Orbital Velocity

For a circular orbit the gravitational force supplies exactly the centripetal force, so $\frac{GMm}{r^2}=\frac{m{v}_o^2}{r}$, which simplifies to $v_o=\sqrt{\frac{GM}{r}}$. For an orbit grazing the surface $r\approx R$, and using $GM=g{R}^2$ gives $v_o=\sqrt{gR}$. Substituting, $v_o=\sqrt{9.8\times 6.4\times 10^6}=\sqrt{6.27\times 10^7}\approx 7.92\times 10^3$ m/s, that is about 7.92 km/s. The option $11.2$ km/s is the escape velocity, larger than the orbital speed by $\sqrt{2}$. The option $5.60$ km/s underestimates by using $r=2R$. The option $9.80$ km/s mistakenly equates the speed with the numerical value of $g$. This is the canonical NCERT first cosmic velocity. A key conceptual point is that orbital speed decreases for higher orbits, since $v_o=\sqrt{GM/r}$ falls as $r$ grows, so distant satellites move more slowly than low-orbit ones. The orbital speed is also independent of the satellite's own mass, because mass cancels when gravitational force is equated to the centripetal requirement, meaning a heavy and a light satellite share the same orbit at the same speed. A plausibility check confirms $v_o=v_e/\sqrt{2}$, so the orbital speed sits sensibly below the escape speed, exactly as expected for a bound circular orbit.