Does the existence of gravity weigh heavily upon you? Does the incessant pulling of planet earth on your shoes get you down? Are you frustrated by your inability to fall towards the ground at high speeds and miss? Perhaps you are beginning to believe that gravity has you trapped like a rat; a flightless rat, with neither a Saturn V, nor a space shuttle to boot.

Fear not, dear reader. For I am about to teach you how to defy the Earth's heavy handed grip on you and your possessions.

Find a balloon. This may be the most difficult part of your task, as balloons are skittish creatures; they are known to explode with little or no notice. Beware.

Inflate the balloon using ordinary air. Using helium would clearly be cheating; you don't want to lie to yourself about this sort of thing.

Rub the balloon in your hair, or in the hair of a nearby dog or child.

Continue doing so until the balloon has accumulated sufficient static charge. Try to avoid accidentally discharging the static electricity by touching metallic objects, or other people.

Now for the moment of truth: take the balloon, and stick it onto the wall. Bask in your own amazing abilities and the wonder of the natural world as it sticks there, defying gravity for the entire world to see.

So, why does it work?

There are two forces acting on the balloon. The first is gravitational force, which is pulling the balloon towards the center of the earth. But, there are clearly other forces at work here, since the balloon isn't falling. When the balloon rubs through your hair, electrons jump from your hairs to the surface of the balloon; this gives the balloon a negative charge since there are now more electrons in the balloon than there are protons (the number was equal before the hair rubbing).

It is a well known fact that opposite charges attract while like charges repel. When you bring the balloon near the wall, the electrons in the wall are repelled by the negatively charged balloon, leaving that area of the wall with a positive charge. The end result is that the negatively charged balloon is attracted to the positively charged wall with more force than gravity can possibly muster.

Both electrical and gravitational force vary inversely with the square of the distance between the two objects/charges; to find the force you simply take the product of the masses or charges, divide by the square of the distance (taken from the centers of the two objects) between them, and multiply by a constant. The mass of the earth is 5.9743 x 10^{24} kg, and its radius is 6,377.83027 km, which is the approximate distance from the balloon to the earth’s center. Let’s say that your average balloon is four grams (.004 kg), and that the charge on the balloon courtesy of your hair is 5*10^{-7} C (coulombs)*. For the sake of argument, assume that the balloon has a radius of .1m, and that the charge induced in the wall by the balloon is also 5*10^{-7} coulombs. Thus:

F_{g }= G(Mm/r_{e}^{2}) and F_{e}=k(Q_{1}Q_{2}/r_{b}^{2}) where G is the universal gravitation constant, 6.670 x 10^{-11} Nm^{2}/kg^{2} and k is Coulomb’s constant, 8.99 x 10^{9} Nm^{2}/C^{2}.

Plugging the above numbers into our respective equations and consulting google tells us that:

**F**_{g}= 0.0490040801 newtons

and **F**_{e}=.22475 newtons.

So in this instance, the gravitational force is five times weaker than the electrical force--and you can keep adding charge to the balloon! In fact, the only thing that keeps the balloon from accumulating an infinite amount of charge is the electrical repulsion between the charged balloon and the new electrons that you're trying to put on its surface.

Balloons: Two. Earth: Zero. Gravity is truly weak.

**Values for mass, charge, and radius of a balloon were lifted from "Back of the Envolope Physics" and are very approximate. Values for mass and radius of earth from E2 node earth.** *
*For those that care, a more in depth explanation of the mechanics of the situation (beware, physics ahead!):* *There are four forces acting on the balloon*

Fs

/\

N <<<<O|w|>>>Fe W is the location of the wall, included for reference.

\/

mg (mass times gravity, g, which is 9.8m/s^{2 }on earth)

Mg is the gravitational force, Fe we are already familiar with, Fs is the force of static friction caused by the balloon "sticking" to the wall, and N is the normal force, or the force that the wall exerts on the balloon; it is equal and opposite to the electrical force.

Since the balloon is not moving in the X or y direction, the net force on it is zero (max and may = 0). Thus:

Fs--mg = ma_{y} = 0

Fs=mg The balloon is actually held in place by the force of friction, which resists its motion down the wall.

The force of friction is equal to the normal force times the coefficient of friction (Us), which varies depending on what two surfaces are being moved across one another.

Fs= Us*N

N=Fe Recall that the normal force is, in this instance, equal to the electrical force.

Fs=Us*Fe

Us*(k(Q1Q2/r^{2})=mg Substitution of Coulomb's law.

Us=mg/(k(Q1Q2/r^{2})) Solve for Us. When we crunch the numbers, we find that for our example balloon, the minimum coeficcient of static friction necessary to keep the balloon in place is .174.

Of course, for any given coefficient of friction and mass of the balloon, you can easily determine the charge necessary to hold the balloon to the wall:

Us*(k(Q1Q2/r^{2})=mg

Q1Q2=mgr^{2}/(Us*k) If we remember our assumption that the charge induced in the wall is equal to the charge of the balloon, Q1=Q2, ergo Q1*Q2 = Q1^{2}. So:

Q_{balloon}=√(mgr^{2}/(Us*k))

TADA!