Two satellites P_"1" and P_"2" are revolving in orbits of radii R and 4R. The ratio of maximum and minimum angular velocities of the line joining P_"1" and P_"2" is ??

1 Answer
Jun 16, 2018

-9/5

Explanation:

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