A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is `V`. Due to the rotation of planet about its axis the acceleration due to gravity `g` at equator is `1//2` of `g` at poles. The escape velocity of a particle on the planet in terms of `V`.

Options

A. `V_(e) = 2V`
B. `V_(e) = V`
C. `V_(e) = V//2`
D. `V_(e) = sqrt(3)V`

Answer

To find the escape velocity of a particle on a rotating spherical planet in terms of the velocity \( V \) at its equator, we can follow these steps:

### Step 1: Understand the relationship between linear velocity and angular velocity
The linear velocity \( V \) at the equator can be expressed in terms of the angular velocity \( \omega \) and the radius \( R \) of the planet:
\[ V = \omega R \]

### Step 2: Analyze the gravitational acceleration at the equator
The gravitational acceleration \( g \) at the equator is affected by the centrifugal acceleration due to the planet's rotation. The gravitational acceleration at the equator can be expressed as:
\[ g = g_0 - \omega^2 R \]
where \( g_0 \) is the gravitational acceleration at the poles.

### Step 3: Use the given condition for gravitational acceleration
According to the problem, the gravitational acceleration at the equator is half of that at the poles:
\[ g = \frac{g_0}{2} \]
Substituting this into the equation from Step 2 gives:
\[ \frac{g_0}{2} = g_0 - \omega^2 R \]

### Step 4: Rearrange the equation to find \( \omega^2 R \)
Rearranging the equation from Step 3:
\[ \omega^2 R = g_0 - \frac{g_0}{2} = \frac{g_0}{2} \]

### Step 5: Substitute \( \omega \) in terms of \( V \)
From Step 1, we know:
\[ \omega = \frac{V}{R} \]
Substituting this into the equation from Step 4 gives:
\[ \left(\frac{V}{R}\right)^2 R = \frac{g_0}{2} \]
This simplifies to:
\[ \frac{V^2}{R} = \frac{g_0}{2} \]

### Step 6: Rearranging to find \( g_0 \)
From the above equation, we can express \( g_0 \):
\[ g_0 = \frac{2V^2}{R} \]

### Step 7: Use the formula for escape velocity
The escape velocity \( V_E \) from a planet is given by the formula:
\[ V_E = \sqrt{2g_0 R} \]
Substituting the expression for \( g_0 \) from Step 6 into this formula:
\[ V_E = \sqrt{2 \left(\frac{2V^2}{R}\right) R} = \sqrt{4V^2} = 2V \]

### Conclusion
Thus, the escape velocity \( V_E \) of a particle on the planet in terms of \( V \) is:
\[ \boxed{2V} \]