A planet of mass `m_(1)` revolves round the sun of mass `m_(2)`. The distance between the sun the planet is `r`. Considering the motion of the sun find the total energy of the system assuming the orbits to be circular.

Explanation

To find the total energy of a system where a planet of mass \( m_1 \) revolves around the sun of mass \( m_2 \) at a distance \( r \), we will consider both the kinetic energy and the gravitational potential energy of the system.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Planet**:
The planet is in circular motion around the sun, which means there is a centripetal force acting on it. This centripetal force is provided by the gravitational force between the planet and the sun.

2. **Write the Gravitational Force Equation**:
The gravitational force \( F_g \) between the planet and the sun is given by:
\[ F_g = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant.

3. **Write the Centripetal Force Equation**:
The centripetal force \( F_c \) required to keep the planet moving in a circular path is given by:
\[ F_c = \frac{m_1 v^2}{r} \]
where \( v \) is the orbital speed of the planet.

4. **Set the Gravitational Force Equal to the Centripetal Force**:
Since the gravitational force provides the necessary centripetal force, we can set the two equations equal to each other:
\[ \frac{G m_1 m_2}{r^2} = \frac{m_1 v^2}{r} \]

5. **Solve for the Orbital Speed \( v \)**:
We can simplify the equation by canceling \( m_1 \) (assuming \( m_1 \neq 0 \)) and multiplying both sides by \( r \):
\[ \frac{G m_2}{r} = v^2 \]
Therefore, we find:
\[ v = \sqrt{\frac{G m_2}{r}} \]

6. **Calculate the Kinetic Energy \( KE \)**:
The kinetic energy of the planet is given by:
\[ KE = \frac{1}{2} m_1 v^2 \]
Substituting the expression for \( v^2 \):
\[ KE = \frac{1}{2} m_1 \left(\frac{G m_2}{r}\right) = \frac{G m_1 m_2}{2r} \]

7. **Calculate the Gravitational Potential Energy \( PE \)**:
The gravitational potential energy of the system is given by:
\[ PE = -\frac{G m_1 m_2}{r} \]
(The negative sign indicates that the gravitational force is attractive.)

8. **Calculate the Total Energy \( E \)**:
The total energy of the system is the sum of the kinetic energy and the potential energy:
\[ E = KE + PE = \frac{G m_1 m_2}{2r} - \frac{G m_1 m_2}{r} \]
Simplifying this gives:
\[ E = \frac{G m_1 m_2}{2r} - \frac{2G m_1 m_2}{2r} = -\frac{G m_1 m_2}{2r} \]

### Final Result:
The total energy of the system is:
\[ E = -\frac{G m_1 m_2}{2r} \]