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

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SUMMARY

The discussion centers on the implications of setting the distance variable \( r \) to zero in the gravitational formula \( F = \frac{(k)mm}{r^2} \). It is established that mathematically, this leads to a singularity where force \( F \) approaches infinity. However, in practical terms, two masses cannot occupy the same space, making the formula inapplicable at \( r = 0 \). Instead, as \( r \) approaches zero, \( F \) grows without bound, but real-world constraints and quantum mechanics must be considered for very small distances.

PREREQUISITES
  • Understanding of gravitational force equations, specifically \( F = \frac{(k)mm}{r^2} \)
  • Basic knowledge of limits in calculus, particularly \( \lim_{r \to 0} F \)
  • Familiarity with concepts of singularities in mathematics
  • Introduction to quantum mechanics and its implications on mass and force
NEXT STEPS
  • Study the implications of singularities in physics and mathematics
  • Learn about the role of quantum mechanics in force calculations at small scales
  • Explore the concept of limits in calculus, focusing on practical applications
  • Investigate the relationship between mass and distance in gravitational and quantum contexts
USEFUL FOR

Students of physics, mathematicians, and anyone interested in the intersection of classical mechanics and quantum physics will benefit from this discussion.

garylau
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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
 
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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.
 
BvU said:
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?
 
garylau said:
What happen if i put r=0 into the formula F=(k)mm/r^2then 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).
 
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garylau said:
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.
 
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BvU said:
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?
 
garylau said:
is it possible that the mass is very large but the r is very small?
Yes. Moreover, in quantum physics mass is, in a certain sense, proportional to ##1/r##.
 
garylau said:
is it possible that the mass is very large but the r is very small?
Yes, then you get a very large force. But it is always finite. And if you want to put the masses too close, you have to consider quantum mechanics where the simple approach with a well-defined distance fails.
 

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