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.

Aren't they different?

Aren't they different?

I read your page because I have a few questions. On your page you have a link what's explaining the Big G. But I don't understand how in the equation F1=F2=G((m1xm2)/r2) F1=F2 with the explaining text "the attractive force (F) between two bodies is proportional to the product of their masses (m1 and m2)". If m1 is earth and m2 is the moon, then both should have the same force? Can't believe that, but may be I'm mixing up the big G with g. I can understand G((m1xm2)/r2), but I think that it will be different for F1 and F2. I'm not sure if I wrote the equation correct in this way. Remember that force and acceleration are different things. To give a simple example, imagine that a Mack truck and a ping-pong ball are connected by a rubber band. The rubber band is trying to pull the Mack truck and the ping-pong ball together with a force of one Newton.

Now try to explain how the force on one end of the rubber band is different than the force on the other end. How would that be possible? The ping-pong ball experiences the force in a different direction, but it's the same amount of force.

We can compute force F, for masses M₁ and M₂, a separation between them of r, and gravitational force G:

$ \displaystyle F = \frac{G M_1 M_2}{r^2}$

Where:

F = force, Newtons

G = universal gravitational constant

M₁ = mass 1, kilograms

M₂ = mass 2, kilograms

r = radius that separates masses M₁ and M₂, meters

The force F in the above equation is the same for both masses, no matter how different they are. The masses experiences the force in an opposite direction, but the amount of force is the same.

But — very important — the acceleration experienced by the ping-pong ball (if it is allowed to move) is much greater than the acceleration experienced by the Mack truck. This is because acceleration depends on mass:

$ \displaystyle a = \frac{F}{M}$

This means that, for a given force, a more massive object M₁ experiences less acceleration than a less massive object M₂. For a given force, the acceleration an object experiences is inversely proportional to its mass.

Here's a thought experiment: imagine a ten-kilogram object M₁ and a one-kilogram object M₂, sitting on perfectly smooth ice, connected by a rubber band. The rubber band is exerting a force of one Newton. If the masses are released from constraint, the less massive object M₂ will move toward the more massive object M₁ at ten times the rate of its partner.

Imagine further that you anchor mass M₁ at position A on the smooth ice, and anchor M₂ at position B. You are required in advance to draw a line on the ice where they will meet when they are released. Don't read ahead — think about it.

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The line should be drawn at one-tenth the distance between M₁ and M₂, nearest to M₁ (the more massive object). When the masses are released, and assuming a lot of things that aren't usually true in a real experiment, like no friction and an ideal rubber band, the two masses will collide at a location at 1/10 the original distance, but nearest to mass M₁.

In the real world, one of planets instead of masses on a smooth sheet of ice, two orbiting planets, regardless of their relative masses, are actually orbiting around a point defined by the difference between their masses. For example, if the solar system consisted only of the sun and Jupiter, the center of their rotation would not be the center of the sun as is commonly thought, but a location near the sun's surface, a location defined by the difference between their masses.

The most important difference between force and acceleration is that acceleration requires motion, which requires time. To expand a bit further, in order to fully understand how force and acceleration differ, you need to learn Calculus, a way to compute motion and dynamic change. There are a number of excellent reasons to learn Calculus, this is just one. (Obligatory link to Calculus primer.)

I hope this helps.