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What is the delta v required to place a satellite in L2 from earth transfer orbit ( Like the James Webb Telescope )? Does orbits closer to L2 different delta v from orbits far from L2?

Ashvin
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  • It is not clear to me whether you are asking for the actual numbers for the JWST (and with the launcher's actual delivery performance) or you are asking for a general theoretical number to reach Sun-Earth L2 from the Earth surface. For the actual numbers we have some ideas given by the actual 2 mid-course correction burns durations, MCC1a and MCC1b (65 minutes and 9+ minutes, resp.). These can be correlated with the analysis by Petersen (cf Table 3). – Ng Ph Dec 30 '21 at 15:03
  • (cont.) Accordingly, my calculation gives approximately: 20 m/s (MCC1b+MCC1a). Recall that this corresponds to the intentional "hold back" launch strategy proper to the JWST (because it cannot brake if it goes too fast). So, it remains to know (calculate) the actual Delta-v the JWST got from the launcher at the point it was released. I don't know whether it can be inferred from the public data released by Arianespace/ESA/NASA. – Ng Ph Dec 30 '21 at 15:18
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    @NgPh I am looking for general delta v for sun-earth L2 insertion. – Ashvin Dec 31 '21 at 17:00
  • Then you should make that explicit in your question. You may consider also to add precisions as from where you consider the "insertion" to start (from Earth? from LEO?, ...), how long to stay around L2, and whether you are looking for minimum transfer time or minimum fuel (maximum payload), etc ... – Ng Ph Jan 02 '22 at 20:57
  • @NgPh edited the question body on your suggestion – Ashvin Jan 03 '22 at 15:39
  • Your edits still leave loose ends in the question. "To place a satellite in L2" is not a clearly-defined goal (does "in" mean mathematically at rest at the equilibrium point or in some 3D box?). Note that the Delta-v to crash a probe on the Moon is not the same as the Delta-v to land it softly on the Moon surface. Secondly, "Earth transfer orbit (like [that taken by] the JWST)" assumes that the transfer from Earth to a point of the first orbit correction is a well-known (unique) one. I don't think there is a consensus on that. 3rdly, you didn't specify the fly-time, nor the loitering time. – Ng Ph Jan 04 '22 at 09:02
  • (cont.) Alternatively, one may consider that the 2 first corrections (MCC1a & MCC1b) are parts of the maneuvers to get into the correct "Earth transfer orbit", although for JWST they are not excuted by the launcher. In this case, the Delta-v you are looking for is that of the MCC2 (which may well be zero, or at least negligible, for the insertion goal of JWST). – Ng Ph Jan 04 '22 at 09:16

1 Answers1

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Getting from LEO to Earth-Sun L2

(source: https://www.researchgate.net/figure/3-Duration-and-delta-v-requirements-of-a-SEL-2-servicing-mission_tbl8_319122341)

Departure: 3.23km/s Arrival: 0.9km/s

Note that this is to the actual L2 point.

Second part of your question.. To match a Halo orbit around L2 such as the JWST uses, would require a small amount more departure delta-v, and very very little (<50m/s) of arrival delta-v.
The JWST will actually not slow down when it gets near its Halo Orbit, but will speed up by another tiny bit, to insert itself into the orbit.

CuteKItty_pleaseStopBArking
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  • "this is to the actual L2 ... to match a Halo orbit around L2 such as the JWST uses, would require a small amount more departure delta-v, and very very little (<50m/s) of arrival delta-v". Can you expand? ... what is "small amount more"? Where did you get the upper bound <50m/s for arrival from? – Ng Ph Jan 05 '22 at 21:00