# The Cosmologies of Galileo and Newton

## Example 3: An apple of my eye...litte $g$ and big $G$ 

**The Question:** 

Place your apple on the ground—notice that it’s distance from the center of the Earth is $R_E$. Let’s calculate the force on that little apple with mass $m$ due to the big Earth, with mass $M_E$. Here's a picture:

<img align="center" src="./appleearth.png">

<BR CLEAR="all">

**The Answer:**

Newton taught us that the force between them is from Newton’s Gravitational law

$$
F = G\cfrac{M_Em}{R_E^2}.
$$

Now isolate the little *m* outside of the other terms:

$$
F = m\left(G\cfrac{M_E}{R_E^2}\right)=ma \nonumber
$$

and can you see that we’ve discovered an acceleration buried in the middle term by recognizing $F=ma$ in it:

$$
a = G\cfrac{M_E}{R_E^2}.
$$

Since this situation is an *apple on the surface of the Earth*, what we’ve really found is a derivation for Galileo’s $g$! So we can just identify:

$$
g = G\cfrac{M_E}{R_E^2}.
$$ (gG)

All constants. Try it: 

* $M_E=5.972 \times 10^{24}$ kg
* $R_E= 6378 \times 10^3$ m
* $G=6.67 \times 10^{-11}$ m$^3$kg$^{-1}$m$^{-2}$

$$
g = G\frac{M_E}{R_E^2}=\left( 6.67 \times 10^{-11} \right) \frac{5.972 \times 10^{24}}{ (6378 \times 10^3)^2} = 9.79 \text{ m/s}^2
$$

Bingo.