Energy of a satellite in circular orbit is `E_(0)`. The energy required to move the satellite to a circular orbit of 3 time the radius of the initial orbit is `(x)/(3)E_(0)`. Find the value of x.

Answer

To solve the problem, we need to find the energy required to move a satellite from a circular orbit of radius \( R \) to a circular orbit of radius \( 3R \). The energy of the satellite in the initial orbit is given as \( E_0 \), and we need to express the energy required to move to the new orbit in terms of \( E_0 \).

### Step-by-step Solution:

1. **Calculate the Total Energy in the Initial Orbit**:
The total energy \( E_0 \) of a satellite in a circular orbit can be expressed as:
\[ E_0 = K + U \]
where \( K \) is the kinetic energy and \( U \) is the potential energy.

The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m v^2 \]
The orbital velocity \( v \) can be calculated using:
\[ v = \sqrt{\frac{GM}{R}} \]
Thus, the kinetic energy becomes:
\[ K = \frac{1}{2} m \left(\sqrt{\frac{GM}{R}}\right)^2 = \frac{GMm}{2R} \]

The potential energy \( U \) is given by:
\[ U = -\frac{GMm}{R} \]

Therefore, the total energy \( E_0 \) is:
\[ E_0 = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R} \]

2. **Calculate the Total Energy in the New Orbit**:
Now, consider the satellite in the new orbit at a radius of \( 3R \). The total energy \( E_0' \) in this orbit can be calculated similarly:
\[ K' = \frac{1}{2} m v'^2 \]
where \( v' = \sqrt{\frac{GM}{3R}} \). Thus, the kinetic energy becomes:
\[ K' = \frac{1}{2} m \left(\sqrt{\frac{GM}{3R}}\right)^2 = \frac{GMm}{6R} \]

The potential energy \( U' \) at this new radius is:
\[ U' = -\frac{GMm}{3R} \]

Therefore, the total energy \( E_0' \) is:
\[ E_0' = \frac{GMm}{6R} - \frac{GMm}{3R} = \frac{GMm}{6R} - \frac{2GMm}{6R} = -\frac{GMm}{6R} \]

3. **Calculate the Energy Required to Move to the New Orbit**:
The energy required to move the satellite from the initial orbit to the new orbit is given by:
\[ \Delta E = E_0' - E_0 \]
Substituting the values we calculated:
\[ \Delta E = -\frac{GMm}{6R} - \left(-\frac{GMm}{2R}\right) = -\frac{GMm}{6R} + \frac{3GMm}{6R} = \frac{2GMm}{6R} = \frac{GMm}{3R} \]

4. **Express the Energy Required in Terms of \( E_0 \)**:
We know that:
\[ E_0 = -\frac{GMm}{2R} \]
Therefore, we can express \( \Delta E \) in terms of \( E_0 \):
\[ \Delta E = \frac{GMm}{3R} = \frac{1}{3} \left(-2E_0\right) = -\frac{2E_0}{3} \]

5. **Relate \( \Delta E \) to the Given Expression**:
According to the problem, the energy required to move the satellite to the new orbit is given as:
\[ \Delta E = \frac{x}{3} E_0 \]
Setting the two expressions for \( \Delta E \) equal gives:
\[ -\frac{2E_0}{3} = \frac{x}{3} E_0 \]
Dividing both sides by \( E_0 \) (assuming \( E_0 \neq 0 \)):
\[ -2 = x \]

### Final Answer:
Thus, the value of \( x \) is:
\[ \boxed{-2} \]