A satellite is revolving in a circular orbit at a height $$h$$ from the earth's surface (radius of earth $$R$$; $$h << R$$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to (Neglect the effect of atmosphere.)

We begin by recalling the two standard formulae that govern motion under gravity around the earth:

1. The magnitude of the gravitational acceleration at the earth’s surface is related to the earth’s mass by $$g \;=\;\dfrac{G\,M}{R^{2}}.$$

2. For any circular orbit of radius $$r$$ about the earth, the orbital speed $$v_\text{orbit}$$ is given by $$v_\text{orbit}\;=\;\sqrt{\dfrac{G\,M}{r}}.$$

3. From the same radius $$r$$, the escape speed $$v_\text{escape}$$ is obtained from the energy condition $$\dfrac{1}{2}m v_\text{escape}^{2}\;=\;\dfrac{G\,M\,m}{r},$$ whence $$v_\text{escape}\;=\;\sqrt{\dfrac{2\,G\,M}{r}}.$$

The satellite in the problem is already moving in a circular orbit of radius

$$r \;=\; R + h,$$

where $$R$$ is the radius of the earth and $$h$$ is the altitude of the satellite. The given condition $$h \ll R$$ will allow a useful approximation later.

Using formula 2, the present orbital speed of the satellite is

$$v_\text{orbit}\;=\;\sqrt{\dfrac{G\,M}{R+h}}.$$

Using formula 3, the escape speed needed from the same point is

$$v_\text{escape}\;=\;\sqrt{\dfrac{2\,G\,M}{R+h}}.$$

The satellite must increase its speed by the difference of these two values. Hence the required increment $$\Delta v$$ is

$$\Delta v \;=\; v_\text{escape} \;-\; v_\text{orbit}$$ $$=\;\sqrt{\dfrac{2\,G\,M}{R+h}} \;-\;\sqrt{\dfrac{G\,M}{R+h}}.$$

We factor out the common square-root:

$$\Delta v \;=\;\sqrt{\dfrac{G\,M}{R+h}}\;\Bigl(\sqrt{2}\;-\;1\Bigr).$$

Next we eliminate $$G\,M$$ in favour of the surface quantity $$g$$ by substituting $$G\,M = g\,R^{2}$$ from formula 1:

$$\sqrt{\dfrac{G\,M}{R+h}} \;=\; \sqrt{\dfrac{g\,R^{2}}{R+h}}.$$

Because $$h \ll R,$$ we may approximate $$R+h \approx R.$$ This gives

$$\sqrt{\dfrac{g\,R^{2}}{R+h}} \;\approx\; \sqrt{\dfrac{g\,R^{2}}{R}} \;=\; \sqrt{g\,R}.$$

Substituting this back, we obtain

$$\Delta v \;\approx\; \sqrt{g\,R}\,\Bigl(\sqrt{2}\;-\;1\Bigr).$$

This expression matches Option B.

Hence, the correct answer is Option B.