Onwards to Uranus and Neptune! Options

[SFJCody]

As soon as MESSENGER gets to Mercury, the most poorly explored planets in the solar system will be Uranus and Neptune. Could this lead to a revival of interest in the ice giants and their retinue, in the same way that the existence of New Horizons is perhaps partly due to the Pluto stamp*?







*via Pluto Fast Flyby and later Pluto Kuiper Express

[dvandorn]

My understanding has always been that gravity assists work by robbing a body of a bit of its rotational velocity by flying along with its rotating gravity field. The Sun rotates -- why ought this process not work with the Sun?

-the other Doug

[ermar]

Actually, this is not how gravity assists work - rather, they take advantage of an orbiting body's orbital motion relative to background space. Say a planet is orbiting with a speed V relative to the sun and a spacecraft approaches it (from the planet's point of view) from one side with a speed v. From the planet's perspective, the spacecraft ought to have the same relative magnitude of speed leaving it as approaching it, so the spacecraft leaves with speed v too. From, say, the Sun's point of view, though, the spacecraft approached the planet (which was moving at speed V) with some other speed z; by the time the spacecraft has left the vicinity of the planet, the planet has "dragged it along," and some significant fraction of the planet's orbital speed V is added to the original spacecraft velocity. (This can also be used to decelerate the spacecraft, too, by setting up the initial encounter differently.) The bottom line is that gravity assists depend on the mass and orbital velocity of the "assisting" body alone (robbing it of orbital, not rotational momentum), and have nothing to do with its rotation rate.

[JRehling]

Little physics tidbit: Imagine a spacecraft orbiting the Sun in an elliptical orbit, and imagine that it (for the time being) will not pass closely enough to any planet for them to have a significant pull on it.

The sum of the gravitational potential energy of the craft and its kinetic energy is utterly constant. (This is true of circular, parabolic, and hyperbolic trajectories as well.) What it loses in kinetic energy from perihelion to aphelion (or at any other time) is exactly equal to what it gains in gravitational potential energy. And this will be true forever.

So imagine a spacecraft that has left Earth, and that proceeds in to some closer distance to the Sun, then passes on the other side back to (and then past) the orbit of the Earth. (Again, assume that Venus and Mercury don't get in the way.) Between launch (once it gets sufficiently outside the Earth's sphere of influence) and perihelion, what it gains in kinetic energy will be lost, exactly, in gravitational potential energy. Between perihelion and crossing, again, the Earth's orbit, what it lost in kinetic energy will be gained, exactly, in gravitational potential energy. And moreover, because gravitational potential energy is determined by the distance from the Sun, the kinetic energy of the craft when it crosses the Earth's orbit on the way out will be EXACTLY what it was when it left the Earth in the first place. You gain a net sum of nothing. You may as well have just launched in the outward direction in the first place (at some different launch window).