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Step-by-Step Solution
Step 1: Identify the Key Data
• Period of nearer satellite, T_1 = 1\text{ hour} .
• Period of farther satellite, T_2 = 8\text{ hours} .
• Orbit radius of the nearer satellite, R_1 = 2\times10^3\text{ km} .
Step 2: Use Kepler’s Third Law to Find Farther Satellite’s Orbit Radius
For satellites orbiting the same planet, Kepler’s third law states T^2 \propto R^3 . Hence,
\displaystyle \frac{T_2^2}{T_1^2} \;=\; \frac{R_2^3}{R_1^3} \quad\Rightarrow\quad \frac{(8)^2}{(1)^2} \;=\; \frac{R_2^3}{(2\times10^3)^3}.
So,
\displaystyle 64 \;=\; \frac{R_2^3}{(2\times10^3)^3} \quad\Rightarrow\quad R_2^3 \;=\; 64\,\bigl(2\times10^3\bigr)^3.
Therefore,
\displaystyle R_2 \;=\; 4\,R_1 \;=\; 4\times (2\times10^3)\;\text{km} \;=\; 8\times10^3\;\text{km}.
Step 3: Compute the Angular Velocities
• Nearer satellite’s angular velocity:
\displaystyle \omega_1 \;=\;\frac{2\pi}{T_1} \;=\; \frac{2\pi}{1} \;=\; 2\pi\;\text{rad h}^{-1}.
• Farther satellite’s angular velocity:
\displaystyle \omega_2 \;=\;\frac{2\pi}{T_2} \;=\;\frac{2\pi}{8} \;=\;\frac{\pi}{4}\;\text{rad h}^{-1}.
Step 4: Express the Relative Position Vector
Let the planet be at the origin. The position vectors are:
\displaystyle \vec{r}_1(t) \;=\; R_1\bigl(\cos(\omega_1 t),\,\sin(\omega_1 t)\bigr), \quad
\vec{r}_2(t) \;=\; R_2\bigl(\cos(\omega_2 t),\,\sin(\omega_2 t)\bigr).
Then the vector from the nearer satellite to the farther satellite is:
\displaystyle \vec{R}(t) \;=\; \vec{r}_2(t)\;-\;\vec{r}_1(t).
Step 5: Find the Instant When They Are Closest
They are closest when they lie on the same radial line from the planet. At one such instant (take t=0 ),
\vec{r}_1(0) = (R_1, 0)
\vec{r}_2(0) = (R_2, 0)
Hence,
\displaystyle \vec{R}(0) = \bigl(R_2 - R_1,\,0\bigr) = \bigl(8\times10^3 - 2\times10^3,\,0\bigr) = (6\times10^3,\,0).
Step 6: Find the Rate of Change of the Angle of \vec{R}(t)
We denote by \alpha(t) the angle of \vec{R}(t) with respect to the x-axis. The instantaneous angular speed is:
\displaystyle \frac{d\alpha}{dt} \;=\; \frac{\vec{R}(t)\,\times\,\dot{\vec{R}}(t)}{\bigl|\vec{R}(t)\bigr|^2}\,\Bigg|_{t=0}.
Step 7: Calculate \dot{\vec{R}}(0)
Differentiate \vec{R}(t) = \vec{r}_2(t) - \vec{r}_1(t) :
\displaystyle \dot{\vec{r}}_1(t)
= \bigl(-R_1 \omega_1 \sin(\omega_1 t),\,R_1 \omega_1 \cos(\omega_1 t)\bigr),
\displaystyle \dot{\vec{r}}_2(t)
= \bigl(-R_2 \omega_2 \sin(\omega_2 t),\,R_2 \omega_2 \cos(\omega_2 t)\bigr).
Thus:
\displaystyle \dot{\vec{R}}(t)
= \dot{\vec{r}}_2(t) - \dot{\vec{r}}_1(t).
At t=0 :
\displaystyle \dot{\vec{r}}_1(0)
= \bigl(0,\;R_1 \omega_1\bigr)
= \bigl(0,\;2\times10^3 \times 2\pi\bigr)
= \bigl(0,\;4000\pi\bigr),
\displaystyle \dot{\vec{r}}_2(0)
= \bigl(0,\;R_2 \omega_2\bigr)
= \bigl(0,\;8\times10^3 \times \frac{\pi}{4}\bigr)
= \bigl(0,\;2000\pi\bigr).
Therefore,
\displaystyle \dot{\vec{R}}(0)
= \bigl(0,\;2000\pi\bigr) - \bigl(0,\;4000\pi\bigr)
= \bigl(0,\;-2000\pi\bigr).
Step 8: Evaluate the Cross Product and Magnitude
• \vec{R}(0) = (6000, 0)
• \dot{\vec{R}}(0) = (0, -2000\pi)
The 2D cross product (scalar value) is:
\displaystyle \vec{R}(0)\,\times\,\dot{\vec{R}}(0)
= 6000\times(-2000\pi) - 0\times 0
= -12\times10^6\,\pi.
The magnitude of \vec{R}(0) squared is:
\displaystyle \bigl|\vec{R}(0)\bigr|^2 = (6000)^2 = 3.6\times10^7.
Thus,
\displaystyle \frac{d\alpha}{dt}\bigg|_{t=0}
= \frac{-12\times10^6\,\pi}{3.6\times10^7}
= -\frac{\pi}{3}\;\text{rad h}^{-1}.
The negative sign indicates the direction of rotation as seen by the nearer satellite, but the magnitude of this observed angular speed is \displaystyle \frac{\pi}{3}\;\text{rad h}^{-1}.
Step 9: Identify the Value of x
We have
\displaystyle \text{Observed angular speed} = \frac{\pi}{x} \;\text{rad h}^{-1},
and we found it to be
\displaystyle \frac{\pi}{3}\;\text{rad h}^{-1}.
Hence,
\displaystyle \frac{\pi}{x} = \frac{\pi}{3} \quad\Rightarrow\quad x = 3.
Therefore, the required value of x is 3.
