A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _________ s. [take $$g = \pi^2$$ m s$$^{-2}$$]

At height $$h = R$$ from surface, distance from center = $$2R$$.

Effective $$g' = \frac{g}{(1 + h/R)^2} = \frac{g}{4}$$.

$$T = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{4}{g/4}} = 2\pi\sqrt{\frac{16}{g}} = 2\pi \times \frac{4}{\sqrt{g}}$$

With $$g = \pi^2$$: $$T = 2\pi \times \frac{4}{\pi} = 8$$ s.

Therefore, the answer is $$\boxed{8}$$.