The gravitational force between two masses \( m_1 \) and \( m_2 \) is:

\[
F = G \frac{m_1 \cdot m_2}{r^2}
\]

To maximize \( F \), we set \( m_1 = x \) and \( m_2 = 100 - x \), then maximize the product \( m_1 \cdot m_2 = x(100 - x) \).

This product is maximized when \( x = 50 \). So, the masses should be split equally.

The ratio of masses is:

\[
\frac{m_1}{m_2} = \frac{50}{50} = 1
\]

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:

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

To find the net gravitational force on the mass \( m \) at the centroid, we follow these steps:

1. Forces due to extra masses \( 2m \):
- Each of the two lower vertices has an extra mass \( 2m \).
- The gravitational force due to one \( 2m \) mass at a distance \( \frac{a}{\sqrt{3}} \) from the centroid is:
\[
F = \frac{G (2m) m}{\left( \frac{a}{\sqrt{3}} \right)^2} = \frac{6 G m^2}{a^2}
\]

2. Net force between the two extra masses:
- The angle between the two forces is \( 120^\circ \).
- The resultant force is given by the vector addition formula:
\[
F_{\text{resultant}} = \sqrt{F^2 + F^2 + 2 F F \cos 120^\circ}
\]
Since \( \cos 120^\circ = -\frac{1}{2} \):
\[
F_{\text{resultant}} = \sqrt{F^2 + F^2 - F^2} = F = \frac{6 G m^2}{a^2}
\]

3. Conclusion:
The net gravitational force on the mass \( m \) at the centroid is:
\[
F_{\text{resultant}} = \frac{6 G m^2}{a^2}
\]
This force is directed vertically downward.

The gravitational force between two masses \( m_1 \) and \( m_2 \) is given by Newton's law of gravitation:

\[
F = \frac{G m_1 m_2}{r^2}
\]

Let the total mass be \( M = 100 \, \text{kg} \), and split it into two parts: \( m_1 = x \) and \( m_2 = 100 - x \).

The gravitational force becomes:

\[
F = \frac{G x (100 - x)}{r^2}
\]

To maximize \( F \), we need to maximize \( x(100 - x) \), which is a quadratic function. The product \( x(100 - x) \) is maximized when \( x = 50 \).

Thus, the ratio of the two masses is:

\[
m_1 : m_2 = 50 : 50 = 1:1
\]

Therefore, the masses should be in a 1:1 ratio for the gravitational force to be maximum.

The additional kinetic energy required to transfer a satellite from a circular orbit of radius \( R_1 \) to \( R_2 \) is the difference in kinetic energy between the two orbits.

Kinetic energy in a circular orbit is:

\[
K = \frac{GMm}{2R}
\]

The kinetic energy difference is:

\[
\Delta K = \frac{GMm}{2R_1} - \frac{GMm}{2R_2}
\]

Simplifying:

\[
\Delta K = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]

So, the additional kinetic energy is:

\[
\Delta K = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]

The acceleration of the earth towards the body is \(a_E = 2g\) and the body towards the earth is \(a_B = g\). The relative acceleration is \(a_{rel} = 3g\). Using \(h = \frac{1}{2} a_{rel} t^2\), we get \(t = \sqrt{\frac{2h}{3g}}\).

The error in force is \(\Delta F = m(g_1 - g_2) \approx m \frac{2g}{R} h\). Since \(g = \frac{4}{3} \pi G \rho R\), we have \(\frac{2g}{R} = \frac{8}{3} \pi G \rho\). Therefore, \(\Delta F = \frac{8}{3} \pi G \rho m h\).

For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).