[meteorite-list] Re: 85 AD -2002 AD

From: Bob Martino <martino.6_at_meteoritecentral.com>
Date: Thu Apr 22 09:54:05 2004
Message-ID: <v03130302b894e978a283_at_[65.24.110.16]>

Orbital Mechanics! Woo Hoo! Let's see if I can oil up my rusty memory of
those bygone days in Physics Class (I'll jump in here and try to be
helpful, since Mr. Nowak has kindly decided to not push the religious
discussion brought up earlier).

>The asteroid belt to my knowledge is 168,000,000
>million miles away from the Earth. If the average
>speed of the meteoroid coming out of the belt where 10
>mph that would mean in 24 hours it traveled 240 miles
>one week 1680 miles one year 87600 miles. In 100 years
>8760000 miles To do 168,000,000 miles would take about
>1917 years. So If I found a meteorite in 2002 the
>meteoroid had to start it=EDs journey in 85 A.D. (if
>using AD - Anno Domini is to heavy for this list let
>me know) I wanted to present this write up as a sales
>pitch to the general public. Is this on or off track??

10 mph is far,far too slow. Without using any serious equations, we can
ballpark this.

Take an asteroid with a semimajor axis of 3 AU (Astronomical Units, the
Earth-Sun distance, 93 million miles). This is about average for the main
belt. So we imagine a circular orbit with a radius of 279 million miles.
The circumference would be 2Pi(r), or 1.75 billion miles.

Kepler's Law gives us the time it takes this critter to orbit the Sun. P =
=3D
Ka^3 where P =3D Period, K is a constant dependent on the mass of the star,
and a =3D the semimajor axis of the orbit. Since our star is the Sun, K is
defined to have the value of 1 (don't worry about units - they all work
out). For an a =3D 3 AU, the period is 27 years (3 cubed).

So our rock has to travel 1.75 billion miles in 27 years. There are 8760
hours in a year, or 237,000 hours in 27 years. 1.75 billion miles/237,000
hours gives us a speed of about 7380 mph.

But this is all just to get us in the ballpark. It's the speed of a rock
with a perfectly circular orbit with a 3 AU radius. It looks like Mr.
Nowak wants the meteoroid to intersect the Earth's orbit and enter the
atmosphere. Now we have elliptical orbits to deal with, since the rock
will cross Earth's orbit and also spend time in the main belt. For this,
the rock will have different speeds at different places in its orbit (in
accordance with Kepler's Law of Equal Areas). To calculate this more
exactly, one will need more information about the rock's orbit.

Typically, meteoroids enter the atmosphere at over 30,000 mph, if that helps=
=2E

>Here is another one that crossed my mind. If light
>travels 186000000 miles per second. How long will it
>take in TIME for light to travel 3/8 of an inch???

You mean 186,000 miles per second for the speed of light.

There are 5280 feet per mile (I think), 12 inches per feet.
(186,000)x(5280)x(12) =3D 1.18 x 10^10 inches per light second. This is
distance per unit time, but you want time per unit length, so just invert
this quantity. (1)/(1.18 x 10^10) =3D 8.48 x 10^-11 seconds to travel one
inch. Multiply this by 3/8 to get 3.18 x 10^-11 seconds.

Unless I dropped a decimal somewhere, that should be correct. But why 3/8
of an inch? Hmmmmm....

------------------------------------------------------------------------
Bob Martino Can you really name a star?
                                   http://home.columbus.rr.com/starfaq/
"I look up to the heavens
 but night has clouded over
 no spark of constellation
 no Vela no Orion." -Enya
Received on Sun 17 Feb 2002 12:35:16 AM PST