Energy of orbits

In this note I will provide the proof of the following formula: Etot=GMm2pE_{tot}=-\dfrac{GMm}{2p} which is mentioned here

From here we know that: T2=cp3..........[1]T^2=\red{c}p^3..........[1] c is constant for all revolving masses\blue{c\space is\space constant\space for\space all\space revolving\space masses} it must be the same in both, circular and noncircular orbits\blue{\therefore it\space must\space be\space the\space same\space in\space both,\space circular\space and\space non-circular\space orbits}  we can convert [1] into [2]\blue{\therefore\space we\space can\space convert\space [1]\space into\space [2]} T2=4π2GMp3..........[2]\therefore T^2=\red{\dfrac{4\pi^2}{GM}}p^3..........[2] s=l1+ecosθ(i^cosθ+j^sinθ)\vec{s}=\dfrac{l}{1+e\cos\theta}(\hat{i}\cos\theta+\hat{j}\sin\theta) v=2kl(i^sinθ+j^(e+cosθ))\vec{v}=\dfrac{2k}{l}(-\hat{i}\sin\theta+\hat{j}(e+\cos\theta)) k=πpqT,l=q2pk=\dfrac{\pi pq}{T},l=\dfrac{q^2}{p} v2=vv=4k2l2(sin2θ+e2+cos2θ+2ecosθ)\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) =4πp4T2q2(1+e2+2ecosθ)=\dfrac{4\pi p^4}{T^2q^2}(\red{1}+e^2+2e\cos\theta) =GMpq2(1+e2+2ecosθ) (using [2])=\dfrac{GMp}{q^2}(1+e^2+2e\cos\theta)\space (\blue{using \space [2]}) EK=12mv2=GMmp2q2(1+e2+2ecosθ)\Rightarrow E_K=\dfrac{1}{2}mv^2=\dfrac{GMmp}{2q^2}(1+e^2+2e\cos\theta) EP=GMms=GMmpq2(1+ecosθ)E_P=-\dfrac{GMm}{|\vec{s}|}=-\dfrac{GMmp}{q^2}(1+e\cos\theta) Etot=EK+EP=GMmpq2(12+e22+ecosθ)GMmpq2(1+ecosθ)\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}) =GMmp2q2(e21)=GMmp2q2×q2p2=\dfrac{GMmp}{2q^2}(e^2-1)=-\dfrac{GMm\cancel{p}}{2\cancel{q^2}}\times\dfrac{\cancel{q^2}}{p^{\cancel{2}}} =GMm2p=-\dfrac{GMm}{2p}

Note by Zakir Husain
2 weeks, 2 days ago

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