An object is allowed to fall from a height $$R$$ above the earth, where $$R$$ is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be:

We have an object falling from height $$R$$ above the Earth's surface (where $$R$$ is the radius of Earth). We use conservation of energy to find the velocity at the surface.

The gravitational potential energy at a distance $$r$$ from the center of Earth is $$U = -\frac{GMm}{r}$$. At the initial position (distance $$2R$$ from center), the object is at rest, so

$$E_i = -\frac{GMm}{2R}$$

At the Earth's surface (distance $$R$$ from center):

$$E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

By conservation of energy, $$E_i = E_f$$, so

$$-\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

Rearranging:

$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$

This gives $$v^2 = \frac{GM}{R}$$. Now since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$, so

$$v^2 = \frac{gR^2}{R} = gR$$

So, the answer is $$v = \sqrt{gR}$$.