This page discusses relatively simple problems involving weight and static forces on Earth's surface — for example, using the gravitational principles from the earlier discussion to compute the weight of masses that aren't moving.

• A 10-kilogram mass is sitting on a table.
• There are two equal, opposing forces — the mass presses down on the table, and the table presses up on the mass.
• The static gravitational force, the "weight", is proportional to:

  (1) $ \displaystyle f = mg$

  Where:

  • f = Force, newtons.
  • m = Mass, kilograms
  • g = Gravitational acceleration, m/s², described in the previous section.

• In this case, the mass presses down on the table with a force of about 98 Newtons, and the table presses against the mass with an equal, opposite force.
• Because the forces are equal, the mass doesn't move.
• Because the mass doesn't move and for reasons to be explained below, regardless of the amount of force involved, no power is required.
• To convert from force to units of weight, one applies a conversion factor. For kilograms of weight, divide the computed force by little-g. And remember that, confusingly, the kilogram is both a mass unit and a weight unit.

At this point, some readers may wonder why little-g exists. Can't we replace it with the Gravitational Force Equation? Doesn't that produce the same results? Well, the answer is yes, this is true — indeed it must be true. Little-g is only a convenience, a shortcut — computing results using the fundamental force equation must produce the same outcome or something is very wrong.

Little-g represents an intermediate result, a convenient acceleration term that takes some subtle issues into account like latitude and centripetal force, but all these could be taken into account in every computation if one wanted to proceed that way. Because of increasingly cheap computer power and advanced mathematical software, one may choose to use only one equation for everything.

Little-g allows us to say $f = mg$, which seems easier to understand than this equivalent equation:

(2) $ \displaystyle f = \frac{G m_1 m_2}{r^2} - \frac{m_2 v^2 cos(\phi)^2}{r} $

Where:

• f = Force, newtons
• G = Universal gravitational constant, described earlier
• $m_1$ = Mass of Earth
• $m_2$ = Small mass being evaluated
• r = Distance between $m_1$ and $m_2$
• v = Earth's equatorial rotation velocity
• $\phi$ = Latitude

On the other hand, equation (2) produces accurate results at any altitude (set r to equal earth's radius plus the desired altitude) and any latitude ($\phi$) on Earth. For everyday gravitational acceleration calculations where only a few decimal places are needed, it might seem to be an overly precise solution.

Some of the described terms have units of acceleration (m/s²), while others have units of force, usually newtons. Little-g expresses gravitational acceleration in units of m/s². It seems that we're using a term with units of acceleration to compute a force. How can we do that? Here is how:

(3) $ \displaystyle f = ma$

And

(4) $ \displaystyle a = \frac{f}{m}$

Where:

• f = Force, newtons
• a = Acceleration, m/s²
• m = Mass, kilograms

In general, if there is only one mass term in an equation (usually Earth's mass), the result has units of acceleration (because of the equivalence principle — which has the effect that different masses fall at the same rate in a gravitational field). But if there are two mass terms, in most cases the result has units of force.