Another question which have taken hours without any success.
Problem is that I don't understand solution to question 1(every information is at the teht.jpg).
Could somebody explain those solution methods, please?
http://img707.imageshack.us/img707/9272/teht.jpg

Going back to basic orbital mechanics: the angular veocity of a planet about a massive star is governed by:


GM = \omega^2 R^3


So for planets 1 and 2 you have


\omega _1 = \sqrt {\frac {GM} {R_1^3}}\\
\omega _2 = \sqrt {\frac {GM} {R_2^3}}


As the inner planet passes the outer on the same side of the star the difference in velocities between the two planets is


\Delta v = \omega_1 R_1 - \omega_2 R_2


And the velocity of angular displacement of the inner star as seen from the outer (based on the planets being distance R_2 - R_1 apart) is:

\omega_{ang\ disp} = \frac {\Delta V} {R_2-R_1} = \frac {\omega_1 R_1 - \omega_2 R_2}{R_2 - R_1}


Similarly, the velocity of angular displacement for when the inner planet is on the far side of the star (i.e. the two planets are distance R2 + R_1 apart) is:


\omega_{ang\ disp} = \frac {- \Delta V} {R_2+R_1} = \frac {-(\omega_1 R_1 - \omega_2 R_2)}{R_2 + R_1}


The ratio of these two velocities of angular displacement is a number, let's call it N:



N = \frac {\frac {\omega_1 R_1 - \omega_2 R_2}{R_2 - R_1}} {\frac {-(\omega_1 R_1 - \omega_2 R_2)}{R_2 + R_1}} = - \frac {\frac {R_2}{R_1}+1}{\frac {R_2}{R_1} -1}


From the graph this ratio N appears to be about -4.6 (i.e, the slope of the line as it rises past the horizontal axis is about 14, and as it falls past the horizontal axis is about -5/2). So:


-4.6 = - \frac {\frac {R_2}{R_1}+1}{\frac {R_2}{R_1} -1}


Solve for \frac {R_2}{R_1} . I get a value of about 1.43.