A body suspended on a spring balance in a ship weighs `W_(0)` when the ship is at rest. When the ship begins to move along the equator with a speed `v`, show that the scale reading is very close to `W_(0) (1+-2omegaV//g`), where `omega` is the angular speed of the earth.

Explanation

To solve the problem, we need to analyze the forces acting on the body suspended in the spring balance when the ship is moving along the equator with a speed \( v \).

### Step-by-Step Solution:

1. **Weight at Rest**:
When the ship is at rest, the weight \( W_0 \) of the body is given by:
\[ W_0 = mg - m\omega^2 R \]
where \( m \) is the mass of the body, \( g \) is the acceleration due to gravity, \( \omega \) is the angular speed of the Earth, and \( R \) is the radius of the Earth. Since \( \omega^2 R \) is much smaller than \( g \), we can approximate:
\[ W_0 \approx mg \]

**Hint**: Remember that the weight of an object is affected by both gravitational force and the centripetal force due to Earth's rotation.

2. **Effective Angular Speed**:
When the ship moves with speed \( v \), the effective angular speed of the body becomes:
\[ \omega' = \omega \pm \frac{v}{R} \]
This accounts for the additional velocity due to the ship's movement.

3. **New Weight Calculation**:
The new weight \( W \) when the ship is moving is given by:
\[ W = mg - m(\omega'^{2})R \]
Substituting \( \omega' \):
\[ W = mg - m\left(\omega \pm \frac{v}{R}\right)^{2}R \]

4. **Expanding the Square**:
Expanding the square:
\[ W = mg - m\left(\omega^2 \pm 2\omega\frac{v}{R} + \frac{v^2}{R^2}\right)R \]
This simplifies to:
\[ W = mg - m\omega^2 R \mp 2m\omega v + m\frac{v^2}{R} \]

5. **Neglecting Small Terms**:
Since \( \frac{v^2}{R} \) is negligible compared to \( g \) and \( m\omega^2 R \), we can simplify further:
\[ W \approx mg - m\omega^2 R \mp 2m\omega v \]

6. **Substituting \( W_0 \)**:
We know that \( W_0 = mg - m\omega^2 R \), so we can substitute:
\[ W \approx W_0 \mp 2m\omega v \]

7. **Final Expression**:
Factoring out \( mg \):
\[ W \approx W_0 \left(1 \pm \frac{2\omega v}{g}\right) \]

Thus, we have shown that the scale reading is very close to:
\[ W \approx W_0 \left(1 \pm \frac{2\omega v}{g}\right) \]