To solve the problem, we need to analyze the motion of the Earth as it orbits the Sun in a circular path with a constant speed of 30 km/s. We will evaluate the statements provided in the question regarding average velocity, average acceleration, average speed, and instantaneous acceleration.
### Step-by-Step Solution:
1. **Understanding Circular Motion**:
The Earth is in uniform circular motion around the Sun. In uniform circular motion, the speed is constant, but the direction of the velocity changes continuously.
2. **Average Velocity Calculation**:
- Average velocity is defined as the total displacement divided by the total time taken.
- If we consider the Earth starting from a point and moving halfway around the Sun (from January 1 to June 30), the displacement is not zero; it is the straight line distance from the starting point to the point directly opposite on the circular path.
- Therefore, the average velocity from January 1 to June 30 is not zero.
3. **Average Acceleration Calculation**:
- The average acceleration in circular motion is given by the centripetal acceleration, which is directed towards the center of the circular path (the Sun).
- The formula for centripetal acceleration \( a_c \) is given by:
\[
a_c = \frac{v^2}{r}
\]
- Here, \( v = 30 \, \text{km/s} \) and \( r = 1.5 \times 10^8 \, \text{km} \).
- Substituting the values:
\[
a_c = \frac{(30 \, \text{km/s})^2}{1.5 \times 10^8 \, \text{km}} = \frac{900}{1.5 \times 10^8} \approx 6 \times 10^{-6} \, \text{km/s}^2
\]
- This value is not equal to 60 km/s², so the statement about average acceleration is incorrect.
4. **Average Speed Calculation**:
- Average speed is defined as the total distance traveled divided by the total time taken.
- Over the course of one year, the Earth travels a distance equal to the circumference of its orbit:
\[
\text{Distance} = 2 \pi r = 2 \pi (1.5 \times 10^8 \, \text{km}) \approx 9.42 \times 10^8 \, \text{km}
\]
- The time taken is one year (in seconds, approximately \( 3.15 \times 10^7 \) seconds).
- The average speed is thus:
\[
\text{Average Speed} = \frac{9.42 \times 10^8 \, \text{km}}{3.15 \times 10^7 \, \text{s}} \approx 30 \, \text{km/s}
\]
- Hence, the average speed is not zero.
5. **Instantaneous Acceleration**:
- The instantaneous acceleration of the Earth is indeed directed towards the center of the circular path (the Sun).
- This is characteristic of centripetal acceleration, which always points towards the center of the circular motion.
### Conclusion:
- The correct statement among the options is that the instantaneous acceleration of the Earth points towards the Sun.
### Summary of Answers:
1. Average velocity from January 1 to June 30 is **not zero**.
2. Average acceleration is **not 60 km/s²**.
3. Average speed from January 1 to December 31 is **not zero**.
4. Instantaneous acceleration of the Earth **points towards the Sun** (this is correct).
