A satellite orbits the earth at a height of R/6 , its orbital speed ?

Options

A. `sqrt((2GM)/R)`
B. `sqrt((GM)/R)`
C. `sqrt((GM)/(7R))`
D. `sqrt((6GM)/(7R))`

Answer

To find the orbital speed of a satellite orbiting the Earth at a height of \( \frac{R}{6} \), we can follow these steps:

### Step 1: Understand the given parameters
- The radius of the Earth is denoted as \( R \).
- The height of the satellite above the Earth's surface is \( h = \frac{R}{6} \).

### Step 2: Determine the distance from the center of the Earth to the satellite
The total distance \( r \) from the center of the Earth to the satellite is the sum of the Earth's radius and the height of the satellite:
\[ r = R + h = R + \frac{R}{6} = R \left(1 + \frac{1}{6}\right) = R \left(\frac{6}{6} + \frac{1}{6}\right) = R \left(\frac{7}{6}\right) = \frac{7R}{6} \]

### Step 3: Use the formula for orbital speed
The formula for the orbital speed \( v \) of a satellite is given by:
\[ v = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 4: Substitute the distance \( r \) into the formula
Now, substituting \( r = \frac{7R}{6} \) into the orbital speed formula:
\[ v = \sqrt{\frac{GM}{\frac{7R}{6}}} = \sqrt{\frac{GM \cdot 6}{7R}} \]

### Step 5: Simplify the expression
This can be further simplified as:
\[ v = \sqrt{\frac{6GM}{7R}} \]

### Final Answer
Thus, the orbital speed of the satellite is:
\[ v = \sqrt{\frac{6GM}{7R}} \]