[meteorite-list] mercury escape numbers? can it be this?

From: MexicoDoug <mexicodoug_at_meteoritecentral.com>
Date: Fri, 05 Aug 2011 15:04:31 -0400
Message-ID: <8CE21B9A049F54B-193C-D662_at_webmail-d053.sysops.aol.com>

Hi Walter and great to see you still active on the list, also Sterling
and Mark and ET and all listers

Rex is referencing my comment on the escape velocity from the Sun, in
addition to the escape velocity from the planet. I.e., the closer you
are to the Sun, the greater the escape velocity at that distance.

Perhaps I was a bit sloppy, since in fact escape velocity is not the
exact measure needed to get from the Mercury to Earth (note: escape
velocity from the Sun's grip at Mercury to the Sun's grip at Earth),
but rather a significant part of it. It is conveniet however to think
of escape velocity. The same is true for Earth, for example, if you
are at the Moon's distance, the escape velocity from the Earth is only
1.4 km/s instead of the often quoted 11.2 km/s from Earth's surface.
Escape velocity depends upon the mass and distance from the center of
mass as well of the primary object, but as you note it is normally
thought of fromm something being launched from the surface. In this
case, thought, it is the Sun's escape velocity that trumps everything,
incluiding Mercury's surface escape velocity.

Since Rex is correctly observing (as I also mentioned in my original
post) that in an eccentric orbit you can play with the gravitational
potential energy so that you are further away from the Sun, the escape
velocity from the Sun at Earth's distance if that is where the
semimajor axis of the ellipse reaches will not be the barrier. The
question to ask at that point is how something smashing into Mercury in
the first place - when all of Mercury's momentum is in the orbital
direction. So, you there is no easy "escape" from this conumdrum in my
opinion.

As was already covered very effectively by Mark (and also covered in my
original post, and to some extent by Sterling when we discussed moving
orbits in the asteroid belt some time ago) the energies required to
change orbits are on a huge scale. So if we treat this as a "delta"
energy and pay close attention to the velocity vectors, the correct
question to ask IMO is in order to change the energy from a Mercury
orbit to deliver it here on Earth, how do you flip the velocity vector
enough to send it here without channelling such a great amount of
energy into the rock that it completely changes form.

How would you throw a tennis ball out the window of your speeding car
(even without air resistance to deal with) and get it to peacefully go
on its way in a perpendicular direction?

As for the Canadian modeling, Sterling I respect the modelers and have
been there and done that myself, well in material property situations
-not orbits. But, letting the computer plug and chug while we all chug
our own cold ones is not enough unfortunately. With all modeling, a
few basic assumptions are made; with all published work on modeling,
even though the authors say generally the assumptions, it is not always
so hermetic that one can reproduce identically.

I welcome their addition to the literature, had not heard of it
before...but what I will say is it is *NOT* an *ab initio* calculation
that they have done, and I would need to see all the fudge factors to
properly criticize such a work. It does bring up good questions and
comments as a tool - but it doesn't change the physics of the
situation. If I had more time I would get into it, but at the moment I
am luck to find time to type this poorly thought out comment.

Kindest wishes
Doug

Regarding ET's specimen, whether or not it is from Mercury has little
impact on its incredible status. Not being from Mercury (IMO) is in
fact more exciting because we aretalking about a completely different
parent body that holds more of a key to unraveling what happened in the
larger bodies in the evolution of the Solar system.




-----Original Message-----
From: Walter Branch <waltbranch at bellsouth.net>
To: rexscates at comcast.net <rexscates at comcast.net>
Cc: meteorite-list at meteoritecentral.com
<meteorite-list at meteoritecentral.com>
Sent: Fri, Aug 5, 2011 1:34 pm
Subject: Re: [meteorite-list] mercury escape numbers? can it be this?


Hi Rex,

Escape velocity depends on the mass of the planet, not it's orbital
parameters.

Maybe I don't understand your question. Is this what you are asking?

-Walter

Not everything that can be counted, counts and not everything that
counts can be
counted. -A. Einstein.

On Aug 5, 2011, at 12:04 PM, rexscates at comcast.net wrote:

> I don't post much.
> I did not do the math. (my calculus has not been used much for many
many
years.
>
> would not if you had a highly highly elliptical orbit not have to
have the
full sun escape velocity as it could do the whip around gravitational
boost
effect from the sun.
>
> also would a highly elliptical orbit have less of a escape velocity?
> the second part could be wrong but the first one is always a
possibility. (at
least in my mind) :)
>
> -Rex Scates
>
> scalecubes.com
>
>
>
> "But what about Mercury. Mercury's escape velocity is 4.3 km/s. But
>> it's downstream from Earth and the Sun is a huge gravitational drain
>> plug that devours material. If you think Earth gets a piece of Mars,
>> imagine what the Sun gets from Mercury. To escape the Sun ... that
is
>> to go upstream towards Earth, at Mercury, any fragment would have to
>> battle an escape velocity of 67.7 km/s. That's greater than Jupiter
!
>> You might say ... ok, you don't have to actually escape the Sun,
only
>> make it from Mercury to Earth. Well, at Earth, the escape velocity
is
>> 42 km/s from the Sun. That's a loss of 25 km/s ... and don't forget
>> the extra 4.3 km/s to get away from Mercury as well ..."
>>
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Received on Fri 05 Aug 2011 03:04:31 PM PDT