View Single Post
Mentor
P: 14,243
 Quote by Natko Thanks for the quick reply! Can you explain this in more understandable terms as I am only in Grade 9 please, especially what you mean by "gradient," "field," and the "1/r3 factor".
The earth as a whole is accelerating toward the moon. This acceleration is given by

$$a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}$$

where G is the universal gravitational constant, Mmoon, and R is the distance between the center of the moon and the center of the earth. Now imagine a drop of water on the surface of the earth directly on the line from the earth to the moon. This drop of water is a bit closer to the moon than is the center of the earth. Denoting the radius of the earth with a lower case r, the distance between the drop and the moon is R-r. This shorter distance means the drop experiences a slightly greater acceleration than does the earth as a whole:

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}$$

Factoring R out of R-r yields $R-r=R(1-r/R)$. Thus another way to write the acceleration of the drop is

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2}$$

The acceleration of the drop relative to the earth as a whole is simply the difference between these:

\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{(R-r)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \end{aligned}

The first term in parentheses can be rewritten as

$$\frac1{(1-r/R)^2} = 1+2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots$$

With this, the acceleration of the drop relative to the earth becomes

\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \\ &= \frac{GM_{\text{moon}}}{R^2} \left(2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots\right) \end{aligned}
Because the distance to the moon or the sun is much greater than the radius of the earth, the first term is going to be the dominant one. Thus

$$a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R = \frac{2GM_{\text{moon}}r}{R^3}$$

So, 1/R3. Substitute the moon for the sun and the same kind of inverse cube relation will hold.