Author Topic: Keplers Law (Read 1481 times) Tweet Share

naved_s9994 · « **on:** May 12, 2009, 08:57:43 pm »

Hey,

I know how to derive for R, but how would you derive for T, and what would it be?

:angel:

/0 · « **Reply #1 on:** May 12, 2009, 09:16:18 pm »

$T^2 = \left(\frac{4\pi^2}{GM}\right)R^3$

Let $\gamma = \left(\frac{4\pi^2}{GM}\right)$

So $2T\frac{dR}{dT} = \gamma \times 3R^2$

$\therefore \frac{dR}{dT} = \gamma \frac{3R^2}{2T}$

and so:

$\frac{dT}{dR} = \frac{2T}{3 \gamma R^2}$

TrueTears · « **Reply #2 on:** May 12, 2009, 09:17:52 pm »

Quote from: /0 on May 12, 2009, 09:16:18 pm

$T^2 = \left(\frac{4\pi^2}{GM}\right)R^3$

Let $\gamma = \left(\frac{4\pi^2}{GM}\right)$

So $2T\frac{dR}{dT} = \gamma \times 3R^2$

$\therefore \frac{dR}{dT} = \gamma \frac{3R^2}{2T}$

and so:

$\frac{dT}{dR} = \frac{2T}{3 \gamma R^2}$

Lol you actually derived for T :2funny: :2funny: :2funny: :2funny: :2funny:

I think he meant solve for T lolz

Mao · « **Reply #3 on:** May 13, 2009, 08:58:11 am »

/0: silly, it should be $2T \frac{dT}{dR} = \gamma 3R^2$ or $2T = \gamma 3R^2 \frac{dR}{dT}$

anyhow, to the original poster:

$T = \sqrt{\frac{4\pi^2}{GM}R^3}$

/0 · « **Reply #4 on:** May 13, 2009, 11:47:57 am »

true dat

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Author Topic: Keplers Law (Read 1481 times) Tweet Share

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