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What is the advantage of keeping the passive in orbit instead of dropping it in an ocean? If for any reason it fell from orbit, later, it could impact an impor

What is the advantage of keeping the passive in orbit instead of dropping it in an ocean? If for any reason it fell from orbit, later, it could impact an important surface (city, nuclear plant), and definitely a larger area.

ocdtrekkie 2y ago
Generally getting things out of space also requires effort/fuel. A graveyard orbit is specifically easier to get to but uninteresting from a "hitting other things" perspective.
- whartung 2y ago
  Folks don’t appreciate how much energy it actually takes to “fall” into the Earth or even the Sun. It’s not really intuitive when most personal experience is how hard it is to keep things from not falling.
  - HappMacDonald 2y ago
    Both Kerbal Space Program and Outer wilds can sometimes do a good job of driving this point home (especially in OW if you're outside of the ship). It even humbled Thor from PirateSoftware during his stream, which I found hilarious. :D
  - stevage 2y ago
    The only way to understand is to play Kerbal.
  - tgsovlerkhgsel 2y ago
    Most of the attractiveness of the concept of "rods from god" relies on this misconception (if you haven't heard the term, it's a theoretical space weapon consisting of placing large tungsten rods in orbit, then dropping them on whatever needs to be destroyed in a purely kinetic strike with the energy of a massive bomb or tiny nuke). The concept isn't entirely infeasible (you could deorbit the rods actively), but the advantages over just hitting the target with an ICBM quickly dwindle once you realize you can't just "drop" them.
    - sebastiennight 2y ago
      Plus the amount of energy needed to place them there in the first place is also probably way more cost-prohibitive with much less stealth than said ICBM. I also imagine you could probably knock these things off-course for much cheaper than it cost to put them there or to keep them there.
  - Majromax 2y ago
    > Folks don’t appreciate how much energy it actually takes to “fall” into the Earth or even the Sun. It's not that bad from the Lagrange points thanks to multi-body gravitational dynamics. The Lagrange points form part of the so-called 'interplanetary transit network' (https://en.wikipedia.org/wiki/Interplanetary_Transport_Network https://en.wikipedia.org/wiki/Interplanetary_Transport_Netwo...) that supports extremely low-energy transfers in the solar system, provided you're willing to wait a very long time. That being said, there's no reason to send a deep-space probe back to Earth for reentry when it's even easier to send it into a deep-space graveyard orbit of some description with no risk of any potential debris-related problems.
JumpCrisscross 2y ago
> If for any reason it fell from orbit, later You’d need 320+ m/s of delta-V to knock something from Sun-Earth L2 to Earth [1]. Not impossible. But akin to the energy required to go from Earth transfer to Mars or Venus transfer [2]. Less likely to fall to Earth than become an orbital pest. (It's a good question.) [1] https://www.researchgate.net/figure/Duration-and-delta-v-requirements-of-a-SEL-2-servicing-mission_tbl8_319122341 https://www.researchgate.net/figure/Duration-and-delta-v-req... [2] https://en.m.wikipedia.org/wiki/File:Solar_system_delta_v_map.svg https://en.m.wikipedia.org/wiki/File:Solar_system_delta_v_ma...
- vkou 2y ago
  You mean 3,200+ m/s of delta-V.
  - JumpCrisscross 2y ago
    I actually made a comedy of errors when I went back and edited my first estimate ("about 300 m/s"). For debris to become a problem to us ground people, it really just needs to make it to LEO–the atmosphere will take care of it from there. So I was trying to estimate the Δv for L2 to LEO, which is about 200 m/s according to Table 3 on that source. So probably not 3 km/s. But 320 is too precise given the handwavy math I am doing.
  - Dylan16807 2y ago
    That's the delta-V to get away from Earth, and 90% of it is spent shifting from a circular low orbit into a transfer/escape orbit. When you're coming back, you don't need to spend that energy. Instead you slightly adjust your aim so your transfer orbit hits the atmosphere instead of missing it.
- perihelions 2y ago
  It's actually much less than 320 m/s—that's a figure for a fast transfer. If you have time for it, three-body dynamics let you get arbitrarily cheap transfers in the regions of the Sun/Earth/Moon Lagrange points, https://nebula.esa.int/sites/default/files/neb_study/1167/C4000107618ExS.pdf https://nebula.esa.int/sites/default/files/neb_study/1167/C4... (Table 1 disposal options: 88.56 m/s to crash Gaia into the Earth—or even just 6.18 m/s to crash Euclid, another SEL-2 dweller).
Aachen 2y ago
To make it less abstract than the argument that it costs "delta v": it's a bit like asking, what if this rock I placed on the floor impacts a nuclear plant one day? It requires energy to move there. This change in velocity, making something drop from orbit by itself, happens across months~years for only the lowest orbits around the earth, caused by stray particles from our atmosphere impacting the spacecraft/satellites there, but at L2 that's afaik not a factor
noselasd 2y ago
Gaia is a about 4 times farther out from Earth than the Moon, there's little reason to fly it back to Earth.
- adastra22 2y ago
  And insufficient fuel.

ocdtrekkie 2y ago

Generally getting things out of space also requires effort/fuel. A graveyard orbit is specifically easier to get to but uninteresting from a "hitting other things" perspective.

whartung 2y ago
Folks don’t appreciate how much energy it actually takes to “fall” into the Earth or even the Sun. It’s not really intuitive when most personal experience is how hard it is to keep things from not falling.
- HappMacDonald 2y ago
  Both Kerbal Space Program and Outer wilds can sometimes do a good job of driving this point home (especially in OW if you're outside of the ship). It even humbled Thor from PirateSoftware during his stream, which I found hilarious. :D
- stevage 2y ago
  The only way to understand is to play Kerbal.
- tgsovlerkhgsel 2y ago
  Most of the attractiveness of the concept of "rods from god" relies on this misconception (if you haven't heard the term, it's a theoretical space weapon consisting of placing large tungsten rods in orbit, then dropping them on whatever needs to be destroyed in a purely kinetic strike with the energy of a massive bomb or tiny nuke). The concept isn't entirely infeasible (you could deorbit the rods actively), but the advantages over just hitting the target with an ICBM quickly dwindle once you realize you can't just "drop" them.
  - sebastiennight 2y ago
    Plus the amount of energy needed to place them there in the first place is also probably way more cost-prohibitive with much less stealth than said ICBM. I also imagine you could probably knock these things off-course for much cheaper than it cost to put them there or to keep them there.
- Majromax 2y ago
  > Folks don’t appreciate how much energy it actually takes to “fall” into the Earth or even the Sun. It's not that bad from the Lagrange points thanks to multi-body gravitational dynamics. The Lagrange points form part of the so-called 'interplanetary transit network' (https://en.wikipedia.org/wiki/Interplanetary_Transport_Network https://en.wikipedia.org/wiki/Interplanetary_Transport_Netwo...) that supports extremely low-energy transfers in the solar system, provided you're willing to wait a very long time. That being said, there's no reason to send a deep-space probe back to Earth for reentry when it's even easier to send it into a deep-space graveyard orbit of some description with no risk of any potential debris-related problems.

JumpCrisscross 2y ago

> If for any reason it fell from orbit, later You’d need 320+ m/s of delta-V to knock something from Sun-Earth L2 to Earth [1]. Not impossible. But akin to the energy required to go from Earth transfer to Mars or Venus transfer [2]. Less likely to fall to Earth than become an orbital pest. (It's a good question.) [1] https://www.researchgate.net/figure/Duration-and-delta-v-requirements-of-a-SEL-2-servicing-mission_tbl8_319122341 https://www.researchgate.net/figure/Duration-and-delta-v-req... [2] https://en.m.wikipedia.org/wiki/File:Solar_system_delta_v_map.svg https://en.m.wikipedia.org/wiki/File:Solar_system_delta_v_ma...

vkou 2y ago
You mean 3,200+ m/s of delta-V.
- JumpCrisscross 2y ago
  I actually made a comedy of errors when I went back and edited my first estimate ("about 300 m/s"). For debris to become a problem to us ground people, it really just needs to make it to LEO–the atmosphere will take care of it from there. So I was trying to estimate the Δv for L2 to LEO, which is about 200 m/s according to Table 3 on that source. So probably not 3 km/s. But 320 is too precise given the handwavy math I am doing.
- Dylan16807 2y ago
  That's the delta-V to get away from Earth, and 90% of it is spent shifting from a circular low orbit into a transfer/escape orbit. When you're coming back, you don't need to spend that energy. Instead you slightly adjust your aim so your transfer orbit hits the atmosphere instead of missing it.
perihelions 2y ago
It's actually much less than 320 m/s—that's a figure for a fast transfer. If you have time for it, three-body dynamics let you get arbitrarily cheap transfers in the regions of the Sun/Earth/Moon Lagrange points, https://nebula.esa.int/sites/default/files/neb_study/1167/C4000107618ExS.pdf https://nebula.esa.int/sites/default/files/neb_study/1167/C4... (Table 1 disposal options: 88.56 m/s to crash Gaia into the Earth—or even just 6.18 m/s to crash Euclid, another SEL-2 dweller).

Aachen 2y ago

To make it less abstract than the argument that it costs "delta v": it's a bit like asking, what if this rock I placed on the floor impacts a nuclear plant one day? It requires energy to move there. This change in velocity, making something drop from orbit by itself, happens across months~years for only the lowest orbits around the earth, caused by stray particles from our atmosphere impacting the spacecraft/satellites there, but at L2 that's afaik not a factor

noselasd 2y ago

Gaia is a about 4 times farther out from Earth than the Moon, there's little reason to fly it back to Earth.

adastra22 2y ago
And insufficient fuel.