[meteorite-list] Geometry and velocity trump gravity

From: Rob Matson <mojave_meteorites_at_meteoritecentral.com>
Date: Sun, 26 Jun 2016 21:01:26 -0700
Message-ID: <000301d1d028$93edf5d0$bbc9e170$_at_cox.net>

Hi E.P.,

Doug's reply pretty much covers it, but I wanted to give you an example so
that you can see how your intuition is failing you on this problem. Let's take
the case of a slower-than-average asteroid closing velocity encounter of
20 km/sec. And let's say that in the absence of any gravitational attraction
from the earth or the Moon, the asteroid would come no closer than the
altitude of geosynchronous satellites -- 35786 km. (This is a rather close
encounter, I'm sure you'd agree. Any trajectory with a more distant
point of closest approach would obviously experience less trajectory
bending than this case.) If we "turn on" earth's gravity, how much will
the trajectory be diverted toward the earth's center? Without getting into
the calculus of actually solving the problem, let's make some simplifying
assumptions that are very ~generous~ in the amount of gravitational
acceleration we're going to apply.

The acceleration due to gravity at the height of geosync satellites is
around 0.224 m/sec^2. Let's assume that that amount of acceleration is
applied during the entire encounter, and that it is always directed
perpendicular to the asteroid's original velocity vector. Define the start
of the encounter as being when the earth is 45 degrees to the left
of the velocity vector (at range 59629 km -- the square root of 2 times
the distance to the earth's center). The midpoint is when the earth
is 90 degrees to the left of the velocity vector (point of minimum
range to the earth center), and the end is when the earth is 135 degrees
to the left of the velocity vector. To first order, the distance travelled
is 2*42164 km = 84328 km. How long would it take in the absence of
gravity? 84328 km / (20 km/sec) = 4216 seconds -- a little over 70 minutes.
Recall your basic distance equation under constant acceleration:
X = 1/2 * A * T^2. Here we've generously allowed A to be 0.224 m/sec^2
for the entire encounter (when in fact A at the start is only half that).
What does X work out to?

X = 0.5 * 0.224 * (4216)^2 = 1.99 million meters or 1990 km.

Compare that to the 35768 km altitude of the GEO belt, and you see
that the amount of diversion is rather small. So the take-away from
all this is that the outcome is dominated by simple geometry. Yes,
very close encounters with the earth will bend a trajectory -- perhaps
quite significantly, but post-encounter the new trajectory is just as
likely to be diverted toward the Moon as away from it. --Rob

-----Original Message-----
From: Meteorite-list [mailto:meteorite-list-bounces at meteoritecentral.com] On
Behalf Of E.P. Grondine via Meteorite-list
Sent: Sunday, June 26, 2016 9:11 AM
To: meteorite-list at meteoritecentral.com
Subject: [meteorite-list] What killed off megafauna?

Hi Rob, Doug -

(Glad to hear that you're doing okay, Doug)

What both of you need to do is to take the perspective of a potential impactor
passing through the inner solar system,
as seen in the first few seconds of this video:
https://www.youtube.com/watch?v=GDrBIKOR01c

The basic geometry of the problem is set out here:
https://www.youtube.com/watch?v=qTDcfI2dabk

If it makes it easier,
think of the Earth-Moon system as a pair of girls at a darkly lit party,
and yourself as a young man.
(A variant of the famous mathematical problem of the drunk's walk at the frat
party.)
While you may be attracted to the "prettier" (in terms of gravity/area) Moon,
you are far more likely to run into her much larger "wingman", the Earth.
Sorry, but that is just the way it is.

At first, you're getting a warm fuzzy feeling (gravity) from the combined
Earth/Moon system.
That warm fuzzy feeling varies by the cubes of the radius of each of the two
bodies,
and is located somewhere amongst them.

On your approach to the Earth/Moon system, you see your landing area (for only
part of the month, see second video)
in terms of area, which varies by the square of the radius.
But of course gravity also varies by distance,
so as you get close to theses two
you're getting far more of that warm fuzzy feeling from the Earth,rather than
the Moon.
You head that way.

You can use any computer language you prefer to model this system,
and run it on any machine you like.
But as a check on your computer model, you have to rely on the data.
In this case, compare the data on smaller impactors from the Moon,
(Apollo seismic and optical astronomy)
with the data from the reconnaissance systems which have been operating
on the Earth since the 1950's,
and "alarming" the "stuff" out of those looking at the data from them.

If there is a difference between your model's results and the data, then your
model is defective.
 
(An entertaining video on phonetic loading,
the diameters of lunar craters,
typing monkeys, and other natural phenomena:
https://www.youtube.com/watch?v=fCn8zs912OE)

good hunting,
E.P.

PS - Will NEOcam make meteorite hunting far easier?
Will the increased supply drive down prices even further?

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Received on Mon 27 Jun 2016 12:01:26 AM PDT