A geostationary satellite is orbiting the earth at a height of `5R` above the surface of the earth, `R` being the radius of the earth. The time period of another satellite in hours at a height of `2R` form the surface of the earth is

Options

A. `6/(sqrt(2))`
B. `5`
C. `10`
D. `6sqrt(2)`

Explanation

To solve the problem of finding the time period of a satellite at a height of \(2R\) above the Earth's surface, we will use Kepler's third law of planetary motion. Here's a step-by-step solution:

### Step 1: Understand the given data
We know that:
- The radius of the Earth is \(R\).
- The height of the geostationary satellite above the Earth's surface is \(5R\).
- The height of the second satellite above the Earth's surface is \(2R\).

### Step 2: Calculate the orbital radius of both satellites
1. **For the geostationary satellite**:
- The total radius \(r_1\) from the center of the Earth is:
\[ r_1 = R + 5R = 6R \]

2. **For the second satellite**:
- The total radius \(r_2\) from the center of the Earth is:
\[ r_2 = R + 2R = 3R \]

### Step 3: Apply Kepler's Third Law
Kepler's Third Law states that the square of the time period \(T\) of a satellite is directly proportional to the cube of the semi-major axis (orbital radius) of its orbit. Mathematically, this can be expressed as:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
Where:
- \(T_1\) is the time period of the geostationary satellite.
- \(T_2\) is the time period of the second satellite.

### Step 4: Substitute the known values
1. The time period \(T_1\) of a geostationary satellite is equal to the Earth's rotational period, which is \(24\) hours.
2. Substitute \(r_1 = 6R\) and \(r_2 = 3R\) into the equation:
\[ \frac{(24 \text{ hours})^2}{T_2^2} = \frac{(6R)^3}{(3R)^3} \]

### Step 5: Simplify the equation
1. The right side simplifies as follows:
\[ \frac{(6R)^3}{(3R)^3} = \frac{216R^3}{27R^3} = 8 \]
2. Thus, we have:
\[ \frac{(24)^2}{T_2^2} = 8 \]

### Step 6: Solve for \(T_2^2\)
1. Rearranging gives:
\[ T_2^2 = \frac{(24)^2}{8} \]
2. Calculate \(T_2^2\):
\[ T_2^2 = \frac{576}{8} = 72 \]

### Step 7: Find \(T_2\)
1. Taking the square root:
\[ T_2 = \sqrt{72} = 6\sqrt{2} \text{ hours} \]

### Final Answer
The time period of the satellite at a height of \(2R\) from the surface of the Earth is \(6\sqrt{2}\) hours.

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