It takes just as much ∆V to throw a rock from the Oort cloud to Earth as to throw the same rock from Earth to the Oort cloud. The aliens don't get an energy advantage by holding the high ground.
> It takes just as much ∆V to throw a rock from the Oort cloud to Earth as to throw the same rock from Earth to the Oort cloud.
Yes, but they clearly have the capability to get to the Oort Cloud, and we (other than a small probe) do not. Having the high ground (both physically and technologically) makes holding the Oort Cloud a pretty good spot to be in.
(It also takes substantially less ∆V to nudge an existing rock into a nastier trajectory, and we're essentially a big easy to hit static target.)
The whole point of the high ground is someone just needs to ‘drop’ something in, which is cheap and easy, but someone on the ‘low ground’ needs to make up all the energy to get up there before they are even at the same level.
The energy involved in de-orbiting something and dropping it to the surface of earth (at very high speed) is orders of magnitude less than it would take to get the same mass to even earth orbit from the surface.
Holding a brick over the opening of a well is a much more credible threat to someone at the bottom, than that someone with the same brick threatening to throw it back out.
All you need is for orbits to intersect with the surface, not for a zero velocity vector at the surface. That is far cheaper to do than getting to orbit in the first place.
This absolutely applies to orbital mechanics.
Atmospheric braking makes it less likely to punch a hole in a skyscraper in manhattan when de-orbiting. And defacto raises ‘the surface’ in some senses.
It doesn’t change that it’s far ‘cheaper’ to go down a gravity well than go out of one.
Though that atmospheres (and liquid oceans) exist at all does prove exactly the point that I am talking about. The Sun and planets too, come to think of it.
If it was cheaper to ‘get out’ than ‘fall in’, none of those could exist.
Edit: I used a Hohmann transfer orbit calculator, and from an orbit of ~200km above sea level to an orbit intersecting ground level (0km), it only takes 203 m/s.
Without an atmosphere, it would be quite a show of course. But in this situation, that’s the point isn’t it?
Yes, I guess you're right, you aren't going from orbit to orbit, you only try to crash something (and actually want to keep the difference to create an impact).
Still, your logic doesn't apply, and hitting something in a higher orbit shouldn't be that much harder. You also only need to nudge it to hit something UP. (though it does get harder with a lesser impact with increasing distances)
I fact there might not be that much of an effect at all when you consider that both Earth and the Oort cloud are in the Sun orbits - you would have to hit something pretty hard in the Oort cloud to create a massive impact on Earth, there would at best be a bit of leverage, which I guess wouldn't outweight the protective effect of Earth's atmosphere, unless you try to send something monumental, at which point you probably can just come down and hit Earth. Consider also the timescales involved.
Did you do it? That maneuvre would take decades to millenia, and when you use something that far, it may be more efficient to push it into a retrograde orbit so that it hits Earth heads on, instead of chasing after it. It's just absurdly impractical.
Anyway, if you really want to destroy a planet, you want sonething small but fast. It penetrates into the planet, and rips surface on the other side.
You fundamentally misunderstand the concepts. The other poster is correct. Given two orbits, it takes just as much energy to get "up" as it does to get "down".
care to provide some math? delta v calcs quite definitely disagree. no one is going to LEO on only 220 m/s. assuming an impact with the surface of course, which is clearly part of the equation.
if landing gently, then sure. but that is an entirely different problem.
That's not actually true; it's an asymmetric problem. You're asking about impacting another orbiting object with nonzero relative velocity. That's not the symmetric question of transferring between two orbits A and B, both ways; it's the asymmetric question of going from orbit A to "some orbit that intercepts B at some point".
It's much less delta-v to go from the Oort cloud to Earth—to a highly-eccentric orbit that intercepts the Earth's orbit, without matching its velocity.
Worse, they could throw bits of the Oort Cloud at us.