What is a period of revolution of the earth satellite ? Ignore the height of satellite above the surface of the earth.
Given,
(i) the value of gravitational acceleration, `g = 10 ms^(-2)`
(ii) radius of the earth, `R_(g) =6400` km (take, `pi = 3.14`)

Options

A. 85 min
B. 156 min
C. 83.73 min
D. 90 min

Explanation

To find the period of revolution of an Earth satellite, we can use the formula for the period \( T \) of a satellite in circular orbit:

\[ T = 2\pi \sqrt{\frac{r^3}{g}} \]

Where:
- \( T \) is the period of revolution,
- \( r \) is the radius of the orbit (which, in this case, is approximately equal to the radius of the Earth since we are ignoring the height of the satellite above the Earth's surface),
- \( g \) is the acceleration due to gravity.

### Step 1: Convert the radius of the Earth to meters
Given the radius of the Earth \( R_g = 6400 \) km, we need to convert this to meters:

\[ R_g = 6400 \text{ km} = 6400 \times 1000 \text{ m} = 6.4 \times 10^6 \text{ m} \]

### Step 2: Substitute the values into the formula
Now, we can substitute \( r = R_g = 6.4 \times 10^6 \) m and \( g = 10 \text{ m/s}^2 \) into the formula for \( T \):

\[ T = 2\pi \sqrt{\frac{(6.4 \times 10^6)^3}{10}} \]

### Step 3: Calculate \( (6.4 \times 10^6)^3 \)
Calculating \( (6.4 \times 10^6)^3 \):

\[ (6.4 \times 10^6)^3 = 6.4^3 \times (10^6)^3 = 262.144 \times 10^{18} \text{ m}^3 \]

### Step 4: Substitute back into the formula
Now, substituting this back into the formula for \( T \):

\[ T = 2\pi \sqrt{\frac{262.144 \times 10^{18}}{10}} = 2\pi \sqrt{26.2144 \times 10^{17}} \]

### Step 5: Calculate the square root
Calculating the square root:

\[ \sqrt{26.2144 \times 10^{17}} = \sqrt{26.2144} \times 10^{8.5} \]

Calculating \( \sqrt{26.2144} \approx 5.12 \):

\[ \sqrt{26.2144 \times 10^{17}} \approx 5.12 \times 10^{8.5} \]

### Step 6: Calculate \( T \)
Now substituting back into the equation for \( T \):

\[ T = 2 \times 3.14 \times 5.12 \times 10^{8.5} \]

Calculating this gives:

\[ T \approx 32.0 \times 10^{8.5} \text{ seconds} \]

### Step 7: Convert seconds to minutes
To convert seconds to minutes, we divide by 60:

\[ T \approx \frac{32.0 \times 10^{8.5}}{60} \approx 83.73 \text{ minutes} \]

### Final Answer
Thus, the period of revolution of the Earth satellite is approximately:

\[ \boxed{83.73 \text{ minutes}} \]