back to article China to test recoverable moon orbiter

China is about to launch a lunar orbiter capable of returning home. The nation's State Administration of Science, Technology and Industry for National Defence announced over the weekend that the craft has been built and taken to the Xichang Satellite Launch Center. When it launches later this year, the un-named craft is …

  1. Voland's right hand Silver badge

    Err... That does not fit my understanding of orbital mechanics

    Moon escape velocity is way lower than 11.2km/s. 11.2km/s is earth. All you need is (if memory serves me right) 3 km/s or thereabouts to escape the moon gravity well. Then you fall into the 11km/s gravity well towards Earth.

    In fact, if you align yourself correctly you can slingshot to > 11.2 and escape the Earth + Moon gravity well this way (long range probes regularly do that by doing a few Earth-Moon slingshots). That is what makes the moon potentially attractive as a space base (though frankly, its Lagrange L1 and L2 are way better than the moon itself - you can nudge yourself out of these using only a minimal amount of propellant).

    1. I ain't Spartacus Gold badge
      Happy

      Re: Err... That does not fit my understanding of orbital mechanics

      You say that, but it's not so easy. Sure you don't need much propellant to escape lunar gravity, but then you have to deal with the Clangers with their tractor-beams and death-rays. What did you think really happened to Apollo 13?

      1. WraithCadmus
        WTF?

        Re: Err... That does not fit my understanding of orbital mechanics

        "Why are these rock samples covered in soup and blue string?"

  2. John Smith 19 Gold badge
    Unhappy

    11.2Km/s is the speed you get "falling" from the Moons orbit to the Earth orbit.

    Lunar return needs much smaller velocity changes, hence much less powerful and lower Isp can do the job in the first place.

  3. DropBear
    Joke

    Oh, I see their cunning plan now...

    1) send probe to orbit moon and return

    2) send probe to return with moon samples

    3) send probe to return with the entire moon, to sell as cheese.

    4) Profit!

  4. Graham Marsden
    Coat

    A grand day out?

    Eh, Grommit, lad...

    ... someone's copying us!

  5. Bill Gray

    Not much velocity required for the return

    (I sent the following to the 'corrections' link, but it looks as if there's some

    interest here, too...)

    If we'd had to get to 11.2 km/s to leave earth, then do it all over again to

    get back, I don't think anyone would have gotten to the moon and back.

    The escape speed for the moon is a mere 2.4 km/s. What you'd want to do,

    assuming a minimum-fuel maneuver, is to apply a bit more speed so that,

    when you go flying away from the moon, you're departing at its orbital

    speed around the earth, which is about one km/s. That is to say, you leave

    the moon, but kill your speed relative to earth in the process, so it's

    as if you're just "hanging" 385000 km or so above the earth.

    Since you are now at rest relative to the earth, you'll fall back toward

    it, re-enter at 11.2 km/s, and someone in China will pick you up.

    I'll spare you the derivation, but it works out to sqrt(2.4^2 + 1^2) =

    sqrt( 6.76) = 2.6 km/s.

    Oh, %^*! it... that assumes you're lifting off from the moon and have

    to get to escape speed plus the extra bit to cancel out the moon's orbital

    speed around the earth. _This_ object is in orbit around the moon, and

    is already moving at 70% of escape speed, or 1.68 km/s. So we just need

    to add 0.92 km/second, and about three days later, we fall back to earth.

    This, incidentally, is a bit like the way the Apollo missions worked.

    They had to get up to about 11.2 km/s to get as high as the moon. If they

    hadn't done anything, the moon would have swept by them at about a km/s.

    Instead, they fell into the moon's gravity well, speeding up to about

    2.6 km/s at perilune or periselene or whatever it's called. They would

    have gone right by it, except they decelerated by 0.92 km/s, and ended

    up in lunar orbit, circling the moon at a low-lunar-orbit speed of 1.68

    km/s.

    Next, two guys crawled into the LEM, fired its engines to cancel

    out that 1.68 km/s, walked around on the moon, got back into the LEM,

    and fired the ascent engine to recover the 1.68 km/s so they could go back

    to the command module. (Which was a nice solution; it meant there was

    a heck of a lot of hardware to get back to earth and re-enter that

    could stay in lunar orbit. You didn't have to use up 3.36 km/s worth

    of fuel ferrying it down to the surface and bringing it back to lunar

    orbit. I must confess, I've never run the numbers before and had

    missed the importance of this point.)

    Anyway. The return trip was just the same thing in reverse --

    speed up by 0.92 km/s to escape the moon and kill speed relative to

    the earth; wait to drop back down to earth, accelerating to 11.2 km/s

    as you do so; then kill that speed... fortunately by atmospheric

    braking, rather than needing more bucket-loads of rocket fuel.

    -- Bill

  6. Anonymous Coward
    Anonymous Coward

    facts are cool

    nice to see something presented in cold hard scientific terms.

    Neatly explains why we have been too scared to go anywhere else since the 1970s .

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