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 solve the problem step by step, we will follow the logical reasoning outlined in the video transcript.

### Step 1: Understand the Given Information
- The velocity of a point on the equator of the planet is \( v = 7.5 \, \text{km/s} \).
- The acceleration due to gravity at the equator \( g_e \) is half of that at the poles \( g_p \):
\[ g_e = \frac{1}{2} g_p \]

### Step 2: Relate Angular Velocity to Linear Velocity
The linear velocity \( v \) of a point on the equator can be expressed in terms of the angular velocity \( \omega \) and the radius \( r \) of the planet:
\[ v = \omega r \]
From the problem, we have:
\[ \omega = \frac{v}{r} = \frac{7.5 \, \text{km/s}}{r} \]

### Step 3: Write the Expression for Effective Gravity at the Equator
The effective gravitational acceleration \( g_e \) at the equator can be expressed as:
\[ g_e = g_p - \omega^2 r \]
Substituting \( g_e = \frac{1}{2} g_p \) into the equation gives:
\[ \frac{1}{2} g_p = g_p - \omega^2 r \]
Rearranging this yields:
\[ \omega^2 r = \frac{1}{2} g_p \]

### Step 4: Substitute for \( \omega^2 \)
From the previous step, we can express \( g_p \) in terms of \( \omega \):
\[ g_p = 2 \omega^2 r \]

### Step 5: Find Escape Velocity
The escape velocity \( v_e \) from the surface of the planet can be derived from the conservation of mechanical energy:
\[ \frac{1}{2} m v_e^2 - \frac{GMm}{r} = 0 \]
This simplifies to:
\[ \frac{1}{2} v_e^2 = \frac{GM}{r} \]
Thus,
\[ v_e = \sqrt{\frac{2GM}{r}} \]

### Step 6: Relate \( GM/r \) to \( g_p \)
From the expression for \( g_p \):
\[ g_p = \frac{GM}{r} \]
We can substitute this into the escape velocity equation:
\[ v_e = \sqrt{2g_p} \]

### Step 7: Substitute for \( g_p \)
Using \( g_p = 2 \omega^2 r \):
\[ v_e = \sqrt{2 \cdot 2 \omega^2 r} = \sqrt{4 \omega^2 r} = 2\omega \sqrt{r} \]
Now substituting \( \omega = \frac{7.5}{r} \):
\[ v_e = 2 \cdot \frac{7.5}{r} \cdot \sqrt{r} = 15 \, \text{km/s} \]

### Conclusion
Thus, the escape velocity from the pole of the planet is:
\[ \boxed{15 \, \text{km/s}} \]