If the escape velocity of the body which is thrown from earth surface is `V_e`, then the escape velocity of the body if it is thrown at an angle `theta` from horizontal earth`s surface is

Options

A. `V_e`
B. `V_ecostheta`
C. `V_esintheta`
D. `V_etantheta`

Answer

To solve the problem, we need to understand the concept of escape velocity and how it is affected by the angle of projection. The escape velocity from the surface of the Earth is given as \( V_e \). We need to determine if this value changes when the body is thrown at an angle \( \theta \) from the horizontal.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
- Escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body without any further propulsion.
- The escape velocity from the Earth's surface is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Conditions**:
- When the body is thrown from the Earth's surface, its initial potential energy (\( U_i \)) and kinetic energy (\( K_i \)) can be expressed as:
\[ U_i = -\frac{GMm}{R} \]
\[ K_i = \frac{1}{2} mv_e^2 \]
where \( m \) is the mass of the body.

3. **Final Conditions**:
- At an infinite distance from the Earth, both the potential energy (\( U_f \)) and kinetic energy (\( K_f \)) are zero:
\[ U_f = 0 \]
\[ K_f = 0 \]

4. **Applying Conservation of Mechanical Energy**:
- According to the conservation of mechanical energy:
\[ U_i + K_i = U_f + K_f \]
- Substituting the values:
\[ -\frac{GMm}{R} + \frac{1}{2} mv_e^2 = 0 \]

5. **Solving for Escape Velocity**:
- Rearranging the equation gives:
\[ \frac{1}{2} mv_e^2 = \frac{GMm}{R} \]
- Dividing through by \( m \) and multiplying by 2:
\[ v_e^2 = \frac{2GM}{R} \]
- Thus, the escape velocity remains:
\[ v_e = \sqrt{\frac{2GM}{R}} \]

6. **Effect of Angle of Projection**:
- The key point is that escape velocity is derived from energy considerations and does not depend on the angle of projection. Therefore, whether the body is thrown vertically or at an angle \( \theta \), the escape velocity remains the same.

7. **Conclusion**:
- Hence, the escape velocity when thrown at an angle \( \theta \) from the horizontal remains \( V_e \).

### Final Answer:
The escape velocity of the body if it is thrown at an angle \( \theta \) from the horizontal Earth's surface is still \( V_e \).

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