Talk:2011 CQ1

Nosediving in

The article currently (2015 Jan 28) carries this precept: "With a relative velocity of only 9.7 km/s,[2] had the asteroid passed less than 0.5 Earth radii from Earth's surface, it would have fallen as a brilliant fireball." I don't believe this is correct. Here is my reasoning:

The asteroid passed at a distance of 5480 km above the surface. If it passed 0.5 earth radii above the surface, that would be 3184 km. Either altitude is way, way, way higher than the atmosphere. There is zero air friction that high. So this is pretty much a standard two-body problem, where one body is much more massive than the other.

A small body approaching the earth from some far distance will have a certain kinetic and a certain potential energy. As it gets closer, potential energy is converted to kinetic energy, ie. it accellerates toward the earth due to gravitational attraction. After it passes the earth, kinetic energy is converted back into potential energy, ie. it decellerates. If it doesn't actually hit the earth, nor pass thru the atmosphere (encountering friction), then it must depart back into the far distance with the same kinetic energy that it had before.

At each point on its path, if the smaller object's velocity relative to the larger object is less than escape velocity at that point, then its path is an ellipse - it's in orbit. Otherwise - if it's going faster than or equal to escape velocity - its path is a hyperbola or parabola, respectively, and it's not in orbit. These curves are modified by other nearby bodies (2011 CQ1 is actually in orbit around the Sun, on a very large ellipse, which looks like a line, modified to a hyperbola as it passes the earth, if we're sufficiently close.)

It really doesn't matter whether the flyby altitude is a million km or just above the atmosphere, these are the curves. It approacheth, it passeth, then it departeth - unless that path intersecteth with the earth - which clearly doesn't happen at 3186 km above the surface.

Escape velocity at 3186 km above the earth's surface is sqrt(2GM/r) = sqrt( 2 x 6.67×10^−11 m^3/kg s^2 x 5.97219×10^24 kg / (6367km + 3186 km) * 1000 m/km) = 9132 m/sec = 9.1 km/sec.

This means that an object drifting in very, very slowly will accellerate (fall) until it's traveling 9.1 km/sec as it reaches 3186 km above the earth's surface.

This object didn't get that close, and it was going faster than that, ie. it was on a hyperbolic path, not in orbit, and even had it passed within 200 km of the earth's surface (that's still well above any significant atmospheric drag) it would have zipped right past and zoomed off into space again.

It occurs to me the intent of the article's statement was that 2011 CQ1 was going pretty slow, just a little above escape velocity. Friendly Person (talk) 21:58, 14 March 2015 (UTC)

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