# On moving planets

From here we get that : $r=\dfrac{l}{1+e\cos\theta}$ $k=\dfrac{dA}{dt}=\dfrac{r^2}{2}\dfrac{d\theta}{dt}$ $\Rightarrow \boxed{\purple{dt = \dfrac{l^2}{2k}\dfrac{d\theta}{(1+e\cos\theta)^2}}}$ $let\space I = \int \dfrac{d\theta}{(1+e\cos\theta)^2}$ $=\int \red{-}\dfrac{\red{d\cos\theta}}{\red{(1+e\cos\theta)^2}\sqrt{1-\cos^2\theta}}$ $\blue{\because \dfrac{d(1+e\cos\theta)^{-1}}{d\cos\theta}=\dfrac{-e}{(1+e\cos\theta)^2}}$ $\blue{\therefore \dfrac{-d\cos\theta}{(1+e\cos\theta)^2}=\dfrac{d(1+e\cos\theta)^{-1}}{e} }$ $\Rightarrow I=\dfrac{1}{e}\int\dfrac{d(1+e\cos\theta)^{-1}}{\sqrt{1-\cos^2\theta}}$ $let\space v=\dfrac{1}{1+e\cos\theta}$ $\blue{\Rightarrow u=\dfrac{1-v}{ev} }$ $\blue{\Rightarrow 1-u^2 = 1-\dfrac{(1-v)^2}{e^2v^2}} = \dfrac{(e^2-1)(v+(e^2-1)^{-1})^2 - \frac{e^2}{e^2-1}}{e^2v^2}$ $\Rightarrow I =\int\dfrac{vdv}{\sqrt{(e^2-1)(v+(e^2-1)^{-1})^2-\frac{e^2}{e^2-1}}}$ $let\space y=v+\dfrac{1}{e^2-1}\Rightarrow v=y-\dfrac{1}{e^2-1}$ $\Rightarrow I=\int \dfrac{\left(y-\frac{1}{e^2-1}\right)dy}{\sqrt{(e^2-1)y^2-\frac{e^2}{e^2-1}}}$ $\blue{\because (e^2-1)y^2-\dfrac{e^2}{e^2-1}=(1-e^2)\left(\dfrac{e^2}{(e^2-1)^2}-y^2 \right) }$ $\Rightarrow I=\dfrac{1}{\sqrt{1-e^2}}\int\dfrac{\left(y-\frac{1}{e^2-1}\right)dy}{\sqrt{\frac{e^2}{(e^2-1)^2}-y^2}}$ $\blue{\because d\arcsin\left(\dfrac{(e^2-1)}{e}y\right) = \dfrac{dy}{\sqrt{ \frac{e^2}{(e^2-1)^2} -y^2 }} }$ $\therefore I=\dfrac{1}{\sqrt{1-e^2}}\int (y-1\dfrac{1}{e^2-1})d\arcsin\left(\dfrac{(e^2-1)}{e}y\right)$ $=\dfrac{1}{\sqrt{1-e^2}}\left(\int yd\arcsin\left(\dfrac{(e^2-1)}{e}y\right)-\dfrac{1}{e^2-1}\arcsin\left(\dfrac{(e^2-1)}{e}y\right)\right)$ $=\dfrac{1}{(1-e^2)^{\frac{3}{2}}}\left( \sqrt{e^2-(e^2-1)^2y^2} + \arcsin\left(\dfrac{e^2-1}{e}y\right) \right)+c..........\red{[1]}$ $but\space if \space you \space will \space graph \space the \space derivative \space of \space \red{[1]} \space and \space the \space integrand \space of \space I$ $They \space will \space overlap \space only \space when \space -\pi\leq \theta \leq 0$ $To \space make \space it \space work \space for \space -\pi\leq \theta\leq \pi \space we \space will \space do \space a \space small \space modification:$ $I=-\dfrac{\theta}{|\theta|}\dfrac{1}{(1-e^2)^{\frac{3}{2}}}\left( \sqrt{e^2-(e^2-1)^2y^2} + \arcsin\left(\dfrac{e^2-1}{e}y\right) \right)+c$

$\large{\purple{CONCLUSION}}$ $If\space t(\theta_0)\space represents \space the \space time \space taken \space by \space the \space planet \space to \space move \space from \space \theta=0 \space to \space \theta=\theta_0$ $\Rightarrow t(\theta)=-\dfrac{\theta}{|\theta|}\times\dfrac{l^2}{2k(1-e^2)^{\frac{3}{2}}}(f(\theta)-f(0))$ $where\space f(x)=\sqrt{e^2 -(e^2-1)^2g(x)^2}+\arcsin\left(\dfrac{e^2-1}{e}g(x)\right)$ $where\space g(x)=\dfrac{1}{e^2-1}+\dfrac{1}{1+e\cos x}$

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• If there is any typing error, math error, etc, please comment so that I can correct that.

Note by Zakir Husain
3 weeks, 3 days ago

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