The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is v. For a satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is

The correct option is V √(2/3).

Step-by-Step Explanation:

1. Understanding Orbital Velocity:
The orbital velocity $v$ of a satellite in a circular orbit around the Earth at a distance $r$ from the Earth's center is given by the formula:

$v=\sqrt{\frac{GM}{r}}$

where:
- $G$ is the universal gravitational constant,
- $M$ is the mass of the Earth, and
- $r$ is the distance of the satellite from the center of the Earth.

2. Case 1: Satellite just above the Earth's surface:
When a satellite orbits just above the Earth's surface, its orbital radius is approximately equal to the radius of the Earth, $R$. Therefore, we can set $r=R$.
Let its orbital velocity be $v$:

$v=\sqrt{\frac{GM}{R}}$ (Equation 1)

3. Case 2: Satellite orbiting at an altitude of half of the Earth's radius:
Here, the altitude of the satellite from the Earth's surface is:

$h=\frac{R}{2}$

The distance of this satellite from the center of the Earth, $r^{\prime }$, is the sum of the Earth's radius and the altitude:

$r^{\prime }=R+h=R+\frac{R}{2}=\frac{3R}{2}$

Let the new orbital velocity at this distance be $v^{\prime }$. Applying the orbital velocity formula:

$v^{\prime }=\sqrt{\frac{GM}{r^{\prime }}}=\sqrt{\frac{GM}{3R/2}}=\sqrt{\frac{2GM}{3R}}$

4. Relating the two velocities:
We can rewrite the expression for $v^{\prime }$ in terms of $v$:

$v^{\prime }=\sqrt{\frac{2}{3}}·\sqrt{\frac{GM}{R}}$

Substituting Equation 1 into this equation, we get:

$v^{\prime }=v\sqrt{\frac{2}{3}}$

Thus, the orbital velocity of the satellite orbiting at an altitude of half of the Earth's radius is $v\sqrt{\frac{2}{3}}$.