A satellite of mass `m` is circulating around the earth with constant angular velocity. If the radius is `R_(0)` and mass of earth is M, then the angular momentum about the centre of the earth is

Options

A. `m sqrt(GM//R_(0))`
B. `Msqrt(GmR_(0))`
C. `m sqrt(GMR_(0))`
D. `M sqrt(GM//R_(0))`

Explanation

To find the angular momentum of a satellite of mass \( m \) revolving around the Earth with a radius \( R_0 \) and mass \( M \), we can follow these steps:

### Step 1: Understand the formula for angular momentum
The angular momentum \( L \) of an object moving in a circular path is given by the formula:
\[ L = m \cdot v \cdot r \]
where \( m \) is the mass of the object, \( v \) is the tangential (orbital) velocity, and \( r \) is the radius of the circular path.

### Step 2: Determine the orbital velocity
For a satellite in circular orbit, the gravitational force provides the necessary centripetal force. The gravitational force acting on the satellite is given by:
\[ F = \frac{G M m}{R_0^2} \]
where \( G \) is the gravitational constant. The centripetal force required to keep the satellite in circular motion is:
\[ F = \frac{m v^2}{R_0} \]
Setting these two forces equal gives:
\[ \frac{G M m}{R_0^2} = \frac{m v^2}{R_0} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{R_0^2} = \frac{v^2}{R_0} \]
Multiplying both sides by \( R_0 \) gives:
\[ \frac{G M}{R_0} = v^2 \]
Taking the square root of both sides, we find the orbital velocity \( v \):
\[ v = \sqrt{\frac{G M}{R_0}} \]

### Step 3: Substitute the orbital velocity into the angular momentum formula
Now we can substitute \( v \) back into the angular momentum formula:
\[ L = m \cdot v \cdot R_0 \]
Substituting for \( v \):
\[ L = m \cdot \left(\sqrt{\frac{G M}{R_0}}\right) \cdot R_0 \]
This simplifies to:
\[ L = m \cdot R_0 \cdot \sqrt{\frac{G M}{R_0}} \]
Rearranging gives:
\[ L = m \cdot \sqrt{G M R_0} \]

### Final Answer
Thus, the angular momentum of the satellite about the center of the Earth is:
\[ L = m \sqrt{G M R_0} \]

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