How come the Voyager probe is leaving the Solar system?

Its velocity is about 14 km per second. Solar escape velocity is 525 km/s.

Escape velocity decreases with distance from the object.

The high escape velocity is from the near vicinity of the Sun. From Earth’s orbit it’s only 42 km/s. From further out it’s even less.

To rephrase what’s already been said, the further you are from something, the lower the escape velocity is. Your quoted escape velocity is probably from the Sun’s surface, whereever your source determined the exact surface to be. Right now, I figure the Sun’s escape velocity at Voyager 1 to be just 3.7 km/s, and at Voyager 2 to be just 4.2 km/s. Sources:

Does that mean it was moving at 42 km/s when it left Earth?

Hardly. The Voyagers were launched on Titan IIIE launch vehicles employing the Centaur-D upper stage. I don’t know the exact speeds at interplanetary injection, but they were probably somewhere around 32 km/s relative to a heliostationary reference frame. The Voyagers both made substantial momentum changes by gravitational swing-by maneuvers of Jupiter and Saturn (and for Voyager 2, Uranus and Neptune, although this actually slowed the radial velocity of the probe). Escape speed is merely the minimum speed required in order to achieve gravitational escape (i.e. a non-closed orbit in parabolic or hyperbolic form) at a given distance, not the speed that you have to be going at any point, provided you have some way to gain momentum.


The probes got much of their escape energy “slingshotting” from their encounters with Jupiter and other planets, as depicted in animations like this one.

Nothing is “for free,” so I suppose Jupiter is a fraction of a micrometer closer to the Sun now because of this.

Huh.:confused:Please explain.

I know Apollo spacecraft did not achieve escape velocity, but merely the TLI burn put them in a high earth orbit where the moon’s gravity would catch them, and I think the reverse was true on the return, but the above is completely over my head.

When I look at the paths, I get the impression that they were mostly trailing flybys, meaning the Voyagers would have slowed the planets, thus moving their orbits nanometers further from the sun. Except, of course, in the case of Saturn for Voyager I and Neptune for Voyager II, where the deflection probably mostly affected the inclination. What puzzles me, though, is how they got Voyager Six going so damn fast …

A slingshot is when a probe dives into a planet’s gravity well on an open path and emerges on the other side at a higher speed. Imagine a person skating up to another person who grabs their outstretched hand and swings them around, flinging the skater out at a higher speed, the principle is the same.

The stationary person who flings the skater will probably stumble a little, because they have delivered some momentum to the skater and will have to account for it. The probe similarly steals some momentum from the planet, but since the planet is so vastly larger, the change in its motion is not measurable.

One way to become even more confused about all this is to read about the curious relationship between Janus and Epimetheus, which illustrates gravitational momentum exchange on an observable scale.

Slowing the planet won’t change it’s position at that part of its orbit. But since it’s going slower, it will be a little closer to the Sun on the other side of its orbit (the far side of the Sun from where the slingshot took place). On average, it’s distance from the Sun will have decreased.

Even though it will, on average, be moving a little faster, its total energy will have decreased.

Errr. It will have less energy but be faster. What?:eek:

Things in orbit have two types of energy:

[li]Gravitational potential energy, which increases with distance from the central mass.[/li][li]Kinetic energy, that increases with speed.[/li][/ol]
So a low-speed object in high orbit will have a lower kinetic energy than a high-speed object in a low orbit, but the increased gravitational potential energy can more than compensate for a higher total energy.

you can’t look at just the speed.

At the point where the slingshot occurred, it has the same potential energy, and less kinetic energy, so we know its total energy is lower.

On the Far side of the Sun, it will be a little closer to the Sun, so it will have less potential energy. It will be moving a little faster than it would have been, so its kinetic energy is a little larger, but not enough to offset the lower potential energy. So its total energy is less, even though it’s moving faster at that point in its orbit.

Anything moving away from the Sun will find that its speed is decreasing, since it’s got a gravitational force pulling it inwards. Alternately, you can say that its gravitational potential energy is increasing at the expense of kinetic energy decreasing. This is fundamentally no different than noting that when you throw a ball in the air, it slows down while it’s moving upward (just on a much larger scale).

If an object starts off at (or above) escape speed at any given location, and is acted on by nothing other than gravity, it will always remain at (or above) the escape speed for its current location. As the object’s speed decreases, so too does the speed it needs.

“Escape velocity” is a slight misnomer. Subtracting all other forces, if you fire off a rocket from a planet such that its fuel runs out exactly at the point that the rocket reaches escape velocity, it will not ultimately escape. As it travels outward, the planet’s gravity will exert a diminishing force on the rocket, decelerating it so that it constantly remains at the local escape velocity. Assuming no other interference (a very lonely planet), the rocket will eventually come to a stop, but space is never flat, there will be the tiniest whisper of gravitational effect from the planet, which will gradually pull the rocket back in until it smashes back into the planet’s surface – at the speed of escape velocity. To actually escape, you have to exceed escape velocity.

Assuming no other interference it will come to a stop at infinity, i.e. never. That’s escape enough for me.

“Escape velocity” is, however, a misnomer, in that the correct term is “escape speed”. The direction doesn’t actually matter, so long as you don’t run into anything first.

A foot or two short of infinity works for me. :slight_smile:

This is not correct. At escape speed (which, as Chronos notes, is independent of direction and therefore a scalar quantity) in an isolateed two body system (the spacecraft and the massive body it is escaping from) the spacecraft will “escape”, i.e. never return once it achieves escape speed. It will still be under the influence of the massive body and continually decelerating, but the trajectory, which is a segment of an open conic section, i.e. a parabola or hyperbola, will never return. If an object does not achieve escape speed, it will fall into a closed orbit (an ellipse). The only way it would “[smash] back into the planet’s surface” is if lateral velocity (that with a vector pointing perpendicular to the diection of the planet) weree so small at it intersects with the planet (including the atmosphere). This need to reduce the lateral velocity is why it is actually very difficult to send a probe to the Sun versus going outward. From an orbital energy energy standpoint, the Sun is essentially the most difficult place for us to reach.

You are essentially correct. All real world systems are multibody systems, e.g. the Earth-Moon-Sun (and the other planets, especially Jupiter, which is often called the “gravitational bully” of our solar system, but not a consideration for the short duration of the Apollo missions) and so it is entirely possible to go from the sphere of influence (SOI) of one body to another without reaching escape speed, or with careful maneuvering through libration points or by transfering momentum from a passing body via gravitational swing-by maneuver, escape the system entirely. (The sphere of influence is the region around a body–and strictly speaking, not exactly spherical–at which one body dominates the trajectory of an object gravitationally for first order calculations.)

In the case of Apollo, the CSM went from the SOI of the Earth to that of Luna while never achieving Earth escape speed and in fact from the standpoint of an Earthbound observer, never actually leaving Earth orbit, as it entered essentially the same orbit as the Moon with an additonal peturbance due to dancing around the Moon. Of course, we are all (Earth and Moon) in orbit of the Sun, and our solar system (and the Voyagers) are orbiting the center of mass of the Milky Way galaxy, which even the Voyagers will never escape.