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RKJP · Jul 15, 2011, 02:37 AM

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?

ebaines · Jul 15, 2011, 08:51 AM

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.