Yes, the acceleration of the satellite \( S \) is always directed towards the center of the Earth. This is because the gravitational force, which provides the acceleration, always points towards the Earth's center, regardless of the satellite's position in its elliptical orbit. This centripetal acceleration is responsible for keeping the satellite in orbit.

For a planet in an elliptical orbit:

- Total energy (E) is the sum of kinetic energy (T) and potential energy (U).
- In a bound orbit like an ellipse, E is always negative. This indicates that the planet is gravitationally bound to the star and cannot escape.
- Kinetic energy (T) is always positive.
- Potential energy (U) is negative due to the attractive gravitational force, and its magnitude is greater than T.
- Angular momentum (L) is constant for elliptical orbits.

Thus, E is always negative for elliptical orbits.

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:
- \( a = [M L T^{-3}] \)
- \( b = [M L T^{-4}] \)

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.

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 \(T' = 8T\).