The gravitational potential at a point above the surface of earth is $$-5.12 \times 10^7 \text{ J kg}^{-1}$$ and the acceleration due to gravity at that point is $$6.4 \text{ m s}^{-2}$$. Assume that the mean radius of earth to be $$6400 \text{ km}$$. The height of this point above the earth's surface is:

Given: Gravitational potential $$V = -5.12 \times 10^7 \text{ J kg}^{-1}$$, acceleration due to gravity $$g = 6.4 \text{ m s}^{-2}$$, and mean radius of Earth $$R = 6400 \text{ km}$$.

Key formulas:

At distance $$r$$ from the centre of Earth:

$$V = -\frac{GM}{r}$$ and $$g = \frac{GM}{r^2}$$

Find $$r$$ from the ratio $$V/g$$: $$\frac{|V|}{g} = \frac{GM/r}{GM/r^2} = r$$

$$r = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \text{ m} = 8000 \text{ km}$$

Find the height above the surface: $$h = r - R = 8000 - 6400 = 1600 \text{ km}$$

The correct answer is Option (1): $$1600 \text{ km}$$.