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 s\vec{s} be the position vector of the planet. s=l1+ecosθ(i^cosθ+j^sinθ)\Rightarrow \vec{s}=\dfrac{l}{1+e\cos\theta}(\hat{i}\cos\theta+\hat{j}\sin\theta) where θ\theta is the angle of s\vec{s} on the xaxisx-axis

Now let dAdA represent the area swept by the planet in time dtdt dAdt=k=constant From Keplers 2nd law\Rightarrow \dfrac{dA}{dt}=k=constant\space\blue{From\space Kepler's \space 2_{nd}\space law} dt=dAk\Rightarrow dt=\dfrac{dA}{k} Also dA=12r2dθAlso\space {dA}=\dfrac{1}{2}r^2{d\theta} dt=r2dθ2k=l2dθ2(1+ecosθ)2k\Rightarrow dt=\dfrac{r^2d\theta}{2k}=\dfrac{l^2d\theta}{2(1+e\cos\theta)^2k} v=dsdt=2(1+ecosθ)2kdsl2dθ\Rightarrow\vec{v}=\dfrac{d\vec{s}}{dt}=\dfrac{2(1+e\cos\theta)^2k\red{d\vec{s}}}{l^2\red{d\theta}} =2(1+ecosθ)2kl2(lesinθ(1+ecosθ)2(i^cosθ+j^sinθ)lesinθ1+ecosθ(i^sinθ+j^cosθ))=\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) =2kesinθl(i^cosθ+j^sinθ)+2k(1+ecosθ)l(i^sinθ+j^cosθ)=\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) =2kesinθcosθi^l2k(1+ecosθ)sinθi^l+2kesin2θj^l+2k(1+ecosθ)cosθj^l=\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} =2ki^l(esinθcosθsinθesinθcosθ)+2kj^l(esin2θ+cosθ+ecos2θ)=\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) =2kl(i^sinθ+j^(e+cosθ))=\dfrac{2k}{l}(-\hat{i}\sin\theta+\hat{j}(e+\cos\theta)) a=dvdt=dvdθ×2(1+ecosθ)2kl2\Rightarrow \vec{a}=\dfrac{d\vec{v}}{dt}=\dfrac{d\vec{v}}{d\theta}\times\dfrac{2(1+e\cos\theta)^2k}{l^2} =2kl×d(i^sinθ+j^e+j^cosθ)dθ×2(1+ecosθ)2kl2=\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} =4k2(1+ecosθ)2l3(i^cosθ+j^sinθ)=-\dfrac{4k^2(1+e\cos\theta)^2}{l^3}(\hat{i}\cos\theta+\hat{j}\sin\theta) Comparing this with rr a=4k2lr2(i^cosθ+j^sinθ)\Rightarrow \vec{a} =-\dfrac{4k^2}{lr^2}(\hat{i}\cos\theta+\hat{j}\sin\theta) a=4k2lr2\Rightarrow |\vec{a}|=\dfrac{4k^2}{lr^2} Now, if p,qp,q are semi-major and semi-minor axes of the ellipse and TT is the time period of the revolution k=πpqT,l=q2p\Rightarrow k=\dfrac{\pi pq}{T},l=\dfrac{q^2}{p} Also from 3rd law of Kepler T2=cp3, c is a constant for all planetsT^2=cp^3,\space\blue{c\space is\space a \space constant\space for\space all\space planets} a=4π2p3q2cp3q2r2=4π2cr2\Rightarrow |\vec{a}|=\dfrac{4\pi^2\cancel{p^3q^2}}{c\cancel{p^3q^2}r^2}=\dfrac{4\pi^2}{cr^2} a1r2\Rightarrow \boxed{|\vec{a}|\propto\dfrac{1}{r^2}}

Note by Zakir Husain
3 months, 2 weeks ago

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