A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options

A. `sqrt(gx)`
B. `sqrt((gR)/(R - x))`
C. `sqrt((gR^(2))/(R - x))`
D. `sqrt((gR^(2))/(R + x))`

Explanation

To find the orbital speed of a satellite of mass \( m \) revolving around the Earth at a height \( x \) from its surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total distance from the center of the Earth**:
The radius of the Earth is \( R \). Since the satellite is at a height \( x \) from the Earth's surface, the total distance \( r \) from the center of the Earth to the satellite is:
\[ r = R + x \]

2. **Set up the equation for centripetal force**:
The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The centripetal force can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital speed of the satellite.

3. **Express the gravitational force**:
The gravitational force acting on the satellite can be expressed as:
\[ F_g = \frac{GMm}{r^2} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

4. **Set the centripetal force equal to the gravitational force**:
Since the gravitational force provides the centripetal force, we can set them equal to each other:
\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

5. **Cancel out the mass \( m \)**:
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \frac{GM}{r^2} \]

6. **Rearrange the equation to solve for \( v^2 \)**:
Multiplying both sides by \( r \):
\[ v^2 = \frac{GM}{r} \]

7. **Substitute \( r \) with \( R + x \)**:
Now substitute \( r \) with \( R + x \):
\[ v^2 = \frac{GM}{R + x} \]

8. **Take the square root to find \( v \)**:
Finally, taking the square root gives us the orbital speed:
\[ v = \sqrt{\frac{GM}{R + x}} \]

9. **Relate \( GM/R^2 \) to \( g \)**:
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
Therefore, we can express \( GM \) as \( gR^2 \):
\[ v = \sqrt{\frac{gR^2}{R + x}} \]

10. **Final expression for orbital speed**:
Thus, the orbital speed of the satellite can be expressed as:
\[ v = \sqrt{gR^2 \frac{1}{R + x}} = \sqrt{\frac{gR^2}{R + x}} \]