MOND

Last modified by Alexey Popov on 2022/05/31 18:21

$\frac{\mathbf a^2}{a_0} = \frac{G M }{r^2}$

$\mathbf a = \sqrt{\frac{G M }{r^2}a_0}$

$\mathbf a = \sqrt{\frac{G M_G }{r^2}\frac{G M_U H_0^2}{c^2}}$

$\mathbf a = \frac{G H_0}{c}\frac{\sqrt{M_G M_U}}{r}$

$\mathbf v^2 = \frac{G H_0}{c}\sqrt{M_G M_U}$

$\mathbf v^2 = \frac{G \omega_n}{c}\sqrt{M_G m_e}$

$\mathbf a = \frac{G \omega_n}{c}\frac{\sqrt{M_G m_e}}{r}$

$\Delta \mathbf a = \frac{G \omega_n}{c}\frac{d \sqrt{M_G m_e}}{dV}$

$\frac{\mathbf v^4}{m_e} \frac{c^2}{G^2\omega_n^2}= M_G$

$\frac{\mathbf v^4}{M_U} \frac{c^2}{G^2 H_0^2}= M_G$

$\frac{1}{m_e} \frac{1}{\omega_n^2}= \frac{1}{M_U} \frac{1}{ H_0^2}$

$m_e\omega_n^2 = M_U H_0^2$

$H_0 = \sqrt{\frac{m_e \omega_n^2}{M_U}}$

$\nabla \cdot \mathbf a = \frac{G}{\omega_n\sqrt{\frac{m_e }{M_U}}}\rho \omega_n\sqrt{\frac{m_e }{M_U}}$

$a = \frac{G}{r_n^2} m_e \pi^{1/3}$

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Created by Alexey Popov on 2022/02/02 07:52

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