Solution:

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:

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