Voyager

It isn't significant.

The total mass of the entire kuiper belt is less than a tenth of the mass of the Earth. The Oort cloud is about 5 Earth masses.

These two things together are about 1/50,000th of the Solar System's mass. This mass don't matter.

The sun has 99.86% of the Solar System's mass. And most of the rest of the mass is in the gas giants.

Did they even do a calculation for escape velocity? Or did they just shoot them out there at the best speed they could get and figure it would be enough.

I assume they did a calculation. Finding escape velocity knowing the mass of the solar system and the position of the craft is a pretty easy calculation.

In fact I given the fact they plotted an orbit that passed by planets at the right time (which is a far more difficult calculation) I am certain . An average high school student can calculate an escape velocity.

Really, they shot them out there to take fly-by images of the gas giants and their moons. Anything after that is gravy. Shooting them out there with enough velocity to escape the sun's gravity well and reach Saturn and Jupiter was enough to assure they would escape the solar system.

You obviously don't know what you are talking about. It is easy to find an orbit that would reach Saturn and Jupiter that wouldn't "escape" the solar system.

"Escaping" the solar system is a very strange term. Would you care to say what you meant by that phrase? By your definition, does Halley's comet "escape" the solar system?

The phrase "escape the Sun's gravity well" is pure nonsense from any scientific standpoint. The sun's gravity will always be acting on Voyager. There is no "gravity well" for it to "escape". And if there was a "gravity" that didn't include Jupiter and Saturn, why do they keep orbiting the sun?

If you mean "escape velocity" (which is a scientific term but doesn't mean what you apparently think it does), it is certainly possible for a craft to reach Jupiter and Saturn without ever having escape velocity relative to the Solar System.

Just like to argue, don't you, quite apart from your obsessive, hysterical hatred of me. Bitch.

A gravity well or gravitational well is a conceptual model of the gravitational field surrounding a body in space. The more massive the body the deeper and more extensive the gravity well associated with it. The Sun has a far-reaching and deep gravity well. Asteroids and small moons have much shallower gravity wells. Anything on the surface of a planet or moon is considered to be at the bottom of that celestial body's gravity well. Entering space from the surface of a planet or moon means climbing out of the gravity well. The deeper a planet or moon's gravity well is, the more energy it takes to achieve escape velocity.



Plot of a two-dimensional slice of the gravitational potential in and around a uniformly dense, spherically symmetric body.

Don't flatter yourself Setanta.

Looking up science terms on Wikipedia does not replace actually studying science. (And this particular wikipedia article is more poorly written than most of them).

I didn't say that "gravity well" isn't a scientific term. It is a scientific term. I said the idea that the Voyager spacecraft will "escape the gravity well" is nonsense and shows a misunderstanding of the Physics involved.

The voyager will never be outside of the influence of gravity of the Sun. The term "escape velocity" doesn't imply that it will.

Knowing scientific terms to throw around is one thing. Knowing what they actually mean and understanding the science behind them is another.

The term "escape velocity" is very often misunderstood. If a space craft has escape velocity from the sun it is still in orbit. This means it will still be under the influence of gravity from the Sun forever.

Escape velocity simply means it is in a different type of orbit. The property of this different type of orbit is that it will no longer return to its current position.

That's the way I remember it. The original mission was to explore the outer planets (photo's). Then they just gave them enough of a boost from their final planet to "fling" them out as far as they could, just to see what else they could get. I doubt the needed to calculate any type of escape velocity. They probably just gave them the maximum boost they could and let the cards play.

And it worked pretty well. A little bit like the Opportunity and Spirit Rovers on Mars who's original mission was only a couple of years long, but they got a LOT more out of them in the end.

Of course they planned this out. The math is simple enough and NASA scientists are there to calculate trajectories. It's not rocket science... actually I suppose it is rocket science, but it is pretty easy. They know exactly where it will be (assuming it doesn't collide with something unexpected or get picked up by curious aliens) in 50,000 years from now.

This incredible little machine has power up to 2025 and they have experiments running now. Why wouldn't they have taken the 15 minutes on the back of an envelope to figure out where it could get to before the power runs out.

The exact math for Voyager 1 is done here

http://en.wikipedia.org/wiki/Specific_orbital_energy

Interestingly enough, while Voyager is on a hyperbolic trajectory with respect to the Sun. It is on an elliptical trajectory with respect to the Milky Way Galaxy.

Scientists make these calculations all the time.

I assume that NASA's intention was to fly by the Gas Giants, has as been quite rightly pointed out anything more is 'gravy'. However the mission is continuing - and so is the pull of the Solar System all the way to infinity unless a neighbouring sun takes a hand.
All I was trying to point out - and I accept that the combined mass of Kuiper and Oort should be quite small ( although the theorectical radius of the Ort sphere is vast ) after all there are no detected wobbles of the Sun ( unlike the Moon/Earth planetary system. BUT if they are symetrical about the Sun there would not be. In fact if the Oort 'shell' was a uniform distribution of mass it would exert zero pull on the Solar System. The significance of these unknown masses ( and a way of measuring them, ) once you have passed them, would be in the deviation of the retardation of the probe ( as measured by doppler effect ) from the calculated value at that distannce.

Nope.

When you calculate the gravitational field of a uniformly distributed shell the mass acts as if all the mass is at the center. Some of the particles of the field will be much closer to the Voyager, but others of the particles will be much further away. It turns out that it balances out and, for the sake of gravity, the mass of the shell acts as if it were in the center of the Sun.

This is a problem that we do as first year Physics students. You consider the gravitational field on an object (in this case the Voyager space craft) from each particle of a uniformly distributed shell. It is fairly easy to set up the math as a Integral (first year calculus). And the result is that the mass of a shell acts exactly the same in every circumstance as a point mass at the center.

This is true anytime you are outside the surface of the shell. Interestingly enough if you are inside the surface of the shell the gravity from each particle cancels out in another way. In this case the sum of gravity is zero. The gravity of a uniformly distributed sphere has zero effect on objects inside it.

This is just the way the math works out.

Of course since the mass we are talking about here is so small. This effect is infintessimally small.

So no, the deviation on the trajectory of the craft is basically zero. There will probably be very small local effects since the distribution is not perfectly uniform. But likely the effect on the overall trajectory would be so small it couldn't be measured.

I found this summary of the math used to calculate gravitational fields, including for uniform shells, for anyone who is interested.

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm