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:

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

\[ g= \frac{GMr}{R^{3}}  \] for internal point

\[ g= \frac{GM}{r^{2}}  \] for external point

Total gravitational potential at a point \( r = \frac{a}{2} \):

1. Potential due to the mass at the center:
The gravitational potential at a distance \( r \) from a point mass \( M \) is given by:
\[
V_{\text{center}} = -\frac{GM}{r}
\]
At \( r = \frac{a}{2} \), this becomes:
\[
V_{\text{center}} = -\frac{GM}{\frac{a}{2}} = -\frac{2GM}{a}
\]

2. **Potential due to the spherical shell**:
Inside a spherical shell, the gravitational potential is constant and equal to the potential at the surface, which is:
\[
V_{\text{shell}} = -\frac{GM}{a}
\]

### Total potential at \( r = \frac{a}{2} \):

The total gravitational potential is the sum of the potentials due to the mass at the center and the shell:
\[
V_{\text{total}} = V_{\text{center}} + V_{\text{shell}} = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the gravitational potential at a distance \( \frac{a}{2} \) from the center is:
\[
V = -\frac{3GM}{a}
\]

At the midpoint, the gravitational forces exerted by the two identical spheres on the mass \( m \) have the same magnitude but act in opposite directions. Since these forces cancel each other out completely, the net gravitational force on the mass \( m \) is:

\[
F = 0
\]

To find the dimensions of the gravitational constant \( G \), use Newton's law of gravitation:

\[
F = \frac{G M_1 M_2}{r^2}
\]

Where:
- \( F \) is force (with dimensions \( [M L T^{-2}] \)),
- \( M_1 \) and \( M_2 \) are masses (with dimensions \( [M] \)),
- \( r \) is distance (with dimensions \( [L] \)).

Rearranging for \( G \):

\[
G = \frac{F r^2}{M_1 M_2}
\]

Substitute the dimensions:

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

Simplify:

\[
G = [M^{-1} L^3 T^{-2}]
\]

Thus, the dimensions of \( G \) are \( [M^{-1} L^3 T^{-2}] \).

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

Acceleration due to gravity is \(g = \frac{GM}{R^2}\). Since mass \(M = \rho \times \frac{4}{3}\pi R^3\), we get \(g = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi \rho GR\). Rearranging gives \(\rho = \frac{3g}{4\pi RG}\).

Escape velocity is given by the formula \( v_e = \sqrt{2gR} \). The ratio of escape velocity of earth to moon is \( \frac{v_{earth}}{v_{moon}} = \sqrt{\frac{g_{earth}}{g_{moon}} \times \frac{R_{earth}}{R_{moon}}} = \sqrt{6 \times 10} = \sqrt{60} \approx 7.75 \approx 8 \).

Acceleration due to gravity decreases with depth as \( g_d = g(1 - \frac{d}{R}) \) and with height as \( g_h = g(1 - \frac{2h}{R}) \). Since the formulas and rates of decrease are different, they decrease at different rates.