# The gravitational field at a mass point produced by an infinite plane

1. Apr 25, 2014

### 3102

http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S4

1. In the link it says:
$2\pi\rho d\rho$ is the area of the ring of the radius $\rho$ and width $d\rho$, if $d\rho \ll \rho$.

Why is this true??

2. A bit further down in the text it says:
Since $r^2 = \rho^2 + a^2$, $\rho\,d\rho = r\,dr$.

How do you get this result??

3. Even further down, after "But we see that", it says:
$r^2 = ... = a^2 + R^2 - 2Rx$ Then: $2r\,dr=-2R\,dx$

How do you get that?

It would be great if you answered my questions and if you recommended me some book or link that shows how to get these mathematical results.

Last edited: Apr 25, 2014
2. Apr 25, 2014

### Staff: Mentor

The area of the ring is the difference between the area of a circle with radius $\rho$ and the area of a circle with radius $\rho+d\rho$; the ring is what's left if I cut the smaller circle out of the bigger circle. The condition that $d\rho \ll \rho$ allows us to ignore the $(d\rho)^2$ terms.

Differentiate both sides with respect to $\rho$, and a bit of algebra will take you the rest of the way.

(Note that Feynman has moved on to a different problem here)
The same trick as in #2 works here.

These results are all based on introductory-level calculus, and any decent intro textbook will give you the necessary background.

However, it takes a fair amount of practice (not hard, just practice) to get to where you can reach into the mathematical toolbox and pull out the right technique without thinking about it. The exercises in a calculus-based but pragmatic mechanics textbook such as Kleppner and Kolenkow would be a good start.

Last edited: Apr 25, 2014
3. Apr 25, 2014

### 3102

Thanks for your fast and great answer, but unfortunately I don't understand everything.

ad question2) I don't understand what's going on here. I've used your advise and differentiated both sides with respect to $\rho$.
On the right side I've got $f(\rho)=\rho^2+a^2$ this is easy to differentiate: $f'(\rho)=2\rho$, since $a^2$ is a constant.

The left side isn't that easy: $f(\rho)=r(\rho)^2$.

$$f'(\rho)=\dfrac{r}{dr}\cdot\dfrac{dr}{d\rho}$$

$$f'(\rho)=1\cdot\dfrac{dr}{d\rho}$$

If I know set both sides together I get: $1\cdot\dfrac{dr}{d\rho}=2\rho$. Then I multiply both sides with $dp$ and divide them by $2$: $\frac{1}{2} \cdot dr = 2\rho \, d\rho$. This is not the same as $r\,dr=\rho\,d\rho$. What did I do wrong?

ad question3) I think: If I'm able to understand question2, I will also understand question3.

4. Apr 25, 2014

### Staff: Mentor

$\frac{d}{d\rho}(r^2) = 2r\frac{dr}{d\rho}$

$\frac{d}{d\rho}(\rho^2+a^2) = 2\rho$