Темная материя 5_0

Last modified by Alexey Popov on 2022/10/18 15:13

$v=\sqrt{\frac{GM}{R}}$

$v=\sqrt{GM R^3}\frac{1}{R^2}$

$\int \mathbf v \cdot d\mathbf S = \int 4\pi \sqrt{GM R^3} \frac{1}{V}dV$

$\int \mathbf v \cdot d\mathbf S = \int 4\pi \sqrt{\frac{3}{4\pi}G\rho} dV$

$\nabla \cdot \mathbf v = \sqrt{12\pi G\rho}$

$\nabla \cdot \mathbf a = \sqrt{12\pi H_0^2 G\rho}$

$\mathbf a=\sqrt{\frac{GH_0^2M}{R}}$

$\mathbf v^2=\sqrt{GH_0^2M R}$

$\mathbf v=\sqrt[4]{GH_0^2M R}$

$\mathbf v^4=GH_0^2M R$

$\mathbf s=GH_0^2\frac{M}{R^2}$

$\nabla \cdot \mathbf s=4\pi GH_0^2\rho$

$\mathbf a = \frac{GM}{R^2} + H_0\sqrt{\frac{GM}{R}}$

$v^2 = \frac{GM}{R} + H_0\sqrt{GMR}$

$\frac{G}{R}\sqrt{M}^2 + H_0\sqrt{GR}\sqrt{M} - v^2 = 0$

$D = G \left(H_0^2R + 4 \frac{1}{R}v^2\right)$

$\sqrt{M} = \frac{R(-H_0\sqrt{GR} + \sqrt{D})}{2G}$

$M = \frac{R^2}{4G^2} \left(-H_0\sqrt{GR} + \sqrt{D} \right)^2$

$\mathbf a = - \frac{GM}{R^2} -\frac{H_0}{\alpha}\sqrt{\frac{GM}{R}}$

$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{H_0}{\alpha}\sqrt{GMR}$

$\mathbf a = - \frac{\omega_n}{\rho_n R^2}\frac{H_0 M}{\alpha} -\frac{H_0}{\alpha}\sqrt{\frac{GM}{R}}$

$\mathbf a = H_0\sqrt{\frac{GM}{R}}-H_0^2\sqrt{\frac{G}{c^2}MR}$

$v^2 = H_0\sqrt{GMR}-H_0^2\sqrt{\frac{G}{c^2}MR^3}$

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Created by Alexey Popov on 2022/10/11 10:58

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