I'm not rocket scientist, but it seems like the author is calculating a decreasing spiral into the sun rather than an impact. Why not slingshot around a planet and barrel into the sun with all of the orbital velocity still intact.
The ~30km/s figure is what you'd need to drop it straight into the sun, no spiraling involved.
Gravitational assists can help, but they can also help you get other places. Getting to the sun would still be ridiculously hard compared to the alternatives.
But that's is expending a whole lot of energy to "remove" earth's orbital velocity[1]. Why remove it when you can just deflect the rocket by the gravity of Venus and redirect the rocket directly towards the sun? Or are we missing something more fundamental?
> "Gravitational assists can help, but they can also help you get other places."
Without gravitational assists, you need ~30km/s to drop into the sun, and far less to go other places. With gravitational assists both can be easier, but going to the sun will still be one of the harder places to go.
That aside, if you are going to loop it around Venus, why not just hit Venus? I'm sure Venus wouldn't mind.
Yes, I did read that. But what I didn't understand is why you were disregarding what seems to be a key part of interplanetary travel. The point that johngalt and I are making is that "falling" isn't the only way to get there.
Now, the real point seems to be that the Sun's gravity doesn't help you get there — in fact it works against you! You must rely on very precise calculations and instrumentation to target yourself there. So why not dive straight into another planet instead of using it for a gravity assist. That I can understand.
Gravity assists are helpful but they're not all-powerful. There's a hard constraint: your orbit is deflected more as you make your point of closest approach lower, but you can't make it too low without hitting the atmosphere. (Unless you actually want to crash into Venus instead of the sun; in which case, knock yourself out, but the moon is closer.)
Math time: a Hohmann transfer orbit from Earth to Venus will have 2.7km/sec of excess velocity when it reaches Venus. That means your orbit has a 44000km semi-major axis. With a closest approach of 6100km, that gives you a minimum eccentricity of 1.137; if I'm doing my trigonometry right, the most orbital speed you could lose in a single flyby is 4.2km/sec, or about 11% of your speed relative to the sun.
You could do better with multiple carefully-orchestrated flybys, but the point is it's not as simple as just heading for the nearest planet and letting it fling you wherever you want to go.
> Why remove it when you can just deflect the rocket by the gravity of Venus and redirect the rocket directly towards the sun?
The fundamental thing you are missing is that to point the rocket directly at the sun required removing all of the orbital velocity. If you were to just point a rocket directly at the sun ignoring its relative motion and burn you would never actually hit the sun. You would just burn forever and never make any progress.
So, getting to the sun is actually really hard, the earth is moving around the sun at ~30km/sec. For comparison, a satelite in low earth orbit is only moving about 8km sec.
30km/sec is a LOT of velocity change, and you'd need to cancel practically all of it to actually fall into the Sun.
Let me put it another way. If it is comparatively cheap to get to any of the other planets. Couldn't we just go to Neptune and stop our orbital velocity from there where it's only 5kms?
How about exiting planetary orbits retrograde? It seems like larger planets would have escape velocities that are larger than their orbital velocity. Why couldn't our space barge leave Jupiter's orbit on a trajectory that would intersect with the sun?
> Couldn't we just go to Neptune and stop our orbital velocity from there where it's only 5kms?
Yes. This is a bi-elliptic transfer, it's a standard technique, and another thread claims it would save about 40% of the required dV.
> How about exiting planetary orbits retrograde? It seems like larger planets would have escape velocities that are larger than their orbital velocity. Why couldn't our space barge leave Jupiter's orbit on a trajectory that would intersect with the sun?
It could. That's a gravity assist, and it's the only practical way to go just about anywhere with current technology (all our interplanetary probes do that).
(There are arguments against using either with nuclear waste, but I don't think the page is seriously talking about that)
If you're interested in this stuff, definitely play Kerbal Space Program.
Because if the orbital velocity is intact you will perputally miss hitting the sun. The only way you can hit a thing you are orbiting is to remove all the orbital velocity (well, remove enough of the orbital velocity such that your orbit drops low enough to scrape the surface anyway).
While the solar wind gets stronger closer to the sun, it's an r-square law that balances the r-squared force of gravity. Fun fact: If a solar sail could be in a fixed position at Mercury orbit then it could also be in fixed position a Neptune orbit.
Quoting from http://en.wikipedia.org/wiki/Solar_sail, the "total force exerted on an 800 by 800 meter solar sail, for example, is about 5 newtons (1.1 lbf) at Earth's distance from the Sun".
The Orion space capsule is ~25 sq meters. I'll round that up to 100 sq meters (I'm being generous), giving a breaking force of 5/64 N on the waste barge.
The capsule mass is 20 tons. Assuming the waste barge were the same mass (I'm being generous) gives a solar gravitational force in Earth orbit of 120 newtons. This far exceeds the push from solar wind, even with wildly optimistic numbers. (Note: there's some 50,000 tons of high-level nuclear waste in the US.)
So no, you couldn't lose a lot of velocity just braking against the solar wind. Not unless you've also developed effective solar sail technology.
Your calculations seem sensible, but you're only talking about radial forces, ie. countering the "downward" force of gravity with "upward" thrust of a solar sail.
But space flight maneuvers typically use tangential or almost tangential thrust. Slowing down or accelerating the orbital motion will cause the altitude to vary, preserving the orbital energy and angular momentum. Placing a solar sail in a 45 degree angle from the sun will change the orbit slowly but surely, as there will be a little tangential and radial thrust.
I still don't think it's viable for throwing anything into the sun, though.
I think it's viable. There are no other forces acting on you. Set up the solar sail at 45 degrees "against" your orbit. You will slowly but inevitably spiral into the sun without burning any fuel.
If you have effective solar sail technology then all my calculations are thrown out the window.
If you don't, then the time to spiral into the sun is large. Figuring 5/64 N on 20 tons and an orbital velocity of 30km/s gives
t = v / a = 30 km/s / ((5/64) N / 20000 kg)
= 245 years to cut the orbital velocity
This assumes all of the force could be used to slow the waste hauler. As http://en.wikipedia.org/wiki/Solar_sail points out, the force is cos^2(theta) so the actual time at 45 degrees is
t = 245 * (sqrt(2)/2)^2 = 490 years.
The actual time will be smaller because you only need to hit the sun, and solar wind will get stronger. So, 350 years? As a wild-ass guess.
That same page points out that sails don't work much inside of 0.25 AU, because the temperature can exceed the material properties of the sail. Though I think if the apogee is inside of 0.20 AU it's good enough.
To make it worse, the solar wind fluctuates, so unless there's active control on the rocket, its orbit will be unpredictable over the centuries. When it's still near Earth orbit, or when it approaches Venus orbit, what are the chances of a gravitational assist leading to an Earth-return?
With 50,000 tons of high-level waste, and 15 tons per rocket => 3,333 rockets in uncertain orbits, the chances become much higher.
its not the only way, its just one way... there are other ways to increase the eccentricity of an orbit - increasing the orbital velocity at the right time, and in a direction that is not radial is less efficient but since the earth's orbit is nearly circular, its not by much.
this is essentially the brute force approach or the bielliptic transfers mentioned in other comments.
he is describing making it fall into the sun directly (if the 'burn' is instantaneous), or if you like, starting to spiral but then just falling in a straight line.
a slingshot changes orbital velocity... thats the whole point. you are right that it is possible to slingshot into an ellipse intersecting the sun... although setting up that manoeuvre itself would be pretty expensive.
