Determine the speed with which the earth would have to rotate on its axis , so that a person on the equator would weigh `(3)/(5)` as much as at present. Take `R = 6400 km`.

Options

A. `7.82×10^(−4) rads^(−1)`
B. `9.82×10^(−4) rads^(−1)`
C. `3.56×10^(−4) rads^(−1)`
D. `2.82×10^(−4) rads^(−1)`

Explanation

To determine the speed with which the Earth would have to rotate on its axis so that a person on the equator would weigh \( \frac{3}{5} \) as much as at present, we can follow these steps:

### Step 1: Understand the relationship between weight and apparent gravity
The apparent weight of a person is given by the formula:
\[ g' = g - r \omega^2 \]
where:
- \( g' \) is the apparent acceleration due to gravity,
- \( g \) is the actual acceleration due to gravity,
- \( r \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth.

### Step 2: Set up the equation for the new weight
Since we want the apparent weight to be \( \frac{3}{5} \) of the actual weight, we can express this as:
\[ g' = \frac{3}{5} g \]

### Step 3: Substitute \( g' \) into the equation
Substituting \( g' \) into the weight equation gives:
\[ \frac{3}{5} g = g - r \omega^2 \]

### Step 4: Rearrange the equation
Rearranging the equation to isolate \( r \omega^2 \):
\[ r \omega^2 = g - \frac{3}{5} g = \frac{2}{5} g \]

### Step 5: Solve for \( \omega \)
Now, we can solve for \( \omega \):
\[ \omega^2 = \frac{2}{5} \frac{g}{r} \]
Taking the square root of both sides:
\[ \omega = \sqrt{\frac{2}{5} \frac{g}{r}} \]

### Step 6: Substitute the known values
Given that \( g \approx 9.8 \, \text{m/s}^2 \) and \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \), we substitute these values into the equation:
\[ \omega = \sqrt{\frac{2}{5} \cdot \frac{9.8}{6400 \times 10^3}} \]

### Step 7: Calculate \( \omega \)
Calculating the values inside the square root:
\[ \omega = \sqrt{\frac{2 \cdot 9.8}{5 \cdot 6400 \times 10^3}} = \sqrt{\frac{19.6}{32000 \times 10^3}} = \sqrt{\frac{19.6}{32000000}} \approx \sqrt{6.125 \times 10^{-7}} \approx 7.8 \times 10^{-4} \, \text{radians/second} \]

### Final Result
Thus, the speed with which the Earth would have to rotate on its axis is approximately:
\[ \omega \approx 7.8 \times 10^{-4} \, \text{radians/second} \]