The escape velocity \( V_e \) from the Earth's surface is given by:

\[
V_e = \sqrt{\frac{2GM}{R}}
\]

At a height \( h = R \) (i.e., the platform is at a distance \( 2R \) from the Earth's center), the escape velocity \( V \) becomes:

\[
V = \sqrt{\frac{2GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

So, the factor \( F \) is:

\[
F = \frac{V}{V_e} = \frac{1}{\sqrt{2}}
\]

Gravitational potential \( V \) at the midpoint is the sum of the potentials due to both masses:

\[
V = -\frac{G \cdot 10^2}{0.5} - \frac{G \cdot 10^3}{0.5}
\]

Simplifying:

\[
V = -2G(10^2 + 10^3) = -2G \cdot 1100
\]

So, the gravitational potential at the midpoint is:

\[
V = -2200G
\]

The velocity required to reach a height equal to the Earth's radius \( R \) is the escape velocity for a total distance of \( 2R \) from the Earth's center.

Escape velocity formula:

\[
v = \sqrt{\frac{GM}{R}}
\]

At height \( h = R \), the velocity needed is:

\[
v = \sqrt{\frac{GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

Thus, the velocity is:

\[
v = \frac{V_e}{\sqrt{2}} = \frac{\sqrt{2GM/R}}{\sqrt{2}} = \sqrt{\frac{GM}{R}} = V_e/\sqrt{2}= \sqrt{\frac{GM}{R}}
\]

The point where the gravitational field is zero lies closer to the smaller mass \( m \). Let the distance of this point from \( m \) be \( x \), and from \( 4m \) be \( r - x \).

At this point:

\[
\frac{Gm}{x^2} = \frac{G \cdot 4m}{(r - x)^2}
\]

Taking the square root:

\[
\frac{1}{x} = \frac{2}{r - x}
\]

Solving:

\[
r - x = 2x \quad \Rightarrow \quad x = \frac{r}{3}
\]

The gravitational potential \( V \) at this point is the sum of the potentials due to both masses:

\[
V = -\frac{Gm}{\frac{r}{3}} - \frac{G \cdot 4m}{\frac{2r}{3}} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}
\]

Thus, the potential at the point is:

\[
V = -\frac{9Gm}{r}
\]

The escape velocity \( V_e \) is the speed needed to escape Earth's gravitational pull. The energy conservation principle applies, where the total mechanical energy at the surface and at height \( h = 3R \) (where \( R \) is the Earth's radius) should be equal.

The total energy at the surface:
\[
E_1 = \frac{1}{2} m V_e^2 - \frac{G M m}{R}
\]

At height \( 3R \), the total energy is:
\[
E_2 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Since \( E_1 = E_2 \), we equate the two:
\[
\frac{1}{2} m V_e^2 - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

We know that the escape velocity is given by:
\[
V_e^2 = \frac{2 G M}{R}
\]

Substitute this into the equation:
\[
\frac{1}{2} m \frac{2 G M}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Simplifying:
\[
\frac{G M m}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

This simplifies to:
\[
0 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Solving for \( v^2 \):
\[
\frac{1}{2} m v^2 = \frac{G M m}{4R}
\]

\[
v^2 = \frac{2 G M}{4R} = \frac{G M}{2R}
\]

Thus, the velocity at height \( 3R \) is:
\[
v = \sqrt{\frac{G M}{2R}} = \frac{V_e}{\sqrt{2}}= 7.92 km/sec
\]

So, the velocity at height \( 3R \) is \( \frac{V_e}{\sqrt{2}} \).= 7.92 km/sec

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:
\[
V = -\frac{3GM}{a}
\]

The gravitational potential inside any spherical shell is constant and equals the potential at its surface. Therefore, the total potential at the common centre is the sum of the potentials: \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).