[meteorite-list] Maximum theoretical Earth impact velocity

From: Chris Peterson <clp_at_meteoritecentral.com>
Date: Mon, 14 Apr 2014 14:16:05 -0600
Message-ID: <534C4205.5000908_at_alumni.caltech.edu>

Also worth noting is that some orbital parameters (such as semi-major
axis) are exquisitely sensitive to the value of the initial velocity.
Unless you know that to within a few percent, there's hardly any point
in calculating the parameters at all (although inclination can be a
useful value).


Chris L Peterson
Cloudbait Observatory

On 4/14/2014 1:54 PM, Matson, Rob D. wrote:
> Hi Chris/Shawn/All,
> There is obviously something squirrely going on in that generated orbit.
> For one thing, look at the error bars on the velocity -- they cover all
> possible velocities (and as Chris points out, impossible velocities as
> well).
> I do have a comment about the oft-quoted 72 km/sec value. It is derived
> from the orbital velocity of the earth around the sun which averages
> 29.78 km/sec encountering a retrograde body at just below solar system
> escape velocity, which is around 42.1 km/sec at the earth's average
> distance from the sun. This gets you to 71.88 km/sec. However, earth's
> maximum orbital velocity (which occurs at perihelion in early January
> each year) is 30.29 km/sec. And solar system escape velocity at earth's
> perihelion distance is 42.48 km/sec. So that increases the closing
> velocity to 72.77 km/sec. However, this still isn't the theoretical
> maximum closing velocity, since I haven't included the acceleration of
> the intercepting body due to earth's gravity well.
> At an initial closing speed of 72.77 km/sec, there isn't a lot of time
> for the earth's gravity to act on the body before it hits the atmosphere,
> so the velocity boost is much smaller than earth's escape velocity (11.2
> km/sec). From conservation of energy considerations, it can be shown
> that the square of the impact velocity is equal to the sum of the
> squares of the escape velocity and the approach velocity. So:
> Max impact velocity = SQRT(11.2^2 + 72.77^2) = 73.63 km/sec
> So in January, it is certainly possible to have a meteor with a
> velocity above 73 km/sec. Anything above 73.63 would require the object
> to have either originated outside the Solar System, or (as Chris
> mentioned) to have been involved in a slingshot encounter with
> another planet prior to encountering earth.
> --Rob
Received on Mon 14 Apr 2014 04:16:05 PM PDT

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