Aaargh!
I am not a rocket scientist.
Yes, exactly. You’ve heard of “gravitational slingshots”, probably; a badly-named maneuver that steals a little bit of a planet or moon’s orbital velocity and transfers it to the spacecraft, speeding up the spacecraft considerably and slowing the planet by an amount too small to measure? This is the opposite, where you transfer a lot of the speed of the probe to the planet, which speeds the planet infinitesimally and slows the probe considerably.
Inspired by this thread, I designed a sun-grazing craft in Kerbal Space Program and launched it. Here it is, on its way out of the sphere of influence of the planet Kerbin. 19 days to escape the Kerbin system, 31 days to apoapsis in its new orbit around the sun. I have other flights active at the moment, so time-warping will be handled judiciously. For interested readers, I’ll post updates about this to the Kerbal Space Program thread.
I’m not planning on working hard to figure out how to use any of the inner planets to help with this; that’s probably just way too complicated for me.
Damn you! You have just opened a new obsession for me! Between this and Minecraft…
In technical literature these are referred to as “swing-by maneuvers” or “gravity assist” rather than “slingshot”, which seems to be a term concocted by pop science journalists as being more visually stimulating but ultimately misleading. Here is a brief introduction to it from JPL. (I had previously written a Mailbag article on the topic a number of years ago but it seems to have disappeared in the interim.) As you note, the swing-by transfers momentum between the spacecraft and the body it swings by to effect a change in course. Although it is commonly thought to be done to “increase speed” the ultimate change in speed cannot be any more than twice the orbital speed of the planet and is typically far less unless additional impulse is applied.
As the spacecraft has to maintain sufficient momentum to leave the body’s sphere of influence, it is not practical (and except for very artificially constructed cases, not possible) to completely null out the speed of the speed such that it is then motionless with respect to the Sun after the pass and is drawn straight in. It is, however, possible to reorient the trajectory of an object with a sufficiently aggressive swing-by such that the final trajectory is pointing straight down at the Sun, but you’ll never get this with flybys of Venus or Mercury. It is possible to fly outward and around a large gas giant (Jupiter, Saturn, or both) in such a way to end up with zero tangential velocity and a radial vector pointing down into the Sun, but such trajectories are unstable–just a little bit of extra momentum either way and it either falls into a highly elliptical orbit that goes past the Sun and on a long, lonely orbit with an apsis out to or beyond the Oort cloud.
Stranger
The Ulysses Spacecraft used a gravity assist from Jupiter to put it in a solar orbit what passed over the North and South poles of the Sun. However, it’s closest approach to the Sun was no closer than the Earth.
Don’t get too upset, putting your MIL into a permanent orbit could be a good thing too. Your SO could wave to her when she comes in sight once a year or so.
I hope this isn’t a hijack, but I’m thinking this is along the same lines. If the Earth were to suddenly disappear (because the Daleks have stolen it to construct a universe destroying machine), then based on this thread, I’m guessing the moon would go into orbit around the sun? Regardless of where it was in its orbit around Earth at the time of Earth’s disappearance? Or am I misunderstanding something?
Undoubtedly someone more knowledgeable will chime in, but let me have a try first.
Yes it will be in orbit around the sun. No it won’t be the same motion it had – with the earth in place the moon’s motion is something of an epicycloid. (Think of flower petals.) The new orbit won’t be the same as the earth’s current orbit either (although it might be close). The exact orbit would depend on its velocity at the moment the earth disappeared. At a full moon for example the moons velocity with respect to the sun is slightly fast. At new moon it is slightly slower. And of course at the quarter moons it has a component of velocity either towards or away from the sun. Also note that the plane of the moon’s orbit is not the same as the ecliptic and it will have a component of velocity in the z direction as well.
All of this would result in the moon heading on an elliptic orbit with a greater eccentricity than earth’s current orbit and not in the same plane as the earth’s current orbit.
How different exactly would it be? I will leave that calculation to the ones that know. My guess is not that great a difference.
If you’re interested, Simplerockets, available on iphone/android/pc is a simpler, 2d version of Kerbal Space Program which is much more accessible and helps with an intuitive understanding of orbital mechanics without all of the fluff of KSP.
You’re welcome. 
Yes, it will. The Moon has an orbital speed around Earth of about 1 kilometer per second, and the Solar escape velocity if you’re starting from Earth’s orbit is about 42 km/s, so the Moon wouldn’t have the velocity to escape. For comparison, the Earth’s orbital speed is about 30 km/s; if we were going even 12 km/s faster, we’d escape from the Sun’s grasp entirely (unless we hit something or were redirected by another planet’s gravity on the way out).
The exact shape of the Moon’s orbit around the Sun would depend on where the Moon was in its orbit around Earth at the time of said Dalek thievery. The moon would retain the 30 km/s velocity around the Sun that it currently has from being part of the Earth/Moon system, and that would change by as much as 1 km/s either faster or slower from the Moon’s current orbital velocity around the Earth*. So yes, if the Earth disappeared, the Moon would continue orbiting the Sun at some average speed between 29 and 31 km/s. If the Moon were in just the right place, the orbit might even be less eccentric (that is, more nearly circular) than the Earth’s orbit is now.
*For simplicity, I’m ignoring the fact that the Moon’s orbit isn’t exactly parallel to the Earth’s orbit around the Sun. It’s only a few degrees off, and that’s too small an effect for the sort of back-of-the-napkin math we’re doing here.
The core question has already been answered, but another way to think of it is that the Moon is already orbiting the Sun. The Earth introduces only a minor perturbation to an otherwise nearly smooth orbit. The Moon’s orbit around the Sun is all places convex. The biggest change the Moon would notice is that its tidal bulge would gradually fade away, no doubt with some accompanying quakes.
Thanks for the responses.
Stupid Daleks! They’re really annoying sometimes.
I second the recommendation for SimpleRockets; it’s an awesome program to play with and it really helps you get a handle for questions like these without the added complications of the third dimension (and for questions like this, the third dimension is just an extra complication that makes the game, and the corresponding visualization, harder). It’s really interesting to play around with your rocket and see what happens to your orbit as you apply thrust in different directions. And since it works on iOS and Android as well as Windows, you can play around with it situations where you don’t have your PC handy.
Once you get a feel for things in SimpleRockets, THEN you should graduate to the much more intricate, complicated, and detailed universe of Kerbal Space Program. 
ETA: I’ve tried to launch stuff into the sun in SimpleRockets many times, incidentally; it’s amazing how much fuel it takes, compared to what you would think.
This question came up on a Twitter account I subscribe to, and an idea occurred to me: instead of trying to cancel the Earth’s velocity around the sun, wouldn’t it take a lot less delta-V to put something into an elliptical orbit around the sun, then cancel its transverse velocity at aphelion, and let it drop into the sun? I don’t have the knowledge of orbital mechanics to calculate the optimum trajectory but since it takes less delta-V to escape the sun entirely than to cancel any (nearly) circular orbit around it, I should think this maneuver would be much more economical.
Congratulations, you just rediscovered the bi-elliptic transfer! Yes, it is much more efficient than a direct burn. I’ll leave the math as an exercise to the reader 
But the object is in a circular (… earth like) orbit around the sun until delta-V’d …
But sure, sling shot the moon or something, which is free energy so you can do that any any part of the orbit
and yes restrict burns to apogee or perigee for least cost … not much cost saving when in a nearly circular orbit, but there is some…
If I did the math right, it looks like you want your second burn to be at “infinity”. The speed at periapsis will be sqrt(2) times that of a circular orbit, so the delta V requirements are only 41% of what they’d normally be (the second burn will be arbitrarily small). Of course this maneuver will take, well, forever to execute, but obviously you could do a more limited version with a bit less savings.
Putting your second burn at infinite distance minimizes the propellant needed for the second burn, but it doesn’t minimize the total propellant needed.