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Re: (meteorobs) P/2000 G1 & Vgeo
John wrote:
> >HOWEVER, Vgeo is a *vector* thingy, and as far as I know the lower limit to
> >speed is more to do with terminal velocity than Earth's gravity, ie what
> >the minimum speed is that still allows enough friction for burn up to occur
> >in the upper atmosphere. Anybody know better?
and Lew responded:
>That's interesting, John - first time I've heard this explanation! I do wonder
>if it fits, though: don't reentering space craft (e.g., the Shuttle) generally
>have initial Vg well below 11 km/s, but still produce very considerable burns?
The relationships between meteoroid heliocentric velocity (V subscript h,
Vh), geocentric velocity (V subscript g, Vg), and apparent velocity (V
subscript a, Va) can be rather confusing with regard to the
nomenclature. Heliocentric velocity, Vh, is easy enough: the velocity of
the particle in the sun's frame of reference (that is, with the sun held
stationary) -- no confusion here. However, the terms Vg and Va will vary
from meteor textbook to textbook and care must be taken to ensure that you
understand the author's proper meaning. In the early 1960's a standard
nomenclature was finally adopted, so that books and papers written since
then should all carry the same meanings for these terms, but some classics,
such as Lovell's "Meteor Astronomy" (1954) actually have these variables
with the reverse nomenclature from the standard form of today. [This is a
lesson that I had to learn the hard way!]
Here is the way that it works now:
Vg = particle geocentric velocity. This is the velocity of the particle in
the Earth's frame of reference (that is, with the Earth held
stationary). If Vh for the particle is known, then Vg can be found by
vectorially subtracting the heliocentric velocity of the Earth, Ve, from
the particle's heliocentric velocity. The result will be the velocity of
the particle as seen from the geocentric frame.
[note: this is treated as a Galilean transformation, even though the
frames are not truly inertial -- the accelerations are generally neglected.]
Va = particle atmospheric entry velocity. This is an addition to the
geocentric velocity, due to a small particle entering the gravity well of a
large body, such as the Earth (NOT taking into account atmospheric
drag). It is the conversion of the particles gravitational potential
energy to kinetic energy as the particle approaches the surface of the
Earth, which results in an increase of the particle's velocity. This
increase can be derived as follows:
Begin with the standard formula for gravity:
F = - (g * m * Me) / r^2 (in the r direction),
where:
F = gravitational force,
g = universal gravitational constant,
m = particle mass,
Me = Earth mass,
r = radial distance of the particle from the center of the Earth, for r > R,
R = Earth radius.
[Note that I have assumed a coordinate system with the center of the Earth
at the origin.]
Since F is a conservative force, this can be converted to a gravitational
potential energy by F = - del(U), where del = gradient:
U = - (g * m * Me) / r .
[U is assumed to be 0 at r = infinity.]
A particle approaching the Earth will experience a gain in its kinetic
energy, K, as its potential energy, U, goes from a very small negative
value (essentially 0) to a much larger negative value (as a function of
1/r). This obeys the conservation of energy law:
K + U = constant, therefore,
delta K = - delta U.
Placing the particles initial U at approximately 0, will give:
delta K = (g * m * Me) / r .
The overall change in the particle's kinetic energy can be expressed by:
Ki + delta K = Kf ,
where:
Ki = initial kinetic energy,
Kf = final kinetic energy,
and since K = 1/2 * m * v^2, and all terms have a 1/2 * m out front:
Vi^2 + Vesc^2 = Vf^2 ,
where:
Vi = initial velocity,
Vf = final velocity,
Vesc = Earth escape velocity.
Vesc is the velocity associated with the delta K alone, that is, beginning
with a particle initially at rest. If the particle begins on the surface
of the earth it is the velocity necessary to escape the Earth's gravity
well (increase U to 0). This is found by:
delta K = (g * m * Me) / R ,
1/2 * Vesc^2 = (g * Me) / R ,
Vesc^2 = (2 * g * Me) / R ,
Vesc = sqrt((2 * g * Me) / R) ,
for:
g = 6.67*10^-11 m^3 / (kg * sec^2),
Me = 5.98*10^24 kg,
R = 6369*10^3 m,
to give:
Vesc = 11.2 km/sec (Earth escape velocity), and
Vesc^2 = 125 km^2/sec^2.
FINALLY, this allows us to find Va from a given Vg.
begin with:
Vi^2 + Vesc^2 = Vf^2 (from above).
Placing in our initial terms gives:
Va^2 = Vg^2 + Vesc^2 ,
Va^2 = Vg^2 + 125 (in units of km/sec),
Va = sqrt(Vg^2 + 125) .
Or from Lovell (1954), but in today's nomenclature:
Va = sqrt(Vg^2 + 124.9) .
Now, to be picky, the meteoroids we are talking about do not generally
reach the surface of the Earth, so that the actual addition to their
kinetic energy will be slightly less. If a value of r = 6469 km (meteor
height of 100 km above the surface) is used, Vesc = 11.1 km/sec, and Vesc^2
= 123 km^2/sec^2, which gives a slightly smaller Va.
Also (although I think that Lew knows this), the space shuttle is already
in Earth orbit when it begins its descent, so it is ALREADY captured in the
Earth's gravity well. It is thus travelling at less than the Earth's
escape velocity (about 7-9 km/sec if I remember correctly), which gives it
enough velocity to "fall" elliptically (orbit), rather than fall
parabolically. On reentry, atmospheric drag is purposefully used to burn
off (literally) the shuttle's kinetic energy which was added to it at
launch, making it once again a surface bound object.
Best regards,
JIm
------------------------------------------------------------------------
James Richardson (graduate student)
Department of Planetary Sciences
Lunar and Planetary Laboratory
University of Arizona
Tucson, AZ 85721
school email: richardson@lpl.arizonadot edu
school webpage: http://www.lpl.arizonadot edu
Operations Manager
American Meteor Society (AMS)
AMS email: richardson@amsmetoers.org
AMS webpage: http://www.amsmeteors.org
------------------------------------------------------------------------
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