
    Jan 8, 2022 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.