To analyze the statements regarding a geostationary satellite, we will evaluate each statement step by step.
### Step 1: Evaluate Statement 1
**Statement 1:** The orbital speed of a geostationary satellite is nearly 3.1 km/s.
- The formula for the orbital speed \( V \) of a satellite is given by:
\[
V = \sqrt{\frac{GM}{R + h}}
\]
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 (\( 5.97 \times 10^{24} \, \text{kg} \)),
- \( R \) is the radius of the Earth (\( 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)),
- \( h \) is the height of the geostationary satellite above the Earth's surface (\( 35800 \, \text{km} = 3.58 \times 10^7 \, \text{m} \)).
- Plugging in the values:
\[
V = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.4 \times 10^6 + 3.58 \times 10^7)}}
\]
- Calculate \( R + h = 6.4 \times 10^6 + 3.58 \times 10^7 = 4.22 \times 10^7 \, \text{m} \).
- Now, calculate \( V \):
\[
V \approx \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{4.22 \times 10^7}} \approx 3.07 \, \text{km/s}
\]
- Therefore, Statement 1 is **correct** as it approximates to 3.1 km/s.
### Step 2: Evaluate Statement 2
**Statement 2:** The rotation of the geostationary satellite around the Earth is from west to east.
- A geostationary satellite orbits the Earth in sync with the Earth's rotation. The Earth rotates from west to east.
- Therefore, the geostationary satellite also moves in the same direction, which is from west to east.
- Hence, Statement 2 is **correct**.
### Step 3: Evaluate Statement 3
**Statement 3:** The total mechanical energy of an orbiting satellite is smaller than its kinetic energy.
- The total mechanical energy \( E \) of a satellite in orbit is given by:
\[
E = K + U
\]
where \( K \) is the kinetic energy and \( U \) is the potential energy.
- The kinetic energy \( K \) is given by:
\[
K = \frac{1}{2} mv^2
\]
- The potential energy \( U \) is given by:
\[
U = -\frac{GMm}{r}
\]
- For a satellite in a stable orbit, the total mechanical energy can be expressed as:
\[
E = K + U = \frac{1}{2} mv^2 - \frac{GMm}{r}
\]
- It is known that for circular orbits, the total mechanical energy is half of the kinetic energy but negative:
\[
E = -\frac{1}{2} K
\]
- Since \( E \) is negative and \( K \) is positive, it follows that \( |E| < K \), meaning the total mechanical energy is indeed smaller than the kinetic energy.
- Thus, Statement 3 is **correct**.
### Conclusion
All three statements are correct:
- **Statement 1:** Correct
- **Statement 2:** Correct
- **Statement 3:** Correct
### Final Answer
All statements are true.
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