When a space agency (NASA, ESA or similar) sends a probe to the outer solar
system (and beyond) it doesn't simply send the probe directly to its
destination; instead, the probe takes a trip via some of the inner planets to
help it on its way. The manoeuvre involved is sometimes called a sling-shot,
gravity assist; the probe dives in close to a planet and
emerges going considerably faster. This is why space vehicles headed for the
outer solar system commonly spend a year or three dashing round the inner solar
system between leaving Earth and setting out on their long journey to the outer
reaches. There are two processes at work here, each of which plays its part:
one is that the probe gets far more effective use of its engines, the other is
that it effectively steals a little of the planet's energy. I want to explain
how these processes work.
When a space vehicle passes close to a planet, it falls towards the planet
– which causes it to speed up considerably – narrowly misses and
then flies away from the planet. If it hasn't used its engines in the course of
that, its speed relative to the planet – once it's far enough from it to
no longer be appreciably affected by the planet's gravity – is the same as
it was before it came under the influence of the planet's gravity. However,
that's its speed relative to the planet – which is moving,
relative to the rest frame that's interesting to the vehicle's mission –
which is typically a non-rotating frame of reference with respect to which the
Sun is stationary. (Consequently, you can't do the
stealing energy part of the trick with the Sun itself – although
the engine efficiency part does work.) So, in passing the planet, the vehicle's
velocity relative to the planet is rotated (that is, it changes
direction but not magnitude). This is what enables the vehicle to steal some
energy off the planet.
Suppose the planet has velocity V (ignoring the fact that, in the few days it takes the vehicle to pass it, this changes because the planet's trajectory is curved around the Sun – V will actually change direction slightly, but this doesn't change the rough form of the process, it would just make the analysis more complicated) and the vehicle has velocity v before coming near the planet. Its velocity relative to the planet is then v−V; for some rotation R, its velocity relative to the planet shall be R(v−V) after the encounter, hence its final velocity is V+R(v−V). The rotation is, due to the details of orbital mechanics, through an angle which is necessarily less than a half turn; however, for simplicitly, let's first see what we'd get if a half turn were possible: V+R(v−V) would be (since the R for a half turn is simply R(x) = −x) V−(v−V) = V+(V−v) = 2.V −v which, if v is in the opposite direction to V, would simply be 2.V bigger than v, but in the reversed direction.
Now, you can't actually do a half turn, but you can get pretty close. Let v's component parallel to V be −u – that is, u is a vector in the same direction as V and v+u is at right angles to V. Then the vehicle's velocity relative to the planet, v−V, is (v+u)−(u+V) – in which v+u is perpendicular to V and u+V is parallel to it. As long as v+u isn't zero (i.e. the vehicle has some movement cross-wise to the planet's trajectory), if you get your orbital mechanics right, you can arrange (by having your point of closest approach just behind the planet, as it moves along its orbit) for the final velocity relative to the planet to simply reverse the component parallel to the planet's velocity, for a final velocity of (v+u) +(u+V) relative to the planet. That gives the vehicle a velocity (relative to the solar system) of (v+u) +(u+V) +V = (v+u) +(u+2.V), so its velocity relative to the solar system has un-changed component across the planet's path, but the component parallel to its path has been reversed and had 2.V added to it.
The inner planets move at higher speeds than the outer ones – Mercury's speed is almost 50 km/s, Venus's is just over 35 km/s, Earth's is almost 30 km/s – but even Jupiter's (at 13.1 km/s) is nothing to turn up your nose at, especially as it gets doubled. So that 2.V bonus is a welcome gain in the vehicle's energy: but, on top of that, it also gets a benefit from the fact that, in the course of passing the planet, it's transiently going much faster – the heavier the planet, the faster it goes in passing. Of course, it loses most of that speed in the process of flying away from the planet again, but the simple fact of moving faster makes its engines more effective !
The engine of a space vehicle works by throwing something – called
reaction mass – out of the back of the vehicle at (usually) high
speed (relative to the vehicle). Whether it does that using a chemical rocket
or electrically, as in an ion drive, is incidental; what matters is that (in the
vehicle's initial frame of reference) the engine gives the reaction mass some
momentum in one direction, thereby gaining an equal and opposite amount of
momentum for the vehicle. The crucial detail here is that the vehicle gains,
for any given amount of reaction mass expended, a definite amount of momentum
and, hence, a definite change in its velocity. The kinetic energy of the
vehicle is proportional to the square of its velocity; so, if the vehicle
expends its fuel at a time when it's moving very fast, a modest change in its
veolcity can produce a much larger change in its energy: if it's going at
initial speed w and adds y to its speed, the square of its speed goes from w.w
to (w+y).(w+y) = w.w +2.w.y +y.y, so it increases by (y+2.w).y, so the bigger we
can make w the bigger is the gain in energy from a given expenditure of the
engine's fuel and reserves of reaction mass.
Let x be the magnitude of v−V, i.e. the vehicle's initial speed relative to the planet: so its initial kinetic energy per unit mass was x.x/2, in the planet's frame of reference. Let p be the velocity for which: the gravitational potential (due to the planet) at the point of closest approach is −p.p/2; so the vehicle's total kinetic energy at closest approach is (x.x +p.p)/2 and the square of its speed, w.w, is thus equal to x.x +p.p. Its engines increase its speed by y so its kinetic energy per unit mass increases by y.(y+2.w)/2, to (y.(y+2.w) +x.x +p.p)/2. In flying away from the planet again it loses p.p/2 from that, leaving it finally with (y.(y +2.w) +x.x)/2, where w = √(x.x +p.p). Its final speed is thus s = √(y.(y+2.w) +x.x); if it had not encountered the planet, but had simply run its engines to the same extent, this would be x+y = √(y.(y+2.x) +x.x).
Now, given that y is typically going to be tiny compared to x (and, as we'll see, p is large), we can approximate this reasonably well: when y is zero, s is x and ds/dy is w/x = √(1 +p.p/x.x). The engine's effectiveness is, in practice, increased by this factor; if p is √3 ≈ 1.732 times as big as x it's a factor of two. For a swing by Venus, p can get up to about 10 km/s, for Earth it can get up to about 11 km/s; but Jupiter can get it up to almost 60 km/s while Saturn, Uranus and Neptune can manage 36, 21 and 24, respectively. (Those numbers are based on the potential at the apparent surface: in practice, the vehicle has to keep outside the atmosphere, so the best it can get is slightly less; but fairly close, all the same.) So, while the inner planets do well on the relative motion trick, the outer planets are better at making the vehicle's engine more effective.
The above analysis ignored the effect of the Sun's gravity; in the course of passing a planet, our vehicle isn't much influenced by the Sun, but the Sun's gravity has significant effects during the vast distances between planets. Although Venus' gravitational potential near its surface is only worth about 10 km/s as p, the difference in solar gravitational potential between Earth's orbit (but clear of Earth's gravity well) and a close fly-by of Venus is worth 28 km/s as p. While Mercury's own gravity well is only worth 4 and a quarter km/s, its proximity to the Sun gives a fly-by of Mercury an effective p of 53 km/s.
Conversely, going outwards from the Sun, our space vehicle is slowed down by solar gravitation. Although Jupiter is so massive that the gravitational potential at its apparent surface is actually lower (that is, more negative) than at Earth's surface (i.e. climbing out of Earth's gravity well and then fighting the Sun's gravity all the way out from Earth's orbit to Jupiter's takes less energy than you recover by dropping down Jupiter's gravity well), you can't simply exploit that – because you have to get out to Jupiter's orbit before you can exploit Jupiter's gravity well. You need most of 40 km/s's worth of kinetic energy to get from Earth's orbit out to Jupiter's, although you can exploit the orbital speed you had when setting off from Earth to cut that down to less than 25 km/s. Fortunately, motion relative to Venus and (once you've gone via Venus at least once) Earth provides ample opportunity to pick up enough energy to reach Jupiter and exploit its mighty gravity well.
A conjecture (alluded
to here, but I can find no other source for it) attributed to Stanislav Ulam
Ulam's conjecture reveal at least two others, but not this
one) asserts that:
Given sufficient time, classical chaotic motion will spontaneously connect two points in phase space with arbitrary precision. In 1956, American mathematician Stanislaw Ulam conjectured that owing to this phase space-filling aspect of chaotic trajectories, a minimal series of energy expenditures would suffice to transfer a body from one point to another much more rapidly than by spontaneous motion.
The idea is that it should take very little energy to get an object, once clear of Earth, to anywhere in the solar system – or beyond – by exploiting the kinds of techniques described above. The trick is to use a little energy extremely carefully, so as to exploit the dynamics of the rest of the system, effectively persuading it to help you out.Written by Eddy.