archive: Re: SETI Re: Doppler Shift

Re: SETI Re: Doppler Shift

MarcusJohn@aol.com
Sat, 22 Aug 1998 22:40:12 EDT

In a message dated 8/22/98 8:11:37 PM Eastern Daylight Time,
dfheli@pacificnet.net writes:

>
>
> Hello David,
>
> I am puzzled. What does this mean?
>
> >>DW
> The catch here is that the moon doesn't actualy orbit the earth, but,
> to a two body approximation, the centre of gravity of the earth moon
> system, which I estimate to be within the earth but fairly near the
> surface.
>
> Especially the moon part?
>
> Would you please explain this again for me?
>
> Thank you.
>
> Walt Williams, 98.08.22
> dfheli@pacificne.net

Walt, imaging a dumbell rotating in gravity free space, it will clearly rotate
around the midpoint of the bar. That is the center of mass of the system.

Imagine 2 planets of equal mass, but no bar. They will clearly rotate around a
point in between them, not within one of the planets. This point is the center
of mass.

Imagine 2 planets, one slightly heavier than the other. The center of mass is
more towards the heavier planet, so the heavier planet moves less, the lighter
one moves more. But the center is still outside the planets alltogether.

Imagine 2 planets, one much heavier than the other. The center of mass will
now be within the larger planet. But both planets still move around their
total center of mass, and even the larger planet moves some, while the smaller
planet does some of the rotating.

This is the Earth and the Moon. The Earth not only spins, and rotates around
the Sun, but it is continuously pulled towards the moon, for a circular motion
with a frequency of 30 days, and a very small amplitude. The amplitude is
equal to the distance of the center of the -Earth- from the center of the
-Earth moon system-, which was estimated to be just below the surface of the
Earth.

Hope that makes it clear

John.