The earth, radius 6400 km, makes one revolution about its own axis in 24 hours. The centripetal acceleration of a point on its equator is nearly

Options

A. 340 `(cm)/(sec^(2))`
B. 3.4 `(cm)/(sec^(2))`
C. `34(cm)/(sec^(2))`
D. 0.34 `(cm)/(sec^(2))`

Answer

To find the centripetal acceleration of a point on the Earth's equator, we can follow these steps:

### Step 1: Identify the given data
- Radius of the Earth (r) = 6400 km = 6400 × 10^3 m = 6.4 × 10^6 m
- Time for one complete revolution (T) = 24 hours = 24 × 3600 seconds = 86400 seconds

### Step 2: Calculate the angular velocity (ω)
The angular velocity (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{86400} \text{ rad/s} \]

### Step 3: Calculate the centripetal acceleration (a_c)
The formula for centripetal acceleration is:
\[ a_c = \omega^2 \cdot r \]
Substituting the value of ω and r:
\[ a_c = \left(\frac{2\pi}{86400}\right)^2 \cdot (6.4 \times 10^6) \]

### Step 4: Perform the calculations
1. Calculate \( \omega \):
\[ \omega \approx \frac{6.2832}{86400} \approx 7.272 \times 10^{-5} \text{ rad/s} \]
2. Calculate \( \omega^2 \):
\[ \omega^2 \approx (7.272 \times 10^{-5})^2 \approx 5.290 \times 10^{-9} \text{ rad}^2/\text{s}^2 \]
3. Now calculate \( a_c \):
\[ a_c \approx 5.290 \times 10^{-9} \cdot (6.4 \times 10^6) \approx 0.0338 \text{ m/s}^2 \]

### Step 5: Convert to cm/s²
Since the question asks for the answer in cm/s²:
\[ 0.0338 \text{ m/s}^2 = 3.38 \text{ cm/s}^2 \]

### Final Answer
The centripetal acceleration of a point on the Earth's equator is approximately **3.38 cm/s²**.

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