The energy required to move an earth satellites of mass m from a circular orbit of radius 2 R to a radius 3 R is `" "` (R is radius of the earth)

Options

A. `(GMm)/(12R)`
B. `(GMm)/(8R)`
C. `(GMm)/(3R)`
D. `(GMm)/(6R)`

Explanation

To solve the problem of finding the energy required to move an Earth satellite of mass \( m \) from a circular orbit of radius \( 2R \) to a radius \( 3R \) (where \( R \) is the radius of the Earth), we will follow these steps:

### Step 1: Calculate the gravitational potential energy at radius \( 2R \)

The gravitational potential energy \( U \) at a distance \( r \) from the center of the Earth is given by the formula:

\[ U = -\frac{GMm}{r} \]

For the orbit at radius \( 2R \):

\[ U_{2R} = -\frac{GMm}{2R} \]

### Step 2: Calculate the gravitational potential energy at radius \( 3R \)

Using the same formula for gravitational potential energy, we calculate it at radius \( 3R \):

\[ U_{3R} = -\frac{GMm}{3R} \]

### Step 3: Calculate the total energy at radius \( 2R \)

The total mechanical energy \( E \) of a satellite in a circular orbit is the sum of its kinetic energy \( K \) and gravitational potential energy \( U \). The kinetic energy can be derived from the gravitational force providing the necessary centripetal force.

The centripetal force is given by:

\[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

From this, we can derive the speed \( v \):

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

At radius \( 2R \):

\[ v^2 = \frac{GM}{2R} \]

Thus, the kinetic energy \( K \) at radius \( 2R \) is:

\[ K_{2R} = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{GM}{2R}\right) = \frac{GMm}{4R} \]

Now, the total energy \( E_{2R} \) at radius \( 2R \) is:

\[ E_{2R} = K_{2R} + U_{2R} = \frac{GMm}{4R} - \frac{GMm}{2R} \]

Combining these terms:

\[ E_{2R} = \frac{GMm}{4R} - \frac{2GMm}{4R} = -\frac{GMm}{4R} \]

### Step 4: Calculate the total energy at radius \( 3R \)

Using the same approach, we can find the kinetic energy at radius \( 3R \):

\[ v^2 = \frac{GM}{3R} \]

Thus, the kinetic energy \( K \) at radius \( 3R \) is:

\[ K_{3R} = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{GM}{3R}\right) = \frac{GMm}{6R} \]

Now, the total energy \( E_{3R} \) at radius \( 3R \) is:

\[ E_{3R} = K_{3R} + U_{3R} = \frac{GMm}{6R} - \frac{GMm}{3R} \]

Combining these terms:

\[ E_{3R} = \frac{GMm}{6R} - \frac{2GMm}{6R} = -\frac{GMm}{6R} \]

### Step 5: Calculate the energy required to move from \( 2R \) to \( 3R \)

The energy required \( \Delta E \) to move the satellite from radius \( 2R \) to \( 3R \) is given by the difference in total energy:

\[ \Delta E = E_{3R} - E_{2R} \]

Substituting the values we found:

\[ \Delta E = \left(-\frac{GMm}{6R}\right) - \left(-\frac{GMm}{4R}\right) \]

This simplifies to:

\[ \Delta E = -\frac{GMm}{6R} + \frac{GMm}{4R} \]

Finding a common denominator (which is \( 12R \)):

\[ \Delta E = \left(-\frac{2GMm}{12R} + \frac{3GMm}{12R}\right) = \frac{GMm}{12R} \]

### Final Answer

The energy required to move the satellite from a circular orbit of radius \( 2R \) to a radius \( 3R \) is:

\[ \Delta E = \frac{GMm}{12R} \]

---