There are two bodies A and B of same mass. A is placed near equator of earth and B is placed at a height `h` above the pole of earth. if both the bodies weighs equally. Find `h` in terms of radius `R` of earth, angular speed `omega` of earth and `g` acceleration due to gravity close to earth.

Options

A. `R omega^2/(2g)`
B. `R^2 omega^2/(2g)`
C. `gR/omega^2`
D. `g/omega^2`

Answer

To solve the problem, we need to compare the weights of the two bodies A and B, which are placed at different locations on Earth.

### Step-by-Step Solution:

1. **Understanding the Weight of the Bodies**:
The weight of an object is given by the formula:
\[ W = m \cdot g \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity at that location.

2. **Weight of Body A**:
Body A is placed at the equator. The effective acceleration due to gravity at the equator, considering the centrifugal force due to Earth's rotation, is given by:
\[ g_A = g - \omega^2 R \]
where \( g \) is the acceleration due to gravity at the surface of the Earth, \( \omega \) is the angular speed of the Earth, and \( R \) is the radius of the Earth.

Therefore, the weight of body A is:
\[ W_A = m \cdot g_A = m \cdot (g - \omega^2 R) \]

3. **Weight of Body B**:
Body B is placed at a height \( h \) above the pole. The acceleration due to gravity at height \( h \) is given by:
\[ g_B = g \left(1 - \frac{2h}{R}\right) \]
Therefore, the weight of body B is:
\[ W_B = m \cdot g_B = m \cdot g \left(1 - \frac{2h}{R}\right) \]

4. **Setting the Weights Equal**:
Since the weights of both bodies are equal, we can set \( W_A = W_B \):
\[ m \cdot (g - \omega^2 R) = m \cdot g \left(1 - \frac{2h}{R}\right) \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g - \omega^2 R = g \left(1 - \frac{2h}{R}\right) \]

5. **Expanding and Rearranging**:
Expanding the right side:
\[ g - \omega^2 R = g - \frac{2gh}{R} \]
Now, we can rearrange the equation to isolate \( h \):
\[ -\omega^2 R = -\frac{2gh}{R} \]
Multiplying both sides by \( -1 \):
\[ \omega^2 R = \frac{2gh}{R} \]

6. **Solving for \( h \)**:
Now, multiply both sides by \( R \):
\[ \omega^2 R^2 = 2gh \]
Finally, solving for \( h \):
\[ h = \frac{\omega^2 R^2}{2g} \]

### Final Result:
\[ h = \frac{\omega^2 R^2}{2g} \]