Two protons in the nucleus of a helium atom are acted upon by three
forces: gravity, electrostatic repulsion, and the strong force.
What is the ratio of the strengths of these forces? Obviously,
the strong force is in equalibrium with the eletrostatic repulsion, or
else nucleii would either blow apart or else collapse. I'm about
to demonstrate that the electrostatic force swamps the gravitational
force.
Quantity |
Value |
Units |
Reference |
Diameter of a proton |
10^{-15} | m |
http://hypertextbook.com/facts/1999/YelenaMeskina.shtml |
Charge on a proton |
1.602x10^{-19} | Coulombs |
http://www.pa.msu.edu/courses/2000spring/PHY232/lectures/coulombslaw/chargedef.html |
Mass of a proton |
1,6726x10^{-27} | Kg |
http://www.newton.dep.anl.gov/askasci/gen01/gen01078.htm |
F_{gravity}=G*M_{1}*M_{2}/R^{2}
G = 6.67 x 10^{- 11} N m
^{2} /kg ^{2}.
F_{electrostic}=K_{c}*C_{1}*C_{2}/R^{2}
K_{c} is the electrostatic constant or Coulomb force
constant, often written as , where is the permittivity of free space,
also called electric constant, an important
physical constant. The value of k_{C}
is approximately 8.988 x 10^{9} F^{−1}·m or C^{−2}·N·m^{2},
F_{gravity}=1.86599*10^{-34} newtons F_{eletrostatic}=2.5126708992*10^{-14} newtons. The ratio of F_{eletrostatic}/ F_{gravity} is 2.5126708992*10^{-14} N / 1.86599^{-34} N or 1.35^{+20}. Thus, the eletrostatic force is roughly 100 billion billion times more powerful than the gravitational force.
However.... the electrostatic force can be attractive or repulsive. Gravity can only be attractive. If large quantities of charge accumulate in a region, there will be a current flow and the charges will tend to neutralize one another. Or else the charges will repel one another and disperse. So it is hard to accumulate really large quantitites of charge. If large quantities of matter come together, then the gravitation attraction will tend to attract still more matter. Eventually, you get to a black hole.