Correct formula for height of a satellite from earth's surface is:

Derive the correct formula for the height of a satellite above Earth's surface.

For a satellite in circular orbit at distance $$r$$ from Earth's center, the gravitational force provides centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2 r $$

where $$\omega = \frac{2\pi}{T}$$.

$$ \frac{GM}{r^2} = \omega^2 r = \frac{4\pi^2}{T^2}r $$

$$ GM = \frac{4\pi^2 r^3}{T^2} $$

Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$. Substituting:

$$ gR^2 = \frac{4\pi^2 r^3}{T^2} $$

$$ r^3 = \frac{gR^2T^2}{4\pi^2} = \frac{T^2R^2g}{4\pi^2} $$

$$ r = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} $$

Since $$r = R + h$$:

$$ h = r - R = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R $$

The correct answer is Option B: $$\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$$.