User:G123106/sandbox

Knowing k2 and solving for μ gives:

$$k_{2}\approx \frac{3/2}{1+\frac{19\mu }{2\rho gR}},\Rightarrow \mu \approx \frac{2pgR}{19}\left( \frac{3}{2k_{2}}-1 \right)$$

It is common to take $$k_{2}\ll 1\Rightarrow \frac{3/2}{1+\frac{19\mu }{2\rho gR}}\ll 1$$

Suppose that $$\alpha =\frac{19\mu }{2\rho gR}$$

than $$k_{2}=\frac{3/2}{1+a}\ll 1$$

which means that α must be a big number. So we can ignore the +1 and write $$k_{2}=\frac{3/2}{a}$$

So we have $$k_{2}\approx \frac{3/2}{\frac{19\mu }{2\rho gR}}\Rightarrow k_{2}\approx \frac{3pgR}{19\mu }$$

and eq(1) can be adjusted in a way that $$\mu \approx \frac{3pgR}{19k_{2}}$$

Also taking into account that

$$T=2\pi \sqrt{\frac{\alpha ^{3}}{GM}}$$

$$q_{tidal}=\frac{E}{t\cdot m}$$

$$n=\frac{2\pi }{T}\Rightarrow n=\sqrt{\frac{GM}{\alpha ^{3}}}=\left( \frac{GM}{\alpha ^{3}} \right)^{\frac{1}{2}}$$

$$q_{tidal}=\frac{63\left( GM \right)^{\frac{3}{2}}r^{5}k_{2}}{6Q}a^{-\frac{15}{2}}e^{2} \rightarrow q_{tidal}=\frac{63\cdot p\left( GM \right)^{\frac{5}{2}}r^{4}e^{2}a^{-\frac{15}{2}}}{38\cdot 3p\frac{GM}{r^{2}}rQ}19k_{2}$$

$$q_{tid}=63\rho n^{5}r^{4}e^{2}/38\mu Q\leftarrow q_{tidal}=\frac{63p\left[ \left( \frac{GM}{\alpha ^{3}} \right)^{\frac{1}{2}} \right]^{5}r^{4}e^{2}}{38\frac{3pgr}{19k_{2}}Q}\leftarrow $$

$$q_{tidal}=\frac{63\cdot p\left( GM \right)^{\frac{5}{2}}r^{4}e^{2}a^{-\frac{15}{2}}}{38\cdot 3p\frac{GM}{r^{2}}rQ}19k_{2}\leftarrow q_{tidal}=\frac{63\left( GM \right)^{\frac{3}{2}}r^{5}k_{2}}{6Q}a^{-\frac{15}{2}}e^{2}$$

προσπαθεια για ΑΠΟΔΕΙΞΗ της σχεσης. αρχιζουμε απο την κομψη καταληγουμε στην συνθετη που ειναι η γνησια και αρχικη.θελει ακομα καποιες τροποποιησεις