Sleeping Satellite

First, we can easily calculate the angular velocity of the satellite: $\omega=\sqrt{\frac{GM_0}{R_0^3}}$ where $M_0$ is the mass of the Earth and $R_0$ is distance between M and the center of the Earth.

Now, we consider the reference frame associate with M, which travel with angular velocity $\omega$ ,

Suppose when the satellite rotate by a very small angle $\theta$ .

Therefore, the total force into each point mass m which associate with rod b is:

$\Delta F_1= \Delta (m\omega^2R-\frac{GM_0m}{R^2})$

$=(m\omega^2+\frac{2GM_0m}{R_0^3})\Delta R=\frac{3GM_0m}{R_0^3} b\theta$ .

The total force into each point mass m which associate with rod a is:

$\Delta F_2=\frac{3GM_0m}{R_0^3} a$ .

The total torque into the system is: $I \theta ''=2\Delta F_1 b-2\Delta F_2 a\theta$ .

$2m(a^2+b^2) \theta ''=\frac{6GM_0m}{R_0^3} \theta (b^2-a^2)$ .

$\theta '' + \theta \frac{3\omega^2(a^2-b^2)}{a^2+b^2}=0$ .

In order to make the period of small oscillations of the satellite equals its orbital period, we have: $\frac{3(a^2-b^2)}{a^2+b^2}=1$ , so $\frac{a}{b}=\sqrt2 \approx 1.414$ .

When the satellite is tilted at an angle $\theta$ , the coordinates of its four masses can be broken down in the original coordinates as diagrammed in the figure below. Let the height of the mass $M$ relative to the center of the Earth be $R_{o}$ , the radius of orbit.

Fig. 1

Each mass is simultaneously pulled toward the Earth by gravity and pushed away from the Earth by the centrifugal force.

Consider the bottom most mass on the rod of length $a$ . It experiences a gravitational pull of magnitude $\displaystyle \frac{GM_Em}{\left(R_o - a\cos\theta\right)^2}$ as well as a centrifugal push of magnitude $m\omega^2 \left(R_o - a\cos\theta\right)$ .

Similarly, the top mass, also on a rod of length $a$ experiences experiences a gravitational pull of magnitude $\displaystyle \frac{GM_Em}{\left(R_o + a\cos\theta\right)^2}$ as well as a centrifugal push of magnitude $m\omega^2 \left(R_o + a\cos\theta\right)$ .

These forces act through the moment arm $a\sin\theta$ . The gravitational force on the bottom mass is contributes torque in the counter clockwise direction (negative) as does the centrifugal force on the top mass. Likewise, the gravitational force on the top mass and the centrifugal force on the bottom mass both contribute positive torque. Multiplying these forces by their moment arms and summing we find the net torque on the top and bottom mass given by

$\displaystyle -m\omega^22a^2\cos\theta\sin\theta + \frac{GM_Ema\sin\theta}{\left(R_o+a\cos\theta\right)^2} - \frac{GM_Ema\sin\theta}{\left(R_o-a\cos\theta\right)^2}$

Expanding the expression about $\theta=0$ to first order, we find that these terms contribute $\displaystyle \left(-2m\omega^2-\frac{4a^2R_o}{\left(R_o^2-a^2\right)^2}\right)\theta \approx -\left(2m\omega^2 + GM_Em\frac{4}{R_o^3}\right)\theta a^2$

A similar calculation yields a contribution of $\displaystyle \left(2m\omega^2 + GM_Em\frac{4}{R_o^3}\right)\theta b^2$ from the two masses on the rods of length $b$ .

The overall torque is then $\tau = \displaystyle \left(2m\omega^2 + GM_Em\frac{4}{R_o^3}\right)\theta \left(b^2-a^2\right)$

The torque on the satellite is balanced by its rotation, i.e. $\displaystyle \tau = \frac{dL}{dt} = I\ddot{\theta}$

The moment of inertia of the four points masses about the center of mass is given by $\displaystyle I = m(2a^2+2b^2)$ . Therefore, we have

$\begin{aligned} \displaystyle \ddot{\theta} &= \left(2\omega^2 + GM_E\frac{4}{R_o^3}\right) \frac{b^2-a^2}{a^2+b^2}\frac{\theta}{2} \\ &= \left(\omega^2 + GM_E\frac{2}{R_o^3}\right) \frac{1-\eta^2}{1+\eta^2}\theta \end{aligned}$

where we have made the replacement $\displaystyle \eta = \frac{a}{b}$ .

This is just the pendulum equation for small angle oscillations and so the frequency of the system is given by the square root of the negative of the coefficient on $\theta$ . We can clearly see that the coefficient is negative for $\eta > 1$ as suggested in the problem, making it a stable oscillator in that regime.

We can make one further simplification in that the gravitational pull on the satellite is balanced perfectly by the centrifugal force upon it, i.e.

$\displaystyle M_{total}\omega R_o = \frac{GM_EM_{total}}{R_o^3}$ and therefore $\displaystyle \omega^2 = \frac{GM_E}{R_o^3}$ .

The torque equation then becomes

$\displaystyle \ddot{\theta} = 3\omega^2 \frac{1-\eta^2}{1+\eta^2}\theta$

and we get a self consistent relation for $\omega$ :

$\displaystyle \omega^2 = 3\omega^2\frac{\eta^2-1}{1+\eta^2} \rightarrow 1+\eta^2 = 3\eta^2-3 \rightarrow 2\eta^2 = 4 \rightarrow \boxed{\eta = \sqrt{2}}$

Let $d$ be the distance between $M$ and center of earth (radius of orbit) , $M_{e}$ be the mass of earth , and $w$ be the angular velocity of satellite. For sake of clarity, let us name the rods as $A_{1}, A_{2} , B_{1}$ and $B_{2}$ , where $A_{1}$ is the top rod and $B_{1}$ is the right rod.

Now, first it should be clear that the small oscillations would be relative to $M$ , and $w^2$ = $\frac{GM_{e}}{d^3}$

Let's rotate the system by a small angle $\theta$ in clockwise direction . In reference frame of $M$ ,

We know that for $x ,, 1$ , ${(1 + x)}^n = 1 + nx$ , hence,

$F_{A_{1}} = mw^2(d + acos\theta) - \frac{GM_{e}m}{{(d + acos\theta)}^2} = \frac{3GM_{e}m a cos\theta}{d^3}$ upwards

$F_{A_{2}} = \frac{GM_{e}m}{{(d - acos\theta)}^2} - mw^2(d - acos\theta) = \frac{3GM_{e}ma \cos\theta}{d^3}$ downwards

$F_{B_{1}} = mw^2(d + bsin\theta) - \frac{GM_{e}m}{{(d + bsin\theta)}^2} = \frac{3GM_{e}m bsin\theta}{d^3}$ upwards.

$F_{B_{2}} = \frac{GM_{e}m}{{(d - bsin\theta)}^2} - mw^2(d - bsin\theta) = \frac{3GM_{e}mb sin\theta}{d^3}$ downwards.

Moment of inertia about axis passing through M and perpendicular to plane of oscillations = $ma^2 + ma^2 + mb^2 + mb^2 = 2m(a^2 + b^2)$

Hence taking torque about $M$ ,

$F_{A}(2asin\theta) - F_{B}(2bcos\theta) = 2m(a^2 + b^2)\alpha$

Now, approximate $sin\theta = \theta$ and $cos\theta = 1$ to get:

$2m(a^2 + b^2)\alpha = \frac{6GM_{e}m}{d^3} (a^2 - b^2)$

Hence, time period of oscillation = $2 \pi \sqrt{\frac{d^3({\eta}^2 + 1)}{3GM_{e}({\eta}^2 - 1)}}$ .

Equate this to time period of satellite ,i.e. $2 \pi \sqrt{\frac{d^3}{GM_{e}}}$ to get $\boxed{\eta = \sqrt{2}}$