The orbital velocity of an artifical satellite in a cirular orbit above the earth's surface at a distance equal to radiu of earth is v. For a satellite orbiting at an altitude half of earth's radius, orbital velocity is

Options

A. `(3)/(2)v`
B. `sqrt(3//2)`v
C. `2//sqrt(3)v`
D. `2/3 v`

Explanation

To solve the problem, we need to determine the orbital velocity of a satellite at an altitude equal to half the radius of the Earth. We will use the formula for orbital velocity and the concept of gravitational acceleration at different heights.

### Step-by-step Solution:

1. **Understanding Orbital Velocity**:
The orbital velocity \( v \) of a satellite in a circular orbit is given by the formula:
\[ v = \sqrt{g \cdot r} \]
where \( g \) is the acceleration due to gravity at the distance from the center of the Earth, and \( r \) is the distance from the center of the Earth to the satellite.

2. **Given Conditions**:
We know that the orbital velocity at a distance equal to the radius of the Earth is \( v \). Therefore, at this distance:
\[ r = 2R \quad \text{(where \( R \) is the radius of the Earth)} \]
Thus, the orbital velocity at this distance is:
\[ v = \sqrt{g \cdot 2R} \]

3. **Finding Gravitational Acceleration at Height**:
The gravitational acceleration \( g' \) at a height \( h \) above the Earth's surface is given by:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]
For the first case, where the satellite is at a distance equal to the radius of the Earth:
\[ h = R \quad \Rightarrow \quad g' = g \left( \frac{R}{2R} \right)^2 = \frac{g}{4} \]

4. **Calculating Orbital Velocity at Height \( R \)**:
Now substituting \( g' \) into the orbital velocity formula:
\[ v = \sqrt{\frac{g}{4} \cdot 2R} = \sqrt{\frac{g \cdot 2R}{4}} = \sqrt{\frac{gR}{2}} \]

5. **Finding Orbital Velocity at Half the Radius**:
Now, for the satellite orbiting at an altitude of half the Earth's radius:
\[ h = \frac{R}{2} \quad \Rightarrow \quad r = R + \frac{R}{2} = \frac{3R}{2} \]
Now, we need to find \( g'' \) at this height:
\[ g'' = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4g}{9} \]

6. **Calculating Orbital Velocity at This Height**:
Now substituting \( g'' \) into the orbital velocity formula:
\[ v' = \sqrt{\frac{4g}{9} \cdot \frac{3R}{2}} = \sqrt{\frac{6gR}{9}} = \sqrt{\frac{2gR}{3}} \]

7. **Relating \( v' \) to \( v \)**:
Now we can relate \( v' \) to \( v \):
\[ v = \sqrt{\frac{gR}{2}} \quad \text{and} \quad v' = \sqrt{\frac{2gR}{3}} \]
To find the ratio:
\[ \frac{v}{v'} = \frac{\sqrt{\frac{gR}{2}}}{\sqrt{\frac{2gR}{3}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
Therefore:
\[ v' = \frac{2}{\sqrt{3}} v \]

### Final Answer:
The orbital velocity of the satellite orbiting at an altitude equal to half the radius of the Earth is:
\[ v' = \frac{2}{\sqrt{3}} v \]