Alireza Jamali's Blog
$$R=\sqrt[3]{\frac{GM}{H_0^2}} $$
$$R=\sqrt[4]{\frac{cGM}{H_0^3}}$$
$$\frac{dF}{da}=m$$
$$\frac{dF}{da}\frac{da}{d\alpha}=m=\frac{dF}{da}\frac{j}{\sqrt{j^2+j_0^2}}$$
$$j_0=\zeta cH_0^3$$
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