Re: Jerry's Latest Blunder

On Tue, 10 Mar 2009 04:14:13 -0700 (PDT), Cephalobus_alienus@xxxxxxxxxxx wrote:

On Mar 9, 10:17 pm, hw@..(Dr. Henri Wilson) wrote:
On Mon, 9 Mar 2009 17:04:19 -0700 (PDT), Cephalobus_alie...@xxxxxxxxxxx wrote:
On Mar 9, 4:44 pm, hw@..(Dr. Henri Wilson) wrote:

Your calculations are complete nonsense and totally useless.
No wonder you are
still silly enough to support Einstein's stupid theory.

In the above, you must calculate the PE from the Earth's
CENTRE not its surface.

I assume you know how to do that. For instance if the earth
was homogeneous,
WD in raising m=1000 from centre to surface = Integral
(4/3pi.rhoGm.r)dr for r =
0 to 6370000
= 4/3pi.rho.Gmr^2/2
= GMm/12740000
= ~3.14E10J

Amazing. I was -attempting- to demonstrate to you why it is that
your insistence on calculating PE from the surface of the Earth
(and presumably other celestial bodies) is a perfectly dreadful
choice.

You turn around and recommend an even WORSE choice!!!

Calculating PE from the centers of the celestial objects has all
the disadvantages that calculating from their surfaces has, plus
the additional one that an accurate calculation is dependent on
a detailed knowledge of their interior mass distributions.

Earth is certainly not homogeneous in composition, and your
computation is off by a considerable percentage.

The correct choice of reference is to set PE = 0 at r = infinity.
This has the desired effect of automatically eliminating the
contribution of distant stars from the PE computation. The only
"disadvantage" is that the calculated PE's are negative...

Your approach gives the positive amount of work that an object
can do in
falling from infinity to a certain height above a reference mass.
Why you claim
it is negative eludes me.
It might be useful when considering how much energy is required
to accelerate
an already orbiting spaceship to the escape velocity. Apart from
that, I cannot
see any logic in your approach.

As I stated before:
"What you want is a number that reflects the fact that all those
billions of other stars in the galaxy don't make a darn bit of
difference to the shape of the equipotential curve."

All your method does is produce a wrong answer.
All other significant objects have to be considered.

But not stars light years away...

Equipotential surfaces can
only be calculated by considering the combined gravitational
forces, using
their C of G's as the bases..

Let's consider the amount of information necessary to compute an
equipotential surface about a set of celestial bodies using
different choices for PE = 0.

Assuming that the bodies exhibit spherical symmetry:

1) PE = 0 at r = oo
Positions and masses of the bodies

2) PE = 0 at r = radius of object
Positions, masses, and radii of bodies

3) PE = 0 at r = center of object
Positions, masses, radii, and detailed mass distribution
within the bodies
..

4) Force = -GM/r^2 at any point r distance from each
Relative positions and masses of bodies.

The shape of a equipotential surface will not vary due to choice
of computation. In principle, all three methods should yield an
equivalent set of surfaces.

1) doesn't tell you anything

However, methods 2 and 3 require progressively greater amounts
of information to compute, not all of which is necessarily
available.

that was not the case in the example you gave. You needed twice the
calculations that I did.

Jerry



Harry Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm.

......
.

Relevant Pages

  • Re: Jerrys Latest Blunder
    ... your insistence on calculating PE from the surface of the Earth ... equipotential surface about a set of celestial bodies using ... Positions, masses, and radii of bodies ... The shape of a equipotential surface will not vary due to choice ...
    (sci.physics.relativity)
  • Re: Intelligence and Statistics
    ... bodies by a set of differential equations. ... of trajectories in gravitationally interacting bodies, ... Moon is calculating anything. ...
    (sci.cognitive)
  • Re: A Ballistic photon theory
    ... |>That's observation (i.e. empirical data) ... |>The separation is found from the masses of the supposed two stars, ... tell us the inclination of the orbit to the celestial ... |>just two bodies. ...
    (sci.physics.relativity)
  • Re: Jerrys Latest Blunder
    ... equipotential surface about a set of celestial bodies using ...   Positions, masses, and radii of bodies ... Positions, masses, radii, and detailed mass distribution ...
    (sci.physics.relativity)
  • Re: A Ballistic photon theory
    ... >|>in the tank are really two fish, and I don't believe there are two stars ... >|>just two bodies. ... Plot, and repeat. ... See how three orbiting bodies interact: ...
    (sci.physics.relativity)
Quantcast