# Proof of Newton's Gravitation Law

The Planets revolving arround the sun forms an ellipse (1st kepler's law). Let the polar equation of the ellipse be $$r=\dfrac{l}{1+e\cos\theta}$$.

Important thing to note that is we are assuming sun as the origin of the sysytem.

Let $\vec{s}$ be the position vector of the planet. $\Rightarrow \vec{s}=\dfrac{l}{1+e\cos\theta}(\hat{i}\cos\theta+\hat{j}\sin\theta)$ where $\theta$ is the angle of $\vec{s}$ on the $x-axis$

Now let $dA$ represent the area swept by the planet in time $dt$ $\Rightarrow \dfrac{dA}{dt}=k=constant\space\blue{From\space Kepler's \space 2_{nd}\space law}$ $\Rightarrow dt=\dfrac{dA}{k}$ $Also\space {dA}=\dfrac{1}{2}r^2{d\theta}$ $\Rightarrow dt=\dfrac{r^2d\theta}{2k}=\dfrac{l^2d\theta}{2(1+e\cos\theta)^2k}$ $\Rightarrow\vec{v}=\dfrac{d\vec{s}}{dt}=\dfrac{2(1+e\cos\theta)^2k\red{d\vec{s}}}{l^2\red{d\theta}}$ $=\dfrac{2(1+e\cos\theta)^2k}{l^2}\left( \dfrac{le\sin\theta}{(1+e\cos\theta)^2}(\hat{i}\cos\theta+\hat{j}\sin\theta) \dfrac{le\sin\theta}{1+e\cos\theta}(-\hat{i}\sin\theta+\hat{j}\cos\theta) \right)$ $=\dfrac{2ke\sin\theta}{l}(\hat{i}\cos\theta+\hat{j}\sin\theta)+\dfrac{2k(1+e\cos\theta)}{l}(-\hat{i}\sin\theta+\hat{j}\cos\theta)$ $=\dfrac{2ke\sin\theta\cos\theta\hat{i}}{l}-\dfrac{2k(1+e\cos\theta)\sin\theta\hat{i}}{l}+\dfrac{2ke\sin^2\theta\hat{j}}{l}+\dfrac{2k(1+e\cos\theta)\cos\theta\hat{j}}{l}$ $=\dfrac{2k\hat{i}}{l}(e\sin\theta\cos\theta-\sin\theta-e\sin\theta\cos\theta)+\dfrac{2k\hat{j}}{l}(e\sin^2\theta+\cos\theta+e\cos^2\theta)$ $=\dfrac{2k}{l}(-\hat{i}\sin\theta+\hat{j}(e+\cos\theta))$ $\Rightarrow \vec{a}=\dfrac{d\vec{v}}{dt}=\dfrac{d\vec{v}}{d\theta}\times\dfrac{2(1+e\cos\theta)^2k}{l^2}$ $=\dfrac{2k}{l}\times\dfrac{d(-\hat{i}\sin\theta+\hat{j}e+\hat{j}\cos\theta)}{d\theta}\times\dfrac{2(1+e\cos\theta)^2k}{l^2}$ $=-\dfrac{4k^2(1+e\cos\theta)^2}{l^3}(\hat{i}\cos\theta+\hat{j}\sin\theta)$ Comparing this with $r$ $\Rightarrow \vec{a} =-\dfrac{4k^2}{lr^2}(\hat{i}\cos\theta+\hat{j}\sin\theta)$ $\Rightarrow |\vec{a}|=\dfrac{4k^2}{lr^2}$ Now, if $p,q$ are semi-major and semi-minor axes of the ellipse and $T$ is the time period of the revolution $\Rightarrow k=\dfrac{\pi pq}{T},l=\dfrac{q^2}{p}$ Also from 3rd law of Kepler $T^2=cp^3,\space\blue{c\space is\space a \space constant\space for\space all\space planets}$ $\Rightarrow |\vec{a}|=\dfrac{4\pi^2\cancel{p^3q^2}}{c\cancel{p^3q^2}r^2}=\dfrac{4\pi^2}{cr^2}$ $\Rightarrow \boxed{|\vec{a}|\propto\dfrac{1}{r^2}}$

Note by Zakir Husain
3 months, 2 weeks ago

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