3rd Kepler's law for Jupiter and the Earth:
\[\left(\frac{R_J}{R_E}\right)^3 = \left(\frac{T_J}{T_E} \right)^2 \quad \Rightarrow \quad R_J = R_E \left( \frac{T_J}{T_E} \right)^{2/3}=778.0\cdot 10^6~\text{km}\]3rd Kepler's law for the Earth and Io:
\[ \frac{4 \pi^2}{T_E^2} = \frac{GM_S}{R_E^3}, \quad \frac{4 \pi^2}{T_0^2} = \frac{GM_J}{R_M^3} \quad \Rightarrow \quad \frac{M_J}{M_S} = \frac{R_M^3}{R_E^3} \frac{T_E^2}{T_0^2} = 9.54 \cdot 10^{-4}\]
Over a time $T_\text{real}$ Jupter covers a small angle $2 \pi T_\text{real}/T_J$ on its orbit. Over the same time Io should cover $2\pi$ plus its this small angle on its oribt, thus
\[ 2 \pi \frac{T_\text{real}}{T_0} = 2\pi + 2 \pi \frac{T_\text{real}}{T_J}\]
The angular velocity of the rotating frame is $\omega_{SJ} = 2 \pi / T_J$. Therefore, the angular of the Earth is
\[ \omega = 2 \pi / T_E - \omega_{SJ} = 2\pi \left( \frac{1}{T_E} - \frac{1}{T_J} \right) = 2\pi \frac{T_J T_E}{T_J-T_E}\]Also, $v_E = R_E \omega$
Using the Pythagorean Theorem,
\[d^2 = R_E^2 \sin^2 \theta + (R_J - R_E \cos \theta)^2 = R_J^2 + R_E^2 - 2 R_J R_E \cos \theta. \]\[ d = R_J \left( 1 - \frac{R_E}{R_J} \cos \theta \right)\]
Delay for the light coming from the eclipse is $d/c$. Therefore to determine $T_\text{obs}$ we should calculate how $d$ changes over one period:
\[T_\text{obs} = T_0 + \frac{\dot{d} T_0}{c} = T_0 \left(1 + \frac{\omega R_E}{c} \sin \omega t \right)\]
Using the derived dependence,
\[ \Delta T = 2 T_0 \frac{\omega R_E}{c}, \quad c = 2 \omega R_E \frac{T_0}{\Delta T} \]