Theories of the microscopic never seem to include reference to gravity in the atom. Should they? What do you think?

Numbers don’t lie: The reality is that gravity inside the atom is minuscule. Let’s take a look at this in terms of scale, and then examine the equations for determining gravitational pull.

## Atoms and our Scale of Reference

It is the human tendency to draw conclusions – with reference to the extremely large and the extremely small – on the basis of what we experience in our scale of reference.

In fact, much good science has been realized using such assumptions. But only much good science – by no means all.

In fact, many of the most incredible discoveries have not been achieved using this method of scientific introspection.

Scientific theories must take into account scale and the principles of quantum mechanics for the microscopic and more for the sub-microscopic.

Fortunately, for ease of discussion, what we are going to discuss in this article lies within the realm of the older scientific techniques of classical physics.

## Earth and Moon Gravity

To calculate the gravitational force between earth and moon, scientists use Newton’s Law of Gravity,

**F = Gm _{1}m_{2}/r^{2}**

**G**is the gravitational constant, whose value is 6.673 x 10^{–}^{11}meters^{3}/kilogram-second^{2.}- The symbols
**m**_{1}and**m**_{2}are the masses of the earth and moon in kilograms. - The
**r**stands for the radial distance between them in meters.

Plugging in the appropriate values gives a force of 1.982 x 10^{20} kg-m/sec^{2} = 1.982 x 10^{20} Newtons. Is this a large quantity? Consider how it looks when that number is written out,

**198,200,000,000,000,000,000 Newtons**. That’s 198.2 quintillion N of force keeping the moon in its orbit!

## An Isolated Hydrogen Atom

To make a fair comparison between electrostatics and gravity in the world of the atom, let’s consider one isolated atom of the simplest sort-hydrogen. We will choose its simplest isotope, with only a one-proton nucleus and a single orbiting electron.

The measured ground-state radius of hydrogen, r, is 5.292 x 10^{-11} m.

The equation for the electrostatic force is:

**F = kq _{1}q_{2}/r^{2}**

In this formula, **q _{1}** and

**q**are the charge of the electron and the charge of the proton (1.602 x 10

_{2}^{-19}Coulombs), and Coulomb’s constant,

**k**, equals 8.988 x 10

^{9}N-m

^{2}/C

^{2}.

Notice how similar this equation appears to Newton’s Law of Gravity. This should not be unexpected, as we are comparing one force with another force.

We enter the appropriate values to get the following equation:

**F = (8.988 x 10 ^{9})(1.602 x 10^{-19}) (1.602 x 10^{-19}) / (5.292 x 10^{-11})(5.292 x 10^{-11}).**

This expression tells us what the force is if it is 100% electrostatic force and none of it is attributable to gravity. Solving gives us:

**F = 8.237 x 10 ^{-8} N.**

This is the electrostatic force holding the electron in its orbit. What if this was all gravitational force? What would the hydrogen atom’s radius be? We calculate,

**F = (6.673 x 10 ^{-11})(1.673 x 10^{-27})(9.109 x 10^{-31}) / (5.292 x 10^{-11})(5.292 x 10^{-11}) **

or:

**F = 3.631 x 10 ^{-47} N.**

Notice the difference between these two numbers. The electrostatic force is,

**F _{(elec)} = 0.00000008243 N**. This is not large, but neither is a hydrogen atom. It is a force to be reckoned with.

The gravity in the atom, on the other hand, is,

**F _{(grav)} = 0.000000000000000000000000000000000000000000000003613 N**.

This is so small in comparison with the electrostatic force that we can entirely dismiss it within the atom. Of course when a tremendous group of atoms comes together, the number above is magnified equally and comes to be of significance.^{
}

## Gravity Within the Atom

There is gravity inside atoms – but the amount of force is very small. The electrostatic force is larger, particularly in proportion to the size of the atom itself – and you should always keep scale in mind.

## Leave a Reply