The force \( F = at + bt^2 \) has dimensions of force \([M L T^{-2}]\).

For \( at \), the dimensions of \( a \) must be:

\[
[M L T^{-2}] = [a][T]
\]

Thus, the dimensions of \( a \) are:

\[
a = [M L T^{-3}]
\]

For \( bt^2 \), the dimensions of \( b \) must be:

\[
[M L T^{-2}] = [b][T^2]
\]

Thus, the dimensions of \( b \) are:

\[
b = [M L T^{-4}]
\]

So, the dimensions are:
- \( a = [M L T^{-3}] \)
- \( b = [M L T^{-4}] \)

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:

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

Using conservation of energy: \(v_\infty = \sqrt{v^2 - v_e^2}\). Given \(v = 22.4\text{ km/s} = 2v_e\), we find \(v_\infty = \sqrt{(2v_e)^2 - v_e^2} = v_e\sqrt{3} = 11.2\sqrt{3}\text{ km/s}\).

For a satellite in circular orbit, Potential Energy is \( U = -\frac{GMm}{r} \) and Total Energy is \( E_0 = -\frac{GMm}{2r} \). Thus, Potential Energy is twice the Total Energy, \( U = 2E_0 \).

As a satellite moves to a lower orbit, total energy decreases (becomes more negative) due to dissipation, but kinetic energy increases, so speed increases. The speed of a satellite is not universally constant.

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 \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).

When rotating oppositely, \(\frac{1}{T_{\text{rel}}} = \frac{1}{T_s} + \frac{1}{T_e} \Rightarrow \frac{1}{8} = \frac{1}{T_s} + \frac{1}{24}\), which gives \(T_s = 12\text{ hours}\). When rotating in the same direction, \(\frac{1}{T_{\text{rel}}'} = \frac{1}{T_s} - \frac{1}{T_e} = \frac{1}{12} - \frac{1}{24} = \frac{1}{24}\), so \(T_{\text{rel}}' = 24\text{ hours}\).

The total energy of a satellite is \(E = -\frac{GMm}{2r}\). The required energy is \(\Delta E = E_f - E_i = -\frac{GMm}{8R} - \left(-\frac{GMm}{4R}\right) = \frac{GMm}{8R}\). Since \(g = \frac{GM}{R^2}\), we get \(\Delta E = \frac{mgR}{8}\).

According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).

By Kepler's Third Law, \(T^2 \propto R^3\). Therefore, \(\frac{T'}{T} = \left(\frac{4R}{R}\right)^{3/2} = 8\), which gives \(T' = 8T\).