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What's the physical intuition here? Is it that you want the de-orbit to hit the atmosphere at a shallow angle instead of plunging to earth?

• jessaustin 7y ago
  A small change in momentum requires less energy than a large change in momentum.
  • jessaustin 7y ago
    Umm, wat? Does someone seriously disagree with this?
• Dylan16807 7y ago
  It doesn't de-orbit at all. For physical intuition, uh, the atmosphere is a hundred times further away than the graveyard orbit.
• pdonis 7y ago
  > Is it that you want the de-orbit to hit the atmosphere at a shallow angle instead of plunging to earth? De-orbiting to hit the atmosphere at a shallow angle would actually take slightly less delta-v than de-orbiting to a straight plunge to Earth. But both are much larger than the delta-v required to get to a graveyard orbit.
• penagwin 7y ago
  Geostationary orbit is REALLY far away from earth (and also the atmosphere) compared to other orbits, but it's very special and so we want to keep as little junk in it as possible. Other types of orbits are so low, that they have to keep maintaining their orbit or they'll decay into the atmosphere. So we're in a stable orbit with a high periapsis (closest part of our orbit to earth). To de-orbit we need to lower the periapsis so that it touches the atmosphere, and on each pass the satellite will slow down, further lowering the apoapsis and periapsis. The problem is that to do this we have to burn all that horizontal velocity that's keeping us in orbit in the first place, which is a crapload. An alternative is to instead change the orbit so it's no longer in geostationary orbit and potentially interfering with live satellites, so instead they just push the satellite further out into the "graveyard orbit". There's 100x difference in the needed energy for the two methods, and I'm sure you can see why most would opt for the graveyard orbit. EDIT: To add, it's actually rather difficult to "plunge into earth". To do this you need to burn all of your horizontal velocity (I think that's around 3.07 km/s in geostationary orbit?). Reaching the sun is actually really difficult as well for the same reason, you have to lose all of your horizontal velocity, and earth has a ton.
  • Aperocky 7y ago
    Wernher von Kerman approves this message.
    • dvtrn 7y ago
      Jeb seems concerned
    • penagwin 7y ago
      My understanding of orbital mechanics is mostly because of KSP (and Scott Manley). I HIGHLY recommend both of them.
  • mark-r 7y ago
    How much of a danger is debris from an exploded satellite? Each fragment is going to be independent and have its own orbit. Any orbit higher or lower than the geostationary belt won't be a problem. Any orbit equal to the original satellite's won't be a problem either. Only those pieces maintaining the same altitude but a different direction will ever intersect the belt, and that will be a rare occurrence.
    • Aperocky 7y ago
      Orbit can be both higher and lower, actually, any deltaV except purely in the axis of movement will result in exactly that. Yes. you can accelerate away from the Earth and have a lower periapsis (conversely, you can accelerate towards the Earth and have a higher apoapsis). The only way you can maintain the same periapsis is by accelerating in the same direction you're going. Highly impossible in an explosion. Also note, it is impossible to raise periapsis in a single event (or single time frame), except by leaving Earth orbit entirely (i.e. eject to planetary space)
    • Floegipoky 7y ago
      You're correct that each fragment will have its own independent orbit, based on the delta-v that it gained from the explosion. But they won't circularize; each of those new orbits will intersect the original orbit at the point where it exploded. The orbital periods of the fragments will also not match the original, meaning that the relative intersection will shift along the original orbit every time. So an explosion in geostationary orbit would be a Really Big Deal and could potentially threaten the whole geostationary fleet.
• willis936 7y ago
  Geosynchronous orbit is a Very high altitude. It’s 6% of the distance to the moon. Inverse square law tells us the surface area that we have to worry about for debris is massive compared to lower altitude orbits. Additionally, it takes a lot of fuel to be able to de orbit high altitude satellites. As orbital altitude increases, de orbit costs start getting out of hand and there is less benefit to deorbiting. So while lower altitude, crowded orbits are required to de orbit to avoid catastrophic cascades from collisions, higher altitude orbits are allowed to simply get out of the way.
  • ska 7y ago
    Additionally, it takes a lot of fuel to be able to de orbit high altitude satellites. I don't think this is so much an a 'additionally', as a 'mostly'.
    • willis936 7y ago
      Forest for the trees. The point is there is a crossover point in terms of cost/utility. That one word doesn’t seriously help in making the point.
• s_tec 7y ago
  Sattelites are always falling to the ground, but they are also going sideways really fast. Imagine tossing a pebble off of a bridge onto a boat - if the sideways velocity is too high, the pebble will miss the boat and land in the water instead. Sattelites are like that. They are going sideways so fast that they miss the Earth entirely as they fall to the ground. If you want a sattelite to come back to earth, you need to get rid of that sideways velocity, which takes fuel.
• jvm_ 7y ago
  Earth orbit is like a funnel. If you're in geo-synchronous orbit, you're at the top edge (Earth is down by the hole). It's easier to jump off the edge of the funnel (and fall into the Sun's funnel) than it is to climb all the way back down to earth.
  • fennecfoxen 7y ago
    The idea that there is an "edge" to the funnel is misleading. And if there were any edge, and it was at geosynchronous orbit, the Moon would have fallen off long ago.
    • jvm_ 7y ago
      I guess the edge is more of the sweet spot where geosynchronous exists, too close or too far are both non-GEO.