1. Distance of satellite from Earth's center

If the satellite is at height $x$ above the Earth's surface, then its distance from the center of the Earth is

$r = R + x$

2. Gravitational force provides centripetal force

For a satellite in circular orbit,

$\frac{GMm}{(R+x)^2} = \frac{mv^2}{R+x}$

Cancel $m$ from both sides:

$\frac{GM}{(R+x)^2} = \frac{v^2}{R+x}$

So,

$v^2 = \frac{GM}{R+x}$

Hence,

$v = \sqrt{\frac{GM}{R+x}}$

3. Use relation between $g$ and $GM$

On the Earth's surface,

$g = \frac{GM}{R^2}$

Therefore,

$GM = gR^2$

Substitute into the expression for $v$:

$v = \sqrt{\frac{gR^2}{R+x}}$

4. Match with options

Thus the orbital speed is

$\boxed{\left(\frac{gR^2}{R+x}\right)^{1/2}}$

This matches Option D.