The earth moves round the sun in a near circular orbit of radius `1.5 xx 10^(11)` m. Its centripetal acceleration is

Options

A. `1.5 xx 10^(-3) m//s^(2)`
B. `3 xx 10^(-3) m//s^(2)`
C. `6 xx 10^(-3) m//s^(2)`
D. `12 xx 10^(-3) m//s^(2)`

Explanation

To find the centripetal acceleration of the Earth as it moves around the Sun, we can follow these steps:

### Step 1: Understand the formula for centripetal acceleration
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear velocity and \( r \) is the radius of the circular orbit.

### Step 2: Relate linear velocity to angular velocity
The linear velocity \( v \) can also be expressed in terms of angular velocity \( \omega \):
\[ v = \omega r \]
Thus, we can rewrite the centripetal acceleration as:
\[ a_c = \frac{(\omega r)^2}{r} = \omega^2 r \]

### Step 3: Determine the angular velocity \( \omega \)
The angular velocity \( \omega \) is related to the time period \( T \) of the Earth's orbit. The formula for angular velocity is:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time taken for one complete revolution around the Sun.

### Step 4: Calculate the time period \( T \)
The Earth takes approximately 365 days to complete one orbit around the Sun. We need to convert this time into seconds:
\[ T = 365 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]
Calculating this gives:
\[ T = 365 \times 24 \times 60 \times 60 = 31,536,000 \text{ seconds} \]

### Step 5: Substitute \( T \) into the formula for \( \omega \)
Now we can calculate \( \omega \):
\[ \omega = \frac{2\pi}{31,536,000} \]

### Step 6: Calculate \( \omega^2 \)
Next, we need to square \( \omega \):
\[ \omega^2 = \left(\frac{2\pi}{31,536,000}\right)^2 \]

### Step 7: Substitute \( \omega^2 \) into the centripetal acceleration formula
Now substitute \( \omega^2 \) back into the centripetal acceleration formula:
\[ a_c = \omega^2 r \]
where \( r = 1.5 \times 10^{11} \text{ m} \).

### Step 8: Calculate \( a_c \)
Now we can calculate the centripetal acceleration:
\[ a_c = \left(\frac{2\pi}{31,536,000}\right)^2 \times (1.5 \times 10^{11}) \]

### Step 9: Final calculation
Calculating this gives:
\[ a_c \approx 6 \times 10^{-3} \text{ m/s}^2 \]

Thus, the centripetal acceleration of the Earth moving around the Sun is approximately \( 6 \times 10^{-3} \text{ m/s}^2 \).

### Answer
The centripetal acceleration of the Earth is \( 6 \times 10^{-3} \text{ m/s}^2 \).