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Think of it this way: If it was flying by slowly, it would get caught by the sun’s gravity and end up in a circular orbit around the sun. If the orbit was perf
by Uehreka 7y ago
Think of it this way:
If it was flying by slowly, it would get caught by the sun’s gravity and end up in a circular orbit around the sun. If the orbit was perfectly circular, it would have an eccentricity of 0.
If it was flying by a little bit faster, it might barely get caught by the sun and end up with an orbit like Halley’s comet where it comes close to the sun once every hundred years and then flies off to the far reaches of the solar system before coming back again. It would have an eccentricity of 0.9-ish.
If it was flying a little bit faster, the sun’s gravity would try to catch it, but it would fail! The object would “swoop” around the sun then go flying back the direction it came, never to return again. It would have an eccentricity of about 1 or a little more (Less than 1 means it’s in orbit, 1 or more means it’s not coming back).
If it was flying REALLY REALLY fast, the sun’s gravity would try to pull it in, but this time the object has other plans. It barely even changes course and rockets through the solar system in an almost perfectly straight line. Its eccentricity would be something like 89 (there’s no upper limit, although at a certain point the object would have to travel at close to lightspeed to acheive certain really high eccentricities if flying close to the inner solar system).
This object (the real-life one we’re talking about) is going faster than the “swoop” object but slower than the “other plans” object. Its path is being bent somewhat by the sun’s gravity, but it is going to leave an never come back. So its eccentricity is 3.
One last thing: eccentricity is PURELY a math thing that describes circles, ellipses and curves. It’s just that when you’re talking about orbital mechanics, it gets interwoven with other aspects of an object’s orbit, like its velocity and its altitude at the closest point in the orbit.
- _kst_ 7y ago"If it was flying by slowly, it would get caught by the sun’s gravity and end up in a circular orbit around the sun. If the orbit was perfectly circular, it would have an eccentricity of 0." An interstellar object cannot be flying by slowly enough to enter a circular orbit. If it were, it would already have been in a circular orbit. Trajectories can be extrapolated both into the future and into the past. It has enough momentum to escape the Solar System if and only if it came from outside the Solar System. This assumes the only interaction is the gravitational pull between the Sun and the object. A close approach to Jupiter, for example, might slow an incoming object into an elliptical orbit, or speed a Solar System object into a hyperbolic escape trajectory.
- davidmurdoch 7y agoIt it not at all possible for an interstellar object to be captured into a stable solar orbit?
- deleted 7y ago[deleted]
- lojack 7y agoCaptured into a stable orbit, sure, but it will never have an eccentricity of 0 (circular). The limit can approach zero, but will never be equal to it. Even the planets don’t have a circular orbit FWIW.
- _kst_ 7y agoAn interstellar object cannot be captured into a stable orbit unless some other force (such as Jupiter's gravity) acts on it. Orbits are time-symmetric, so an interstellar object being captured into a stable elliptical orbit would be equivalent to an object in a stable elliptical orbit escaping the Solar System (again, ignoring other influences).
- tempestn 7y agoIs that true of comets as well? It seems conceivable to me that the sun could spin it around such that the comet's off-gassing could reduce its speed and allow it to be captured. Obviously unlikely, and highly dependent on conditions (in particular incoming velocity) but possible.
- lojack 7y agoIf you're ignoring all other influences then there is no such thing as interstellar objects. By definition, every object in a two body system are gravationally bound to each other.
- ravar 7y ago
- maroonblazer 7y agoThank you, great explanation. Do we have any sense of how large this object is? Or is that a silly question that only non-astronomers like me ask?
- radeklew 7y agoAccording to this tweet[0] the object's coma[1], the cloud of ice and dust surrounding the object, makes it too difficult to measure. [0]https://twitter.com/AscendingNode/status/1171913641647493120 https://twitter.com/AscendingNode/status/1171913641647493120 [1]https://en.wikipedia.org/wiki/Coma_(cometary) https://en.wikipedia.org/wiki/Coma_(cometary)
- drevil-v2 7y agoThank you. I learnt something very cool today!
- noneeeed 7y agoThank you, that was a great explanation with lovely examples.