When a projectile attains the escape velocity, then on surface of planet its

Options

A. KE is greater than PE numarically
B. PE is greater than KE numerically
C. Both PE and KE are numerically equa
D. none

Answer

To solve the question regarding the relationship between kinetic energy and potential energy when a projectile attains escape velocity on the surface of a planet, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
Escape velocity is the minimum velocity an object must have to break free from the gravitational attraction of a planet without any additional propulsion.

2. **Energy Considerations**:
At escape velocity, the total mechanical energy (kinetic energy + potential energy) of the projectile must be zero. This is because the projectile needs to have enough kinetic energy to overcome the gravitational potential energy of the planet.

3. **Formulas**:
- Kinetic Energy (KE) of the projectile:
\[ KE = \frac{1}{2} mv^2 \]
- Gravitational Potential Energy (PE) at the surface of the planet:
\[ PE = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the projectile, and \( r \) is the radius of the planet.

4. **Setting Up the Equation**:
For the projectile to escape, the total energy must be zero:
\[ KE + PE = 0 \]
Substituting the expressions for KE and PE:
\[ \frac{1}{2} mv^2 - \frac{GMm}{r} = 0 \]

5. **Solving for Escape Velocity**:
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{GMm}{r} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{GM}{r} \]
Multiplying both sides by 2:
\[ v^2 = \frac{2GM}{r} \]
Thus, the escape velocity \( v \) is:
\[ v = \sqrt{\frac{2GM}{r}} \]

6. **Conclusion**:
When the projectile reaches escape velocity, the kinetic energy is equal in magnitude to the gravitational potential energy, but opposite in sign. Therefore, we conclude that:
\[ KE = -PE \]
This means that numerically, both the potential energy and kinetic energy are equal.

### Final Answer:
The correct option is that both potential energy and kinetic energy are numerically equal when a projectile attains escape velocity on the surface of the planet.