The gravitational potential \( V \) at height \( h \) from the Earth's surface is given by:

\[
V = -\frac{GM}{R + h}
\]

The acceleration due to gravity \( g \) at height \( h \) is:

\[
g = \frac{GM}{(R + h)^2}
\]

Given:
- \( V = -5.4 \times 10^7 \, \text{J/kg} \)
- \( g = 6.0 \, \text{m/s}^2 \)
- \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)

From the first equation:

\[
-5.4 \times 10^7 = -\frac{GM}{R + h}
\]

From the second equation:

\[
6.0 = \frac{GM}{(R + h)^2}
\]

Dividing the two equations to eliminate \( GM \):

\[
\frac{V}{g} = \frac{-(R + h)}{(R + h)^2}
\]

Simplifying:

\[
\frac{-5.4 \times 10^7}{6.0} = -(R + h)
\]

Solving for \( h \):

\[
R + h = 9 \times 10^6 \, \text{m}
\]

\[
h = 9 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m} = 2600 \, \text{km}
\]

Thus, the height is \( 2600 \, \text{km} \).

The gravitational potential at the center due to a semicircular rod is given by:

\[
V = - \frac{GM}{R}
\]

Where \( R \) is the radius of the semicircle, which is related to the length of the rod:

\[
L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}
\]

Substituting \( R \) into the potential formula:

\[
V = - \frac{GM}{\frac{L}{\pi}} = - \frac{\pi GM}{L}
\]

Thus, the gravitational potential at the center is:

\[
V = - \frac{\pi GM}{L}
\]