[meteorite-list] Meteorite impact angles

From: Matson, Robert <ROBERT.D.MATSON_at_meteoritecentral.com>
Date: Tue Dec 7 19:35:15 2004
Message-ID: <AF564D2B9D91D411B9FE00508BF1C86904EE6222_at_US-Torrance.mail.saic.com>

Hi All,

Earlier I wrote:

> This series of photos made the rounds a few weeks ago -- it definitely
> is not a meteorite fall. Simplest reason: it doesn't fall vertically.

Ron replied:

> Peekskill and Nakhla didn't fall straight down either.

I guess it would have been better to simply say that a small meteorite
could not travel in a straight line at such a shallow angle.

> Peekskill traveled an additional 50 km AFTER ablation had ended.
> Nakhla was observed to hit the ground 30 degrees from the horizontal.

I seriously doubt the latter observation unless the particular Nakhla
meteorite was experiencing significant lifting forces and/or gale
force winds. Consider a meteoroid traveling parallel to the ground
at an altitude of 10 km (rather unlikely). Give it a very high
velocity at this point of 1200 kph (greater than Mach 1), and to
simplify the problem remove the atmosphere. Assuming a ballistic
trajectory and a flat earth, at what angle will it hit the ground?

X = 0.333 km/sec * T
Y = 10 km - 0.5 * 0.0098 km/sec^2 * T^2

where T is time in seconds. Y will be zero (ground level) at
T = 45.2 seconds.

DX/DT = .333
DY/DT = -0.0098 * T

DY/DX = -.0294 * T; at 45.2 seconds, DY/DX = -1.329

Impact angle from horizontal = arctan(-DY/DX) = ~53 degrees

If you add back the atmosphere, the angle will be much steeper than
this since terminal velocity will cause the meteoroid to decelerate
and take much longer than 45.2 seconds to fall from 10-km altitude.

Best,
Rob
Received on Tue 07 Dec 2004 07:29:38 PM PST