The orbital velocity of a satellite close to the earth is v. Then the orbital velocity at a height `(1//4)^("th")` of earth's radius is

Options

A. 2 v/5
B. 5 v/s
C. `2v//sqrt5`
D. `sqrt5v//2`

Explanation

To find the orbital velocity of a satellite at a height of \( \frac{1}{4} \) of the Earth's radius, we can follow these steps:

### Step 1: Understand the formula for orbital velocity
The orbital velocity \( v \) of a satellite close to the Earth is given by the formula:
\[ v = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Determine the new radius at the height
When the satellite is at a height of \( \frac{1}{4} \) of the Earth's radius, the total distance from the center of the Earth to the satellite becomes:
\[ R' = R + h = R + \frac{1}{4}R = \frac{5}{4}R \]

### Step 3: Write the formula for the new orbital velocity
The new orbital velocity \( v' \) at this height can be expressed as:
\[ v' = \sqrt{\frac{GM}{R'}} \]
Substituting \( R' \) into the equation gives:
\[ v' = \sqrt{\frac{GM}{\frac{5}{4}R}} = \sqrt{\frac{4GM}{5R}} \]

### Step 4: Relate the new orbital velocity to the original velocity
Now, we can express \( v' \) in terms of the original velocity \( v \):
\[ v' = \sqrt{\frac{4}{5}} \cdot \sqrt{\frac{GM}{R}} = \sqrt{\frac{4}{5}} \cdot v \]

### Final Answer
Thus, the orbital velocity at a height of \( \frac{1}{4} \) of the Earth's radius is:
\[ v' = \sqrt{\frac{4}{5}} \cdot v \]

---