Orbital Velocity

As explained in NCERT Class 11, Chapter 8 (Gravitation), a satellite in a circular orbit needs the gravitational force to supply exactly the centripetal force required for circular motion: $\frac{m{v}_o^2}{r}=\frac{GMm}{r^2}$. Cancelling the mass and one factor of $r$ gives $v_o=\sqrt{\frac{GM}{r}}$. For a near-surface orbit the radius $r\approx R$, and using the surface relation $g=\frac{GM}{R^2}$ this simplifies neatly to $v_o=\sqrt{gR}$, avoiding any need for $G$ or the Earth's mass. Substituting $g=9.8{\text{m/s}}^2$ and $R=6.4\times 10^6\text{m}$: $v_o=\sqrt{9.8\times 6.4\times 10^6}=\sqrt{6.272\times 10^7}\approx 7920\text{m/s}=7.92\text{km/s}$. The option $11.2\text{km/s}$ is the escape velocity, which carries an extra factor of $\sqrt{2}$ and would let the satellite leave orbit entirely. The option $5.60\text{km/s}$ uses an incorrectly halved radius. The option $3.96\text{km/s}$ simply halves the correct answer. As a plausibility check, orbital speed should be smaller than escape speed by a factor of $\sqrt{2}$, and indeed $7.92\times \sqrt{2}\approx 11.2\text{km/s}$, confirming the value. This makes physical sense because escaping requires enough energy to reach zero total energy, whereas orbiting only requires balancing gravity with circular motion, so an orbiting body is always slower than one escaping from the same radius.