These orbital speed of satellite revolving very close to earth

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 revolving very close to the Earth, we can derive the expression using the concepts of gravitational force and centripetal force. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Satellite
When a satellite is in orbit around the Earth, two main forces act on it:
1. Gravitational Force (F_g)
2. Centripetal Force (F_c)

The gravitational force pulls the satellite towards the Earth, while the centripetal force is required to keep the satellite in circular motion.

### Step 2: Write the Expression for Gravitational Force
The gravitational force acting on the satellite can be expressed using Newton's law of gravitation:
\[ F_g = \frac{G \cdot M \cdot m}{r^2} \]
Where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite (approximately equal to the radius of the Earth when the satellite is very close to the surface).

### Step 3: Write the Expression for Centripetal Force
The centripetal force required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m \cdot v^2}{r} \]
Where:
- \( v \) is the orbital speed of the satellite.

### Step 4: Set Gravitational Force Equal to Centripetal Force
For the satellite to remain in orbit, the gravitational force must equal the centripetal force:
\[ F_g = F_c \]
Substituting the expressions for \( F_g \) and \( F_c \):
\[ \frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot v^2}{r} \]

### Step 5: Simplify the Equation
We can cancel the mass of the satellite \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G \cdot M}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{G \cdot M}{r} = v^2 \]

### Step 6: Solve for the Orbital Speed \( v \)
Taking the square root of both sides gives us the expression for the orbital speed:
\[ v = \sqrt{\frac{G \cdot M}{r}} \]

### Step 7: Substitute \( r \) with \( R \) (Radius of the Earth)
Since the satellite is very close to the Earth's surface, we can approximate \( r \) as \( R \) (the radius of the Earth):
\[ v = \sqrt{\frac{G \cdot M}{R}} \]

### Final Answer
Thus, the orbital speed of a satellite revolving very close to the Earth is:
\[ v = \sqrt{\frac{G \cdot M}{R}} \]