Hacker News Reader

Comment by saagarjha | Hacker News Reader

This doesn't answer my question of whether the orbiter will remain in its highly elliptical orbit or not.

“On entering Moon's sphere of influence, on-board thrusters will slow down the spacecraft for Lunar Capture. Subsequently the orbit of Chandrayaan-2 around the moon will be circularised to a 100x100 km orbit through a series of orbital maneuvers.”

Thanks!

The series of 4 planned orbital maneuvers are outlined here https://www.isro.gov.in/update/14-aug-2019/chandrayaan-2-suc...

You have to imagine the eccentric orbit, 114 km perigee on the far side and a long wait going out to 18k km, toward and away from Earth, looping past the moon's poles.

Over time the eccentric orbit will swing away from pointing at Earth. They will do burns at perigee and the eccentricity will gradually contract until the orbit is about circular. After a week it has Earth in view full- time, a good time to do the final burns; after 2 weeks it has swung around to going behind the moon again, still going over the poles. They will land during the first period when the orbiter is in view full-time.

Congratulations to everyone from India. This has to be a proud moment.

Actually I was wrong about the initial and later direction of the orbit axis. It started with the orbiter swinging way to the left, as seen from earth (axis pointing at us), and is rotating now to be behind the moon, most of the time, come early next week, and in sight when it needs to do burns to circularize its orbit. By the time it lands, the orbit should have swung further and be coming up close to its axis pointing not too far from us, again. It should be in line of sight most of the time, by then.

It should have.

A 100 km X 30 km looks like elliptical orbit to me.

That's not really all that elliptical, especially compared to the eccentricity that the original orbit has. Also note that that figure was for the lander, not the orbiter.

EVERY orbit is elliptical.

100km X 30km are ELEVATIONS, meaning the real orbit is diameter of Moon + 100km X diameter of Moon + 30km.

If you had this scaled and drawn on a sheet of you would need very accurate instrument to detect any deviation from perfect circle.

> real orbit is diameter of Moon + 100km X diameter of Moon + 30km.

3,574km X 3,504km

That helped me a bit.

So maybe not so precise instrument, but still, if this was 15cm ellipse the difference would be about 1mm.

Yes, that makes sense.

100km x 30km orbit around moon, what was I thinking - facepalm (－‸ლ)

Reply on news.ycombinator.com