A satellite of mass m is orbiting the earth at a height h from its surface. If M is the mass of the earth and R its radius, the kinetic energy of the stellite will be

Options

A. `-(GmM)/((R+h)^(2))`
B. `(GmM)/(2(R+h)^(2))`
C. `(GmM)/((R+h))`
D. `(GmM)/(2(R+h))`

Explanation

To find the kinetic energy of a satellite of mass \( m \) orbiting the Earth at a height \( h \) from its surface, we can follow these steps:

### Step 1: Understand the parameters
- Let \( M \) be the mass of the Earth.
- Let \( R \) be the radius of the Earth.
- The distance from the center of the Earth to the satellite is \( r = R + h \).

### Step 2: Write the expression for gravitational force
The gravitational force acting on the satellite is given by Newton's law of gravitation:
\[ F = \frac{GMm}{(R+h)^2} \]
where \( G \) is the universal gravitational constant.

### Step 3: Relate gravitational force to centripetal force
For a satellite in circular motion, the gravitational force provides the necessary centripetal force:
\[ F = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite and \( r = R + h \).

### Step 4: Set the forces equal
Setting the gravitational force equal to the centripetal force, we have:
\[ \frac{GMm}{(R+h)^2} = \frac{mv^2}{(R+h)} \]

### Step 5: Cancel \( m \) and rearrange
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{(R+h)^2} = \frac{v^2}{(R+h)} \]
Multiplying both sides by \( (R+h) \):
\[ \frac{GM}{R+h} = v^2 \]

### Step 6: Solve for \( v^2 \)
Thus, we find:
\[ v^2 = \frac{GM}{R+h} \]

### Step 7: Calculate the kinetic energy
The kinetic energy \( K \) of the satellite is given by:
\[ K = \frac{1}{2} mv^2 \]
Substituting \( v^2 \) from the previous step:
\[ K = \frac{1}{2} m \left( \frac{GM}{R+h} \right) \]
\[ K = \frac{GMm}{2(R+h)} \]

### Final Answer
The kinetic energy of the satellite is:
\[ K = \frac{GMm}{2(R+h)} \]

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