A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution is :

Answer

To solve the problem of determining the time period of a satellite that moves in a circular orbit around the Earth, where the radius of its orbit is half that of the Moon's orbit, we can follow these steps:

### Step 1: Understand the relationship between the radius and the time period
According to Kepler's Third Law of planetary motion, the square of the time period (T) of a satellite is directly proportional to the cube of the radius (r) of its orbit. Mathematically, this is expressed as:
\[ T^2 \propto r^3 \]

### Step 2: Define the variables
Let:
- \( R_m \) = radius of the Moon's orbit
- \( R_s \) = radius of the satellite's orbit
- \( T_m \) = time period of the Moon
- \( T_s \) = time period of the satellite

From the problem, we know that:
\[ R_s = \frac{1}{2} R_m \]

### Step 3: Set up the equation using Kepler's Third Law
Using Kepler's Third Law, we can write:
\[ \frac{T_s^2}{T_m^2} = \frac{R_s^3}{R_m^3} \]

### Step 4: Substitute the value of \( R_s \)
Substituting \( R_s = \frac{1}{2} R_m \) into the equation gives:
\[ \frac{T_s^2}{T_m^2} = \frac{\left(\frac{1}{2} R_m\right)^3}{R_m^3} \]

### Step 5: Simplify the equation
This simplifies to:
\[ \frac{T_s^2}{T_m^2} = \frac{\frac{1}{8} R_m^3}{R_m^3} = \frac{1}{8} \]

### Step 6: Solve for \( T_s \)
Taking the square root of both sides, we find:
\[ \frac{T_s}{T_m} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]
Thus,
\[ T_s = T_m \cdot \frac{1}{2\sqrt{2}} \]

### Step 7: Express the final answer
To express the time period of the satellite in terms of the Moon's time period:
\[ T_s = 2^{-\frac{3}{2}} \cdot T_m \]

This is the required time period of the satellite.