For a satellite escape velocity is 11km /s . If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be

The correct option is 11 km/ s.

To understand why the escape velocity remains unchanged, we must look at the physical definition and derivation of escape velocity.

Escape velocity is defined as the minimum speed required for an object to escape the gravitational pull of a massive body (like Earth) and reach infinity with zero kinetic energy. We can find this speed using the law of conservation of mechanical energy.

The total mechanical energy ($E$) of a satellite of mass $m$ at a distance $R$ from the center of a planet of mass $M$ is the sum of its kinetic energy and gravitational potential energy:

$E=E_{\mathrm{kinetic}}+E_{\mathrm{potential}}$

$E=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

where $G$ is the universal gravitational constant. For the satellite to just escape the gravitational field, its total energy at infinity must be at least zero. Setting the total energy to zero gives us the boundary condition for escape:

$\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0$

Solving for the escape velocity ($v_e$), we get:

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

Key observations from this formula:
1. Kinetic energy is a scalar quantity ($\frac{1}{2}m{v}^2$), which depends only on the magnitude of the velocity (speed) and not on its direction.
2. Potential energy is also a scalar quantity depending only on the distance from the center of the planet.
3. Consequently, the escape velocity formula does not contain any directional or angular components.

Therefore, the escape velocity is completely independent of the angle of projection. Launching the satellite at an angle of 60° with the vertical does not alter the energy requirements for escape. The escape velocity remains the same, which is 11 km/s.