According to Kepler's third law, #T^2 propto R^3 implies omega propto R^{-3/2}#, if the angular velocity of the outer satellite is #omega#, that of the inner one is #omega times (1/4)^{-3/2} = 8 omega#.
Let us consider #t=0# to be an instant when the two satellites are collinear with the mother planet, and let us take this common line as the #X# axis. Then, the coordinates of the two planets at time #t# are #(R cos (8omega t),R sin (8omega t))# and #(4R cos (omega t),4R sin (omega t))#, respectively.
Let #theta# be the angle the line joining the two satellites makes with the #X# axis. It is easy to see that
#tan theta = (4R sin(omega t)-Rsin(8 omega t))/(4R cos(omega t)-Rcos(8 omega t)) = (4 sin(omega t)-sin(8 omega t))/(4 cos(omega t)-cos(8 omega t)) #
Differentiation yields
#sec^2 theta (d theta)/dt = d/dt (4 sin(omega t)-sin(8 omega t))/(4 cos(omega t)-cos(8 omega t))#
# = (4 cos(omega t)-cos(8 omega t))^-2 times #
#qquad [(4 cos(omega t)-cos(8 omega t))(4 omega cos(omega t)-8omega cos(8 omega t))-#
#qquad (4 sin(omega t)-sin(8 omega t))(-4omega sin(omega t)+8 omega sin(8 omega t)) ]#
Thus
# (4 cos(omega t)-cos(8 omega t))^2[1+((4 sin(omega t)-sin(8 omega t))/(4 cos(omega t)-cos(8 omega t)))^2](d theta)/dt#
# = 4 omega [(4 cos^2(omega t)-9 cos(omega t)cos (8 omega t) + 2 cos^2 (omega t))#
#qquad qquad + (4 sin^2(omega t)-9 sin(omega t)cos (8 omega t) +2sin^2 (omega t))]#
# = 4 omega [6-9cos(7 omega t)] implies #
#(17 -8 cos (7 omega t)) (d theta)/dt = 12 omega (2 - 3 cos (7 omega t)) implies#
#(d theta)/dt = 12 omega (2 - 3 cos (7 omega t))/(17 -8 cos (7 omega t)) equiv 12 omega f(cos(7 omega t))#
Where the function
#f(x) = (2-3x)/(17-8x) = 3/8 - 35/8 1/(17-8x)#
has the derivative
# f^'(x) = -35/(17-8x)^2<0#
and is hence monotonically decreasing in the interval #[-1,1]# .
Thus, the angular velocity #(d theta)/dt # is maximum when #cos(7 omega t)# is minimum, and vice versa.
So,
#((d theta)/dt) _"max" = 12 omega (2 - 3 times (-1))/(17-8 times (-1)) #
#qquad qquad qquad qquad = 12 omega times 5/25 = 12/5 omega#
#((d theta)/dt) _"min" = 12 omega (2 - 3 times 1)/(17-8 times 1) #
#qquad qquad qquad qquad = 12 omega times (-1)/9 = -4/3 omega#
and so the ratio between the two is :
#12/5 omega : -4/3 omega = -9:5#
Note The fact that #(d theta)/dt# changes sign is the cause for so called apparent retrograde motion
