What will be velocity of a satellite revolving around the earth at a height h above surface of earth if radius of earth is R :-

Options

A. `R^(2) sqrt(g/(R+h))`
B. `R g/((R+h)^(2))`
C. `R sqrt(g/(R+h))`
D. `R sqrt((R+h)/g)`

Answer

To find the velocity of a satellite revolving around the Earth at a height \( h \) above the surface of the Earth, we can follow these steps:

### Step 1: Understand the Forces Acting on the Satellite
The satellite is in circular motion, which means it requires a centripetal force to keep it moving in that circular path. This centripetal force is provided by the gravitational force acting on the satellite.

### Step 2: Write the Expression for Centripetal Force
The centripetal force \( F_c \) required for the satellite of mass \( m \) moving with velocity \( v \) at a radius \( r + h \) (where \( r \) is the radius of the Earth and \( h \) is the height above the Earth's surface) is given by:
\[ F_c = \frac{mv^2}{r + h} \]

### Step 3: Write the Expression for Gravitational Force
The gravitational force \( F_g \) acting on the satellite is given by Newton's law of gravitation:
\[ F_g = \frac{G M m}{(r + h)^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 4: Set Centripetal Force Equal to Gravitational Force
Since the centripetal force is provided by the gravitational force, we can set these two forces equal to each other:
\[ \frac{mv^2}{r + h} = \frac{G M m}{(r + h)^2} \]

### Step 5: Cancel Out Mass \( m \)
We can cancel the mass \( m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r + h} = \frac{G M}{(r + h)^2} \]

### Step 6: Rearrange the Equation to Solve for \( v^2 \)
Now, we can rearrange the equation to solve for \( v^2 \):
\[ v^2 = \frac{G M}{(r + h)} \]

### Step 7: Take the Square Root to Find \( v \)
Taking the square root of both sides gives us the velocity \( v \):
\[ v = \sqrt{\frac{G M}{(r + h)}} \]

### Step 8: Substitute \( g \) for Gravitational Acceleration
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{r^2} \]
So we can express \( G M \) in terms of \( g \):
\[ G M = g r^2 \]

### Step 9: Substitute Back into the Velocity Equation
Substituting \( G M \) back into the equation for \( v \):
\[ v = \sqrt{\frac{g r^2}{(r + h)}} \]

### Final Expression for the Velocity
Thus, the final expression for the velocity of the satellite is:
\[ v = \sqrt{\frac{g r^2}{(r + h)}} \]