With what speed should the earth rotate in order that the weight of a man on the equator is reduced to half his present weight , if the radius of the earth is 6400km.

Explanation

To solve the problem of determining the speed at which the Earth should rotate in order to reduce a man's weight at the equator to half, we can follow these steps:

### Step 1: Define the Present Weight
Let the present weight of the man be represented as:
\[ W = mg \]
where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 2: Define the Reduced Weight
According to the question, the reduced weight \( W' \) at the equator is given as:
\[ W' = \frac{W}{2} = \frac{mg}{2} \]

### Step 3: Understand the Effect of Earth's Rotation
The effective weight of the man when the Earth is rotating is given by:
\[ W' = W - m\omega^2 r \]
where \( \omega \) is the angular velocity of the Earth and \( r \) is the radius of the Earth.

### Step 4: Substitute Values at the Equator
At the equator, the angle \( \theta = 0 \) degrees, so \( \cos(0) = 1 \). Thus, we can rewrite the equation as:
\[ W' = mg - m\omega^2 r \]

### Step 5: Set Up the Equation
Substituting for \( W' \):
\[ \frac{mg}{2} = mg - m\omega^2 r \]

### Step 6: Simplify the Equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{g}{2} = g - \omega^2 r \]

### Step 7: Rearrange to Solve for \( \omega^2 \)
Rearranging gives:
\[ \omega^2 r = g - \frac{g}{2} \]
\[ \omega^2 r = \frac{g}{2} \]

### Step 8: Solve for \( \omega^2 \)
Thus, we have:
\[ \omega^2 = \frac{g}{2r} \]

### Step 9: Relate Angular Velocity to Linear Velocity
We know that the linear velocity \( v \) is related to angular velocity by:
\[ v = r \omega \]
So, substituting \( \omega = \frac{v}{r} \) into the equation gives:
\[ \left(\frac{v}{r}\right)^2 = \frac{g}{2r} \]

### Step 10: Solve for \( v \)
This simplifies to:
\[ \frac{v^2}{r^2} = \frac{g}{2r} \]
Multiplying both sides by \( r^2 \):
\[ v^2 = \frac{g r}{2} \]
Taking the square root:
\[ v = \sqrt{\frac{g r}{2}} \]

### Step 11: Substitute the Values
Now, substituting the values for \( g \) and \( r \):
- \( g \approx 9.8 \, \text{m/s}^2 \)
- \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Calculating:
\[ v = \sqrt{\frac{9.8 \times (6400 \times 10^3)}{2}} \]

### Step 12: Final Calculation
Calculating the above expression:
\[ v = \sqrt{\frac{9.8 \times 6400 \times 10^3}{2}} \approx 5600 \, \text{m/s} \]

### Final Answer
The speed at which the Earth should rotate to reduce a man's weight at the equator to half is approximately:
\[ v \approx 5600 \, \text{m/s} \]

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