# What happen if i put r=0 into the formula F=(k)mm/r^2

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## Main Question or Discussion Point

What happen if i put r=0 into the formula F=(k)mm/r^2
So 1/0=infinity ?

then F= infinite large in mathematical sense?
is that statement correct ?

thank

BvU
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That's what happens with the formula. In reality you can't have two masses in exactly one place, so the formula isn't endlessly applicable...

Mathematically you have a singularity.

That's what happens with the formula. In reality you can't have two masses in exactly one place, so the formula isn't endlessly applicable...

Mathematically you have a singularity.
But In reality i can have two mass put in r=0.000001m
So the force still become very large?

Math_QED
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What happen if i put r=0 into the formula F=(k)mm/r^2

then F= infinite large in mathematical sense?
is that statement correct to say F is infinitely large?
You cannot put in r = 0. That does not make sense. It does make sense to consider:

$\lim_{r \to 0}F = +\infty$. This means, intuitively, when $r$ approaches $0$, $F$ becomes larger and larger (F grows without bound).

BvU
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But In reality i can have two mass put in r=0.000001m
So the force still become very large?
Well, the masses become smaller too for such small particles. One micron diameter with a reasonable density gives a very small mass !

On even smaller scales still other things happen. Protons, for example have a mass of only 1.67 10-31 kg and a diameter of 0.88 10-15 m. At such small scales other kinds of forces are much stronger.

Well, the masses become smaller too for such small particles. One micron diameter with a reasonable density gives a very small mass !

On even smaller scales still other things happen. Protons, for example have a mass of only 1.67 10-31 kg and a diameter of 0.88 10-15 m. At such small scales other kinds of forces are much stronger.
is it possible that the mass is very large but the r is very small?

Demystifier
Yes. Moreover, in quantum physics mass is, in a certain sense, proportional to $1/r$.