A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is `7.5 kms^(-1)`. Due to the rotation of the planet about its axis, the acceleration due to gravity g at equator is `1//2` of g at poles. What is the escape velocity `("in km s"^(-1))` of a particle on the planet from the pole of the planet?

Answer

To find the escape velocity of a particle on the planet from the pole, we can follow these steps:

### Step 1: Understand the relationship between gravity at the poles and equator
Let the acceleration due to gravity at the pole be \( g \). According to the problem, the acceleration due to gravity at the equator \( g' \) is half of that at the pole:
\[ g' = \frac{g}{2} \]

### Step 2: Relate the gravitational acceleration at the equator to the centripetal acceleration
The centripetal acceleration due to the planet's rotation at the equator is given by:
\[ g' = g - R \omega^2 \]
where \( R \) is the radius of the planet and \( \omega \) is the angular velocity.

### Step 3: Substitute \( g' \) into the equation
Substituting \( g' = \frac{g}{2} \) into the equation gives:
\[ \frac{g}{2} = g - R \omega^2 \]
Rearranging this, we find:
\[ R \omega^2 = g - \frac{g}{2} = \frac{g}{2} \]

### Step 4: Relate linear velocity to angular velocity
The linear velocity \( V \) at the equator is related to the angular velocity by:
\[ V = R \omega \]
Squaring both sides gives:
\[ V^2 = R^2 \omega^2 \]

### Step 5: Substitute \( R \omega^2 \) into the equation
From the previous step, we can express \( R \omega^2 \) as:
\[ R \omega^2 = \frac{g}{2} \]
Substituting \( R \omega^2 \) into the equation \( V^2 = R^2 \omega^2 \) gives:
\[ V^2 = R \cdot \frac{g}{2} \]

### Step 6: Solve for \( gR \)
From the equation \( V^2 = R \cdot \frac{g}{2} \), we can express \( gR \):
\[ gR = 2V^2 \]

### Step 7: Use the escape velocity formula
The escape velocity \( V_e \) from the surface of a planet is given by:
\[ V_e = \sqrt{2gR} \]
Substituting \( gR = 2V^2 \) into this formula gives:
\[ V_e = \sqrt{2 \cdot 2V^2} = \sqrt{4V^2} = 2V \]

### Step 8: Substitute the value of \( V \)
Given that the velocity of a point on the equator is \( V = 7.5 \, \text{km/s} \):
\[ V_e = 2 \times 7.5 = 15 \, \text{km/s} \]

### Final Answer
The escape velocity of a particle on the planet from the pole is:
\[ \boxed{15 \, \text{km/s}} \]