A satellite revolving in a circular equatorial orbit of radius `R = 2.0 xx 10^(4)` km from west to east appears over a certain point at the equator every `11.6 h`. From these data, calculate the mass of the earth. `(G = 6.67 xx 10^(-11) N m^(2))`

Explanation

To calculate the mass of the Earth based on the given data about the satellite, we can follow these steps:

### Step 1: Convert the radius from kilometers to meters
Given:
- Radius \( R = 2.0 \times 10^4 \) km

Convert kilometers to meters:
\[ R = 2.0 \times 10^4 \text{ km} \times 1000 \text{ m/km} = 2.0 \times 10^7 \text{ m} \]

### Step 2: Convert the time period from hours to seconds
Given:
- Time period \( T = 11.6 \) hours

Convert hours to seconds:
\[ T = 11.6 \text{ hours} \times 3600 \text{ s/hour} = 41760 \text{ s} \]

### Step 3: Calculate the angular velocity of the satellite
The angular velocity \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{41760} \approx 1.505 \times 10^{-4} \text{ rad/s} \]

### Step 4: Use the formula for gravitational force and centripetal force
The gravitational force providing the centripetal force for the satellite is given by:
\[ F = \frac{GMm}{R^2} = m\omega^2 R \]
Where:
- \( G \) is the gravitational constant \( 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the Earth
- \( m \) is the mass of the satellite (which will cancel out)

### Step 5: Rearrange the equation to solve for the mass of the Earth \( M \)
From the equation:
\[ \frac{GMm}{R^2} = m\omega^2 R \]
Cancelling \( m \) from both sides:
\[ \frac{GM}{R^2} = \omega^2 R \]
Rearranging gives:
\[ M = \frac{\omega^2 R^3}{G} \]

### Step 6: Substitute the known values into the equation
Substituting \( \omega \), \( R \), and \( G \):
\[ M = \frac{(1.505 \times 10^{-4})^2 \times (2.0 \times 10^7)^3}{6.67 \times 10^{-11}} \]

### Step 7: Calculate the mass of the Earth
Calculating each part:
1. \( (1.505 \times 10^{-4})^2 \approx 2.260 \times 10^{-8} \)
2. \( (2.0 \times 10^7)^3 = 8.0 \times 10^{21} \)
3. \( 6.67 \times 10^{-11} \) remains the same.

Putting it all together:
\[ M = \frac{2.260 \times 10^{-8} \times 8.0 \times 10^{21}}{6.67 \times 10^{-11}} \approx 6.0 \times 10^{24} \text{ kg} \]

### Final Answer
The mass of the Earth is approximately \( 6.0 \times 10^{24} \) kg.

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