Solution:

The total mechanical energy \( E \) of a satellite in a circular orbit is the sum of its kinetic energy \( K \) and potential energy \( U \).

1. **Kinetic Energy (K):**

\[
K = \frac{1}{2} m v^2
\]

2. **Potential Energy (U):**

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

For a satellite orbiting closely around the Earth, its velocity \( v \) is related to the Earth's gravitational constant by:

\[
v = \sqrt{\frac{GM}{r}}
\]

Substitute \( v^2 = \frac{GM}{r} \) into the kinetic energy expression:

\[
K = \frac{1}{2} m \left( \frac{GM}{r} \right) = \frac{GMm}{2r}
\]

Now, the **total energy** is:

\[
E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}
\]

Thus, the total energy of the satellite is:

\[
E = -\frac{GMm}{2r}= -\frac{1}{2}mv^{2}
\]