Question on gravitation law derivation

by Thiafon
Tags: derivation, gravitation
Thiafon is offline
Feb2-13, 07:01 PM
P: 2
Hi, guys.
I was wondering on Newton's Gravity Law derivation, and I found this page:
Everything seems clear, but the first step is just killing me, because I can't get it.
Assuming small incremental changes in s; [tex]
\lim_{t\rightarrow 0} {s} \rightarrow 0
we have the following ratios
[itex]\frac{\omega}{\nu}[/itex]=[itex]\frac{s}{r}[/itex], and [itex]\frac{t}{T}[/itex]=[itex]\frac{s}{2πr}[/itex]
Could someone help me out? Explain, or just say, which part of math do I have to cover in order to understand that?
(btw, I did pre-calculus, and calculus, so concept of limits is familiar to me)
Thanks in advance.
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Drakkith is offline
Feb3-13, 10:52 AM
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It's just telling you the ratios of everything. The vector changes at the same rate that the distance traveled does since v and r are equal. If you double w you double s as well.

In the 2nd ratio, T is the total time of one orbital period and 2*Pi*R is the total distance of the orbit. As t changes, which is the time it takes to transverse the incremental time period s, s changes as well. If you double t you double s. Does that make sense?
Thiafon is offline
Feb3-13, 11:03 AM
P: 2
Yes, thank you a lot! I had kind of intuitive feeling about it, but I wasn't sure if it is correct.

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