The time period of the earth's satellite revolving at a height of 35800 km is

Options

A. 24 h
B. 100 min
C. 12 h
D. 48 h

Explanation

To find the time period of the Earth's satellite revolving at a height of 35800 km, we can follow these steps:

### Step 1: Understand the formula for the time period of a satellite
The time period \( T \) of a satellite can be derived from the formula:
\[ T = 2\pi \sqrt{\frac{(R + H)^3}{GM}} \]
where:
- \( R \) is the radius of the Earth,
- \( H \) is the height of the satellite above the Earth's surface,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth.

### Step 2: Gather the necessary values
- Radius of the Earth \( R \approx 6400 \) km \( = 6.4 \times 10^6 \) m
- Height of the satellite \( H = 35800 \) km \( = 3.58 \times 10^7 \) m
- Mass of the Earth \( M = 6 \times 10^{24} \) kg
- Universal gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

### Step 3: Calculate \( R + H \)
Convert both values to meters and add them:
\[ R + H = 6.4 \times 10^6 \, \text{m} + 3.58 \times 10^7 \, \text{m} = 4.22 \times 10^7 \, \text{m} \]

### Step 4: Substitute values into the time period formula
Now substitute \( R + H \), \( G \), and \( M \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{(4.22 \times 10^7)^3}{(6.67 \times 10^{-11})(6 \times 10^{24})}} \]

### Step 5: Calculate \( (R + H)^3 \)
Calculate \( (4.22 \times 10^7)^3 \):
\[ (4.22 \times 10^7)^3 = 7.52 \times 10^{22} \, \text{m}^3 \]

### Step 6: Calculate the denominator
Calculate \( G \times M \):
\[ (6.67 \times 10^{-11})(6 \times 10^{24}) = 4.002 \times 10^{14} \, \text{N m}^2/\text{kg} \]

### Step 7: Substitute and simplify
Now substitute these values back into the formula for \( T \):
\[ T = 2\pi \sqrt{\frac{7.52 \times 10^{22}}{4.002 \times 10^{14}}} \]

### Step 8: Calculate the final value
Calculate the fraction:
\[ \frac{7.52 \times 10^{22}}{4.002 \times 10^{14}} \approx 1.88 \times 10^{8} \]
Then take the square root:
\[ \sqrt{1.88 \times 10^{8}} \approx 13710 \, \text{s} \]
Finally, multiply by \( 2\pi \):
\[ T \approx 2\pi \times 13710 \approx 86164 \, \text{s} \approx 86400 \, \text{s} \text{ (which is 24 hours)} \]

### Conclusion
The time period of the Earth's satellite revolving at a height of 35800 km is approximately **24 hours**.

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