A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.

Explanation

To find the time period of a satellite at a height of 2.5R above the Earth's surface, we can follow these steps:

### Step 1: Determine the distance of the satellite from the center of the Earth
The radius of the Earth is denoted as \( R \). The height of the satellite above the Earth's surface is \( 2.5R \). Therefore, the distance \( R_B \) from the center of the Earth to the satellite is:
\[ R_B = R + 2.5R = 3.5R \]

### Step 2: Determine the distance of the geostationary satellite from the center of the Earth
The geostationary satellite is at a height of \( 6R \) above the Earth's surface. Therefore, the distance \( R_A \) from the center of the Earth to the geostationary satellite is:
\[ R_A = R + 6R = 7R \]

### Step 3: Use Kepler's Third Law
According to Kepler's Third Law, the square of the time period \( T \) of a satellite is directly proportional to the cube of the semi-major axis (distance from the center of the Earth):
\[ T^2 \propto R^3 \]
This can be expressed as:
\[ \frac{T_B^2}{T_A^2} = \frac{R_B^3}{R_A^3} \]
Where \( T_B \) is the time period of the satellite at height \( 2.5R \) and \( T_A \) is the time period of the geostationary satellite, which is \( 24 \) hours.

### Step 4: Substitute the values into the equation
Substituting the distances we found:
\[ \frac{T_B^2}{(24 \text{ hours})^2} = \frac{(3.5R)^3}{(7R)^3} \]

### Step 5: Simplify the equation
We can simplify the right side:
\[ \frac{T_B^2}{576} = \frac{(3.5)^3}{(7)^3} \]
Calculating the cubes:
\[ (3.5)^3 = 42.875 \quad \text{and} \quad (7)^3 = 343 \]
Thus,
\[ \frac{T_B^2}{576} = \frac{42.875}{343} \]

### Step 6: Solve for \( T_B^2 \)
Cross-multiplying gives:
\[ T_B^2 = 576 \times \frac{42.875}{343} \]
Calculating the right side:
\[ T_B^2 \approx 576 \times 0.124 = 71.424 \]

### Step 7: Calculate \( T_B \)
Taking the square root:
\[ T_B \approx \sqrt{71.424} \approx 8.45 \text{ hours} \]

### Final Answer
The time period of the satellite at a height of \( 2.5R \) from the Earth's surface is approximately \( 8.45 \) hours.

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