View Single Post
D H
D H is offline
#6
Feb1-12, 07:54 PM
Mentor
P: 14,476
Quote Quote by Natko View Post
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

[tex]a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}[/tex]

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:

[tex]a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}[/tex]

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

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

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

[tex]\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}[/tex]

The first term in parentheses can be rewritten as

[tex]\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[/tex]

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

[tex]\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}[/tex]
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

[tex]
a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R
= \frac{2GM_{\text{moon}}r}{R^3}
[/tex]

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