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 in orbit experiences a gravitational force that provides the necessary centripetal force to keep it moving in a circular path. The gravitational force \( F_g \) acting on the satellite of mass \( m \) at a distance \( r + h \) from the center of the Earth is given by:

\[ F_g = \frac{GMm}{(r + h)^2} \]

where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Set the gravitational force equal to the centripetal force
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:

\[ F_c = \frac{mv^2}{r + h} \]

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

Setting the gravitational force equal to the centripetal force gives us:

\[ \frac{GMm}{(r + h)^2} = \frac{mv^2}{r + h} \]

### Step 3: Simplify the equation
We can cancel the mass \( m \) of the satellite from both sides (assuming \( m \neq 0 \)):

\[ \frac{GM}{(r + h)^2} = \frac{v^2}{r + h} \]

Multiplying both sides by \( (r + h)^2 \) gives:

\[ GM = v^2 (r + h) \]

### Step 4: Solve for the orbital speed \( v \)
Rearranging the equation to solve for \( v^2 \):

\[ v^2 = \frac{GM}{r + h} \]

Taking the square root of both sides gives us the orbital speed:

\[ v = \sqrt{\frac{GM}{r + h}} \]

### Step 5: Relate \( g \) at height \( h \) to the formula
The acceleration due to gravity \( g \) at height \( h \) is given by:

\[ g = \frac{GM}{(r + h)^2} \]

We can express \( GM \) in terms of \( g \):

\[ GM = g(r + h)^2 \]

### Step 6: Substitute \( GM \) back into the orbital speed equation
Substituting \( GM \) into the orbital speed equation:

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

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

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