How many times is escape velocity (vₑ) , of orbital velocity (vₒ) for a satellite revolving near earth

The correct option is √2 times.

To understand the relationship between the escape velocity and the orbital velocity of a satellite revolving near the Earth's surface, we can look at their respective mathematical formulas.

1. Orbital Velocity:
Orbital velocity is the minimum velocity required by a satellite to remain in its orbit around the Earth. For a satellite revolving very close to the Earth's surface, the radius of its orbit is approximately equal to the radius of the Earth.

The orbital velocity is given by the formula:

$v_o=\sqrt{\frac{GM}{R}}$

where:
• $G$
is the universal gravitational constant,
• $M$
is the mass of the Earth, and
• $R$
is the radius of the Earth.

2. Escape Velocity:
Escape velocity is the minimum velocity required for an object to break free from the gravitational pull of the Earth without any further propulsion.

From the surface of the Earth, the escape velocity is given by the formula:

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

3. Finding the Relationship:
Now, we can relate the escape velocity to the orbital velocity by rewriting the escape velocity equation:

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

By substituting the expression for orbital velocity into the equation above, we get:

$v_e=\sqrt{2}{v}_o$

Thus, the escape velocity is
$\sqrt{2}$
times the orbital velocity for a satellite revolving near the Earth.