i'm talking about droping a ball into a free fall under assumption that it can pass thru the earth's center all the way to minus height point.(adsbygoogle = window.adsbygoogle || []).push({});

case 1:if you say that the force at the dropping point is F=G(/m/R)(M/R)<>0 then the equilibrium point is the center of the earth.but when the ball gets at the equilibrium point then the force is zero.

case 2:if you say that the droping point is the equilibrium one then F at that moment is zero.but when the ball gets in the center of the earth then F=G(/m/R)(M/R)<>0 where R is the distance that the ball has passed to get there.

actually there is another way of explaining this:

E = F x D = energy = force x (equlibrium distance) = const.

F x dD = - dF x D = D x dF > 0

this means that all three vectors E,F,D rotate around the equilibrium point and the energy vector in the direction of the force.normatively all the vectors remain the same.only if you push the ball (i.e. invest some force in the system) then the normas change.pure and simple dynamics.the projections of the vectors depend on the starting conditions.let n be the vector of your aspect

(if when the projection of D is max the projection of F is zero then n is parallel with F

(if when the projection of D is max the projection of F is also max then n is parallel with E))

then you should project all the vectors E,F,D on the plane normal with n.

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# The G force is not infinity at distance zero(in free fall)

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