"Based on their observations, and those of other astronomers who began tracking the comet's highly elongated orbit, it was calculated that Swift-Tuttle would make its next appearance during the 1980s. They were close. Japanese astronomer Tsuruhiko Kiuchi rediscovered the comet in 1992.
Aside from its unusual orbit, Swift-Tuttle is also significant as the host body of the Perseids meteor shower, one of the most prominent in the northern sky.
Oh, and there's one more thing.
Comets come and go, literally, but Swift-Tuttle's orbit is of particular interest to us earthlings since astronomers calculate that it is very likely to strike either the Earth or the moon on its next pass. They've even zeroed in on a date: Aug. 14, 2126."
http://www.wired.com/science/discoveries/news/2007/07/dayintech_0716
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http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=109p;orb=1;cov=0;log=0;cad=0#orb
It works out if the meteor is deflected towards the earth such that the angle between it’s velocity vector and the earth’s orbital velocity vector is 24 degrees (or 156 degrees, depending on how you measure it), close to a head on collision. I was assuming that the meteor’s velocity vector is almost perpendicular to earth’s and was ignoring the added kinetic energy that the meteor gets from falling towards earth. The earth will deflect the meteor off its unperturbed orbit, so there is enough energy available to get up to 70 km/sec.
V = 42.1219*sqrt( 1/r - 1/(2*a)) km/sec
regardless of eccentricity.
r = distance from the sun in AU,
a = object's semimajor axis in AU
If you set a = 1.0 and r = 1.0, you get 29.8 km/sec, not surprisingly, earth's orbital velocity. If you set a = Inf and r = 1.0, you get 42.1219 km/sec, escape velocity from the sun at the distance of earth's orbit.