# Re: IDE freeze for seconds

Riccardo Facchetti (fizban@tin.it)
Thu, 3 Dec 1998 12:20:15 +0100 (MET)

On Wed, 2 Dec 1998, Roeland Th. Jansen wrote:
> On Wed, Dec 02, 1998 at 03:25:24PM -0000, Simon Kenyon wrote:
> > On 02-Dec-98 Alex Buell wrote:
> > > What is the first thing that ceases when you spin down a planet? That's
> > > right, gravity. Everything not tied down would just fly out into space,
> > > and that includes the atmosphere. Simple.
> > i don't see a smiley, but i have to assume that that was a little joke! right?
> well, the problem is that... it's true. so don't use ower management on the
> earth spin...

No. The problem is that it's _not_ true. Sorry but I can not resist.

Gravity between two bodies is a function like this (Universal Gravity
Law, Newton, formulated long ago):

F=f(m1, m2, x)

where:
F force
m1, m2 masses of the two bodies involved
x distance of the two bodies

On the surface of earth we have:

(M * m)
F = K * ---------
r^2

where
M is the mass of earth
m is the mass of body on earth surface
r is the radius of earth

We know what is the number g (gravity acceleration):

g = 9.81 m/s^2

But confronting g with the Law on the surface, we can write:

g = K * M/(r^2), where K, M and r are constants

so that we obtain the well known:

F = m * g

that is the force exercised by the earth on a body on her surface.

Now, this force is _not_ dependent of earth rotation.
The rotation simply balances part of the F force giving, for vectorial
addition, the well known "1 gravity" on the surface. This one is simple to
demonstrate:

/|\
| Force (centripetal, vector) exercised by earth rotation:
| Fc = m * w^2 * r
| where
| w = earth spinning in 1/sec
| r = earth radius
| m = mass of body on her surface
*
[|] <----- body
/ \
----- [earth surface] ----> earth spinning (w)
|
| Gravity force calculated with Newton's law:
| F = m * g
| where
| m = mass of body
| g = gravity acceleration
|
\|/

Now you can compose the two vectors into the total force exercised by the
earth:

Ft = F - Fc = m * g - m * w^2 * r

We can now, for example calculate what is the w that is necessary to fly:

To fly the Ft force _must_ be 0 (no forces acting on the body) so that:

m * g = m * w^2 * r

g is a constant (oh yes ... here I must note that it is not really a
constant, but for this approximate calculation we can
assume it to be equal to 9.81 m/sec^2)
r is a constant equal approximatively 6378 Km = 6.378 * 10^6 m

so that:

w(fly) = sqrt(g/r) = 1.24 * 10^-3 sec^-1

so that the earth should make 1 spin in:

w(fly) = 1.24 * 10^-3 [1/sec] * 3600 [sec/h] = 4.464 h^-1

For a body on her surface to fly, the earth must spin at more than 4 spins
in an hour (or, if you prefer, 0.22 hours for one single spin) !

This result contradict your statement that if earth spins down all the
body on her surface will fly (it is exactly the countrary ... if the
earth spins down the force acting on bodies is maximum).

Note that the centripetal force is not so high, and it varies with the
position on the surface of earth you are calculating that vector.
For example at the poles the centripetal force is zero because you are on
the axis of rotation of earth (approximatively) while on the equator is
maximum (still approximatively, all the calculations done here made
considering the body at the equator). Of course making these statements I
have considered the earth an omogeneus body: g varies with density
variation of earth too.

Note that this was a very approximate calculus, in any way you look at
it. Anyway the result, knowing its approximation, is a quite well
description of the reality, given that the considered body is stationary
with respect of a reference system blocked on the earth, because if it is
in movement you must account for many other forces that change a lot the
resultant Ft).

Ciao,
Riccardo.

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