# Energy of orbits

In this note I will provide the proof of the following formula: $E_{tot}=-\dfrac{GMm}{2p}$ which is mentioned here

From here we know that: $T^2=\red{c}p^3..........[1]$ $\blue{c\space is\space constant\space for\space all\space revolving\space masses}$ $\blue{\therefore it\space must\space be\space the\space same\space in\space both,\space circular\space and\space non-circular\space orbits}$ $\blue{\therefore\space we\space can\space convert\space [1]\space into\space [2]}$ $\therefore T^2=\red{\dfrac{4\pi^2}{GM}}p^3..........[2]$ $\vec{s}=\dfrac{l}{1+e\cos\theta}(\hat{i}\cos\theta+\hat{j}\sin\theta)$ $\vec{v}=\dfrac{2k}{l}(-\hat{i}\sin\theta+\hat{j}(e+\cos\theta))$ $k=\dfrac{\pi pq}{T},l=\dfrac{q^2}{p}$ $\Rightarrow v^2=\vec{v}\cdot\vec{v}=\dfrac{4k^2}{l^2}(\red{\sin^2\theta}+e^2+\red{cos^2\theta}+2e\cos\theta)$ $=\dfrac{4\pi p^4}{T^2q^2}(\red{1}+e^2+2e\cos\theta)$ $=\dfrac{GMp}{q^2}(1+e^2+2e\cos\theta)\space (\blue{using \space [2]})$ $\Rightarrow E_K=\dfrac{1}{2}mv^2=\dfrac{GMmp}{2q^2}(1+e^2+2e\cos\theta)$ $E_P=-\dfrac{GMm}{|\vec{s}|}=-\dfrac{GMmp}{q^2}(1+e\cos\theta)$ $\Rightarrow E_{tot}=E_K+E_P=\dfrac{GMmp}{q^2}\left(\dfrac{1}{2}+\dfrac{e^2}{2}\cancel{+e\cos\theta}\right) - \dfrac{GMmp}{q^2}(1\cancel{+e\cos\theta})$ $=\dfrac{GMmp}{2q^2}(e^2-1)=-\dfrac{GMm\cancel{p}}{2\cancel{q^2}}\times\dfrac{\cancel{q^2}}{p^{\cancel{2}}}$ $=-\dfrac{GMm}{2p}$

Note by Zakir Husain
2 weeks, 2 days ago

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