A satellite is orbiting around earth with its orbit radius `16` times as great as that of parking satellite. What is the period of this satellite.

Answer

To find the period of the satellite orbiting the Earth with an orbit radius 16 times that of a parking satellite, we can use Kepler's Third Law of planetary motion. Here is the step-by-step solution:

### Step 1: Understand Kepler's Third Law
Kepler's Third Law states that the square of the period of revolution (T) of a satellite is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this can be expressed as:
\[ T^2 \propto r^3 \]
This means:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

### Step 2: Define Variables
Let:
- \( T_p \) = period of the parking satellite
- \( r_p \) = radius of the parking satellite's orbit
- \( T_s \) = period of the satellite we want to find
- \( r_s \) = radius of the satellite's orbit, which is given as \( r_s = 16 r_p \)

### Step 3: Apply Kepler's Third Law
Using Kepler's Third Law for both satellites, we can write:
\[ \frac{T_s^2}{T_p^2} = \frac{r_s^3}{r_p^3} \]

### Step 4: Substitute the Radius
Substituting \( r_s = 16 r_p \) into the equation gives:
\[ \frac{T_s^2}{T_p^2} = \frac{(16 r_p)^3}{r_p^3} \]

### Step 5: Simplify the Equation
This simplifies to:
\[ \frac{T_s^2}{T_p^2} = \frac{16^3 r_p^3}{r_p^3} = 16^3 \]
\[ \frac{T_s^2}{T_p^2} = 4096 \]

### Step 6: Solve for \( T_s \)
Taking the square root of both sides:
\[ \frac{T_s}{T_p} = \sqrt{4096} = 64 \]
Thus:
\[ T_s = 64 T_p \]

### Step 7: Substitute the Period of the Parking Satellite
The period of the parking satellite \( T_p \) is given as 1 day. Therefore:
\[ T_s = 64 \times 1 \text{ day} = 64 \text{ days} \]

### Final Answer
The period of the satellite is **64 days**.

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