(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am studying gravitation and I have been trying to derive Kepler's Third Law using Kepler's Second Law.

2. Relevant equations

The second law :

[tex]\frac{dA}{dt}[/tex] = [tex]\frac{L}{2m}[/tex]

3. The attempt at a solution

To start with, I thought if we take [tex]\int[/tex] [tex]\frac{L}{2m}dt[/tex] from 0 to T; we would find Area of the ellipse.

We also know that [tex]A = \Pi * a * b[/tex]. So, to find T, I could use these two equations.

Moreover, to eliminate b, eccentricity could be used.

[tex]e^{2} = 1 - \frac{b^{2}}{a^{2}}[/tex]

substituting b would make

[tex]A = \Pi * a^{2} * \sqrt{1 - e^{2}}[/tex]

[tex]\frac{L.T}{2m} = \Pi * a^{2} * \sqrt{1 - e^{2}} [/tex]

and we know that

[tex] L = m.Vtan.R[/tex]

where Vtan is tangential component of V. (to the radius R)

And we also know that since T is period, [tex]R = ea + a[/tex]

Thus we can write

[tex] \frac{m.Vtan.(ea + a)}{2m} .T= \Pi * a^{2} * \sqrt{1 - e^{2}} (1)[/tex]

Now, we need to express Vtan. If we write centripetal force, we can conclude that

[tex]Fnet = m.\frac{Vtan^{2}}{R} = \frac{G.m.Msun}{R^{2}}[/tex]

where

[tex]R = ea + a[/tex]

and Msun is the mass of the sun.

[tex] Vtan = \sqrt{\frac{G Msun}{ea + a} [/tex]

and finally, we can rearrange equation (1) as

[tex] T = \frac{2.\Pi.a^{3/2}\sqrt{1-e^{2}}}{\sqrt{GMsun}.\sqrt{1+e}}[/tex]

simplfying a little bit would yield to the result

[tex] T = \frac{2.\Pi.a^{3/2}\sqrt{1-e}}{\sqrt{GMsun}}[/tex]

I seems to be close, but wrong!

Kepler's third law :

[tex] T = \frac{2.\Pi.a^{3/2}}{\sqrt{GMsun}}[/tex]

So the term with squareroot and 1-e shouldn't have been there... I have been trying to find what I'm doing wrong, and I thought my assumption that

[tex] Vtan = \sqrt{\frac{G Msun}{ea + a} [/tex]

seems to be wrong. Although it sounds logical to me... I am stuck here and need your help.

Thanks in advance.

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# Homework Help: Derivation of Kepler's Third Law

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