Re: Gravity and the Sun

From: Greg Neill (gneillREM_at_OVE.THIS.netcom.ca)
Date: 08/19/04 

Date: Wed, 18 Aug 2004 22:02:17 -0400

"JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message
news:2ohuksFb04c0U1@uni-berlin.de...

> You are right Old Man, now I made a stupid arithmetic error with
> the sign of the distance between Earth and Sun, as you have
> pointed out.
>
> Now I get the picture.
> You have assumed that the Earth is free falling around the Sun
> with an acceleration of 0.006 m/s^2.
> Apparently it doesn't matter what is Earth mass, since that all the
> objects fall at the same rate in a gravity field.
>
> The problem is that Earth speed is a fundamental parameter
> in order that Earth orbit be stable.

Well, not exactly. What's important (what is called "an
integral of the motion") is the total mechanical energy
of any given orbit, E. This is the sum of the kinetic
energy, KE, and the potential energy, PE.

The kinetic energy depends upon the velocity of the body
in orbit, while the potential energy depends upon its
position in the gravitational field (distance from the
primary). E = KE + PE.

A few equations:

  If the mass of the primary is M,
     the mass of the body is m, with m << M,
     the speed of the body is v,
     the distance of the secondary from the primary is r,

  and considering the specific quantities for KE and PE
  (that's where we take the values per unit mass of the
   secondary body, ie., kinetic and potential energy per
   unit mass):

     KE = (1/2)*v^2
     PE = -GM/r

  so that E = (1/2)v^2 - GM/r

  (potential energy is referenced from infinity, so will
   be a negative quantity)

Now, so long as E is negative the body will be in a bound
orbit. In fact:

   E < 0 ==> elliptical or circular orbit
   E = 0 ==> escape velocity; parabolic or straight line
   E > 0 ==> hyperbolic trajectory

> An object stopped at Earth distance from the Sun will fall to the
> Sun along a straight line and will not have a "circular" orbit.
> Earth mass and its kinetic energy is fundamental.
> Without the right mass and the right speed the Earth will not
> orbit the Sun, but will either go away or fall straight to the Sun.

So long as the mechanical energy is negative, it will just assume
a new bound orbit.

[snip]

>
> I bet that if the Earth is at a bigger orbit radius with the same
> speed (same angular momentum), when the gravity of the
> Sun is switched on again it must...
>
> What's your bet ?
> 1 - It will spiral inside up to the previous position (the initial orbit).
> 2 - It will spiral away and forever.

It won't spiral if the gravitational field remains constant
once it resumes. The orbit will be a fixed conic section
depending upon the total energy and the body's "initial"
conditions (postion, velocity).

>
> I found this to be a difficult question.
> Need to draw a picture on a computer graphic software.
>
> Applying the angular momentum conservation Law, my bet
> is that Earth will go away due to the inverse square term on
> the radius, which means that inertia really exists, being the
> central force (in the centrifugal force direction) always bigger
> then the central gravity force.
>
> Motion do cares on inertia, not on the equivalence principle
> written on a paper (a mathematical formula).
>
> The fact is that unbalanced satellites will fall, or will go away.
> Satellites don't care about the mathematical formula.
> Satellites must have the right speed at the right distance,
> or else they will fall or will go away.
> The same with the Earth.

This is a common misconception, the idea that orbiting
bodies are somehow balanced on a knife-edge in their
paths. They aren't. You can give them a mighty shove,
and unless you impart enough energy to send them off on
an escape trajectory (make E >= 0), they will simply take
on a new, bound, orbit.

[snip]

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