RE: [Algorithms] rather curious

[Jeff L. <je...@di...>]

Man, we are all working late tonight.  On a sunday, that's kinda sad.
But to drag this back to algorithms, your use of Newton's gravitation law is getting you into trouble.  

Realize that the only thing that effects when two dropped objects hit the ground is the acceleration of those objects.  Force doesn't enter into it yet.  Here is why:

Gravity is as you say F = (Gm1*m2)/r^2, also thanks to Newton we have F=ma.

Solve for the acceleration using those two equations:

m1a =  (Gm1*m2)/r^2     // note the m in f=ma is for the dropped body which I am calling m1.

a = Gm2/r^2

The acceleration of body 1 is only dependent on the gravitation constant, distance, and the mass of the earth (or whatever).  On most of the surface of the earth as you know that a = 9.8 m/s^2.  I guess at airplane height, it is a bit different.

So you can see that a body of any mass will hit the earth at the same time.  The difference due to the original gravitation law is that a more massive body will hit the earth with more FORCE and make a bigger dent.

Of course this ignores the fact that the formulas need to be applied to the earth as it is attracted to the other object.  If the object being dropped was very massive, it would move the Earth also.  This would change the time of collision.  Of course for most objects you will ever be able to drop, this doesn't matter.

-Jeff

At 01:21 AM 9/11/2000 -0500, you wrote:
>But,
>what i brought up is that if we go by Newton's law of universal gravitation
>there is a *difference*
>
> >All bodies, in a vacuum, fall at the same rate-
>if we take this formula
>F = G ((m1*m2)/r^2)
>
>and we run this for 2 separate objects with *very similar* masses,
>F will be different.
>
>F = ma;
>a = F/m;
>
>even at a *tiny precision* they still fall at different rates, thus hitting
>the surface at a different time.
>hence i feel that this should be taught in schools at least for a
>"technical" viewpoint cause it's the *right thing* to do.