If the earth stops rotating about its axis , then what will be the change in the value of g at a place in the equitorial plane ? Radius of the earth = 6400 km.

Answer

To solve the problem of how the value of \( g \) changes at a place in the equatorial plane if the Earth stops rotating, we can follow these steps:

### Step 1: Understand the value of \( g \)
The acceleration due to gravity \( g \) at the surface of the Earth is given by the formula:
\[ g = g_0 - \omega^2 r \cos^2 \lambda \]
where:
- \( g_0 \) is the standard acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \omega \) is the angular velocity of the Earth,
- \( r \) is the radius of the Earth,
- \( \lambda \) is the latitude (which is \( 0 \) at the equator).

### Step 2: Set the parameters for the equatorial plane
At the equator:
- \( \lambda = 0 \)
- Therefore, \( \cos^2 \lambda = \cos^2(0) = 1 \).

The formula simplifies to:
\[ g' = g_0 - \omega^2 r \]

### Step 3: Calculate the angular velocity \( \omega \)
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of rotation (24 hours). Converting 24 hours into seconds:
\[ T = 24 \times 60 \times 60 = 86400 \, \text{s} \]
Thus,
\[ \omega = \frac{2\pi}{86400} \, \text{rad/s} \]

### Step 4: Calculate \( \omega^2 \)
Now, we calculate \( \omega^2 \):
\[ \omega^2 = \left(\frac{2\pi}{86400}\right)^2 \]

### Step 5: Substitute the values
The radius of the Earth \( r \) is given as \( 6400 \, \text{km} \) which is \( 6400 \times 10^3 \, \text{m} \).

Now substituting \( \omega^2 \) and \( r \) into the equation:
\[ g' = g_0 - \left(\frac{2\pi}{86400}\right)^2 \times (6400 \times 10^3) \]

### Step 6: Calculate the change in \( g \)
The change in \( g \) when the Earth stops rotating is given by:
\[ \Delta g = \omega^2 r \]
Calculating \( \Delta g \):
1. Calculate \( \omega^2 \):
\[ \omega^2 = \left(\frac{2\pi}{86400}\right)^2 \approx 7.272 \times 10^{-5} \, \text{rad/s}^2 \]
2. Then calculate \( \Delta g \):
\[ \Delta g = 7.272 \times 10^{-5} \times (6400 \times 10^3) \approx 0.034 \, \text{m/s}^2 \text{ or } 3.4 \, \text{cm/s}^2 \]

### Final Result
Thus, if the Earth stops rotating, the change in the value of \( g \) at a place in the equatorial plane will be approximately \( 3.4 \, \text{cm/s}^2 \).

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