The initial velocity $$v_i$$ required to project a body vertically upward from the surface of the earth to reach a height of $$10R$$, where $$R$$ is the radius of the earth, may be described in terms of escape velocity $$v_e$$ such that $$v_i = \sqrt{\frac{x}{y}} \times v_e$$. The value of $$x$$ will be

We use conservation of energy for a body projected from Earth's surface to height $$10R$$. At the surface, kinetic energy is $$\frac{1}{2}mv_i^2$$ and gravitational potential energy is $$-\frac{GMm}{R}$$. At height $$10R$$ from the surface (i.e., distance $$11R$$ from Earth's centre), the velocity is zero and potential energy is $$-\frac{GMm}{11R}$$.

Applying conservation of energy: $$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$$

This gives $$\frac{1}{2}mv_i^2 = \frac{GMm}{R} - \frac{GMm}{11R} = \frac{GMm}{R}\left(1 - \frac{1}{11}\right) = \frac{GMm}{R} \cdot \frac{10}{11}$$

So $$v_i^2 = \frac{2GM}{R} \cdot \frac{10}{11}$$. We know the escape velocity is $$v_e = \sqrt{\frac{2GM}{R}}$$, so $$v_e^2 = \frac{2GM}{R}$$.

Substituting: $$v_i^2 = \frac{10}{11} v_e^2$$, which gives $$v_i = \sqrt{\frac{10}{11}} \times v_e$$.

Comparing with $$v_i = \sqrt{\frac{x}{y}} \times v_e$$, we get $$x = 10$$ and $$y = 11$$. Therefore, the value of $$x$$ is $$10$$.