1. Concept of escape velocity

Escape velocity is the minimum speed required for a body to escape from Earth's gravitational field without returning.

It is obtained from energy conservation:

$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$

So,

$v_e = \sqrt{\frac{2GM}{R}}$

This depends only on:

• mass of Earth $M$
• radius of Earth $R$
• gravitational constant $G$

It does not depend on the direction of projection.

2. Effect of angle of projection

If the body is projected at $45^\circ$ with the vertical, its velocity has both vertical and horizontal components. But escape depends on the total initial kinetic energy, i.e.

$K = \frac{1}{2}mv^2$

Since kinetic energy depends on the magnitude of velocity $v$, not on its direction, the required escape speed remains unchanged.

3. Given value

Escape velocity from the surface of Earth when projected vertically upward is given as:

$v_e = 11\,\text{km/s}$

Therefore, even when projected at $45^\circ$ with the vertical,

$v_e = 11\,\text{km/s}$

4. Option check

• A: $11\sqrt{2}\,\text{km/s}$ ❌
• B: $22\,\text{km/s}$ ❌
• C: $11\,\text{km/s}$ ✅
• D: $\dfrac{11}{\sqrt{2}}\,\text{km/s}$ ❌

Therefore, the correct answer is C.