# Differences between the gravitational constants $G$ and $g$?

There's a formula (described by Sir Isaac Newton) that gives the force acting between two objects:

$$F = \frac{Gm_1m_2}{r^2}$$

And then there's a formula for weight of an object

$$w = mg$$

My question is, what's the difference between $g$ and $G$ (or force of gravity, just gravity and acceleration due to gravity). Analogies would help!

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"described by Sir Isaac Newton" This is not really true and it would be better if the question is edited by moderators to delete what is in the parentheses. The way it is written it implies that Newton wrote the well-known equation F=Gmm/rr. He did not. – Zeynel Oct 11 '12 at 1:19
@Zeynel: he certainly came up with the notion that gravitational force is proportional to the two masses and inversely proportional to their separation. That is 100% of the content of that equation. – Jerry Schirmer Jan 21 '14 at 14:41

Your weight $w$ on the surface of the Earth is the force $F$ that the Earth exerts on you. So, your weight is $w = F = \frac{G m_{Earth} m_{you}}{R^2}$, where $R$ is the radius of the Earth. To calculate someone else's weight, you'd have to replace the $m_{you}$ with their mass and repeat the calculation. You may notice that $\frac{G m_{Earth}}{R^2}$ remains constant. You can call that $g$ and it evaluates to about $9.8 \frac{meters}{sec^2}$.

It is also called the acceleration due to gravity because from $a = \frac{F}{m}$, using the above expression for $F$ and $m_{you}$ for mass, you are left with $g$. So, in free-fall, this is the rate at which you are accelerating close to the surface of the Earth.

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For one thing, they have different units. $g$ has units of acceleration while $G$ has units of $(length)^3 / mass / (time)^2$. Physically, $g$ really is the acceleration of a falling object. Usually this is specifically for an object falling toward the Earth (and near the Earth's surface), and $g \sim 9.8 m/s^2$. In fact, $$g = {G m_E \over R_E^2}$$ where $m_E$ and $R_E$ are the Earth's mass and radius. So you see that $g$ is really calculated from Newton's law of gravity applied to objects near the Earth's surface (so that $r$ is approximately $R_E$).

I think of $G$ as more or less just a numerical and unit constant. If we used different units for mass or length, then there wouldn't even be a need for $G$.

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To connect the two you need to know: Outside of a (homogenous) sphere the gravitational force on a massive body is the same as that of a point mass, located in the center of the sphere. So to good approximation the force of the Earth exerted on a body of mass $m$ is

$$F = \frac{G m_{Earth}m}{R^2}$$

where $R$ is some approximate value of the earth radius, the prefactor of the mass $m$ is summarized in the constant you call $g$. Gravity on earth is discussed in more detail on wikipedia.

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## protected by Qmechanic♦Jan 21 '14 at 19:05

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