The orbital velocity of a satellite at a height h above the surface of earth is v. The value of escape velocity from the same location is given by

The correct answer is √2 v (corresponding to the option √2 v).

Let us understand the relation between the orbital velocity and the escape velocity step-by-step.

Step 1: Formula for Orbital Velocity ($v_o$)
The orbital velocity $v$ of a satellite orbiting at a height $h$ above the surface of the Earth is the speed required to keep it in a stable circular orbit. It is determined by equating the gravitational force to the required centripetal force:
$v=\sqrt{\frac{GM}{R+h}}$
where:
- $G$ is the universal gravitational constant,
- $M$ is the mass of the Earth,
- $R$ is the radius of the Earth, and
- $R+h$ is the distance from the center of the Earth to the satellite's orbital position.

Step 2: Formula for Escape Velocity ($v_e$)
The escape velocity from a given location (at a distance $R+h$ from the center of the Earth) is the minimum speed needed for an object to escape the Earth's gravitational pull completely. This is found using the principle of conservation of energy (setting the total mechanical energy to zero):
$v_e=\sqrt{\frac{2GM}{R+h}}$

Step 3: Finding the Relationship between Escape Velocity and Orbital Velocity
We can rewrite the expression for the escape velocity by factoring out the term representing the orbital velocity:
$v_e=\sqrt{2}\cdot \sqrt{\frac{GM}{R+h}}$
Substituting $v$ for $\sqrt{\frac{GM}{R+h}}$ in the equation, we get:
$v_e=\sqrt{2}v$

Thus, the value of the escape velocity from the same location is $\sqrt{2}v$.