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-by-Step Solution:

1. **Understand the Formula for Orbital Speed**:
The orbital speed \( v_c \) of a satellite is given by the formula:
\[ v_c = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth to the satellite.

2. **Determine the Distance \( r \)**:
The distance \( r \) from the center of the Earth to the satellite is the sum of the Earth's radius \( R \) and the height \( h \) of the satellite above the Earth's surface. Given that the satellite is at a height of \( \frac{R}{6} \), we can express \( r \) as:
\[ r = R + h = R + \frac{R}{6} \]
To combine these, we can write:
\[ r = R + \frac{R}{6} = \frac{6R}{6} + \frac{R}{6} = \frac{7R}{6} \]

3. **Substitute \( r \) into the Orbital Speed Formula**:
Now substituting \( r = \frac{7R}{6} \) into the orbital speed formula:
\[ v_c = \sqrt{\frac{GM}{\frac{7R}{6}}} \]

4. **Simplify the Expression**:
To simplify the expression, we can rewrite it as:
\[ v_c = \sqrt{\frac{GM \cdot 6}{7R}} = \sqrt{\frac{6GM}{7R}} \]

5. **Final Result**:
Therefore, the orbital speed of the satellite is:
\[ v_c = \sqrt{\frac{6GM}{7R}} \]

### Final Answer:
The orbital speed of the satellite is \( v_c = \sqrt{\frac{6GM}{7R}} \).

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