A satellite is revolving in a circular orbit of round the earth at a height of h. If the acceleration due to gravity at that height is g. The orbital speed of the satellite .

Options

A. `sqrt(gh)`
B. `sqrt(2gh)`
C. `sqrt(g(R+h))`
D. `sqrt(gR)`

Answer

To find the orbital speed of a satellite revolving in a circular orbit around the Earth at a height \( h \), we can follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
The satellite is in circular motion, which means the gravitational force acting on it provides the necessary centripetal force for its motion. The gravitational force \( F_g \) acting on the satellite is given by:

\[ F_g = \frac{G M m}{(R + h)^2} \]

where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( R \) is the radius of the Earth,
- \( h \) is the height of the satellite above the Earth's surface.

The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:

\[ F_c = \frac{m v^2}{R + h} \]

where \( v \) is the orbital speed of the satellite.

### Step 2: Set the gravitational force equal to the centripetal force
Since the gravitational force provides the centripetal force, we can set these two equations equal to each other:

\[ \frac{G M m}{(R + h)^2} = \frac{m v^2}{R + h} \]

### Step 3: Cancel out the mass of the satellite
We can cancel the mass \( m \) of the satellite from both sides of the equation:

\[ \frac{G M}{(R + h)^2} = \frac{v^2}{R + h} \]

### Step 4: Rearrange the equation to solve for \( v^2 \)
Multiply both sides by \( (R + h) \):

\[ \frac{G M}{(R + h)} = v^2 \]

### Step 5: Take the square root to find the orbital speed \( v \)
Taking the square root of both sides gives us the orbital speed:

\[ v = \sqrt{\frac{G M}{R + h}} \]

### Step 6: Relate \( g \) to the height \( h \)
It is given that the acceleration due to gravity at height \( h \) is \( g \). The formula for acceleration due to gravity at height \( h \) is:

\[ g = \frac{G M}{(R + h)^2} \]

### Step 7: Substitute \( g \) into the orbital speed formula
From the equation for \( g \), we can express \( G M \):

\[ G M = g (R + h)^2 \]

Now substitute this into the orbital speed equation:

\[ v = \sqrt{\frac{g (R + h)^2}{R + h}} = \sqrt{g (R + h)} \]

### Final Result
Thus, the orbital speed of the satellite is:

\[ v = \sqrt{g (R + h)} \]