While launching something directly into the sun costs a lot of energy, couldn't we just launch it opposite earths orbit?

So, according to this question, it's pretty hard to launch something into the sun directly, it takes a lot of energy to cancel out earths orbital speed.

That being said, I'm not understanding why this is required. From my understanding of orbital mechanics (which admitted is only available from video games that involve orbital mechanic fiddling), I understand that something needs to be maintaining a fairly constant speed or else it either spirals into or away from the central orbiting object (kind of like the moon is for us).

So, assuming that the earth is orbiting the sun at a direction of (1,0,0), couldn't we launch something in the direction of (-1,0,0) such that by the very fact of it's launch it is going slower than earth and spirals into the sun over a very long (centuries? Millenia?) period of time?

• I imagine that the biggest cost is just getting the waste off of the Earth to begin with, given that the Earth exerts a large gravitational force on everything near its surface. – march Dec 20 '16 at 18:29

The problem is that you suggest an object moving slightly slower than the Earth would spiral into the Sun, but this doesn't happen.

An object orbitting the Sun moves in an ellipse. This diagram shows what happens if we launch objects from the Earth as you suggest:

The black circle shows the orbit of the Earth, which is close to circular (the Earth actually moves in a slightly elliptical orbit). If we fire our projectile so it ends up with a slightly lower velocity than the Earth then we end up with the blue orbit. This is slightly more elliptical than Earth's orbit, and it comes closer to the Sun at perihelion. But this is a perfectly stable orbit and the projectile does not spiral inwards.

If we now give our projectile a bit more oomph so it ends up moving a lot slower than the earth we get the green orbit. This is now a pronounced ellipse, and it comes a lot closer to Sun at perihelion. But again it remains in a stable orbit and won't hit the Sun.

If we now fire our projectile at roughly Earth's orbital speed we get a very elliptical orbit like the red one, and this is what we have to achieve to hit the Sun. I've exaggerated the size of the Sun (if it was to scale it would be about half a millimetre wide) so in fact the orbit would have to be very, very elliptical for our projectile to actually hit the Sun. We'd have to reduce it's tangential velocity to effectively zero.

Ad this is the problem. The orbital velocity of the Earth round the Sun is about 30 km/sec and we need our projectile to have this velocity relative to us to get such a tight orbit. Actually you have to add the escape velocity of the Earth on as well, making the launch velocity about 41 km/sec. If you want to hit the Sun there just isn't any way round this.

Well, there is a way round it but it's complicated. We're used to thinking of using gravity assists to boost the speed of a spacecraft so it can reach the outer planets. However we can use gravity assists to slow down a spacecraft and make it plunge into the Sun. The Mercury probe used exactly this method to put it into an orbit round Mercury. But these are complicated manoeuvres. Mercury required an Earth flyby, two Venus flybys and three Mercury flybys to get it safely to Mercury. That's fine for a high budget space probe but not great for disposing of waste into the Sun.

• Very detailed and easy to follow explanation. +1 would ask inane laymen questions again. :) – Sidney Dec 20 '16 at 19:06
• A Δv of 30 km/s from a 300 km orbit above the Earth (typical launch insertion) is not needed to attain that 30 km/s $v_\infty$ relative to the Earth because the vehicle is already going about 7.73 km/s. An impulsive Δv of 24.2 km/s, directed along the velocity vector, will suffice. Apply this impulsive Δv at the right time from the right low Earth orbit and the vehicle will drop straight into the Sun. On the other hand, waiting half an orbit and applying a Δv of 8.82 km/s puts the vehicle on an escape trajectory from the solar system. – David Hammen Dec 21 '16 at 13:22
• One doesn't need to apply that 24.2 km/s Δv to hit the Sun. One can use gravity assists, as the answer mentions (but these also can be used to reduce the Δv cost of sending something out of the solar system). Another option: Put the vehicle on a solar system escape trajectory, wait, and cancel the velocity. For example, waiting 214 years brings the vehicle to 200 AU from the Sun, at which point the velocity will have fallen to 3 km/s. This reduces the total Δv (from low Earth orbit) needed to hit the Sun to about 11.8 km/s. – David Hammen Dec 21 '16 at 14:21