Orbital Velocity

Question

An imaging satellite is to be placed in a circular orbit skimming just above the Earth's surface for a low-altitude survey. Using g = 9.8 m/s^2 and R = 6.4 × 10^6 m, what orbital speed must it maintain?

RankGuru · Accepted Answer

For a circular orbit the gravitational force supplies exactly the centripetal force, so $\frac{GMm}{r^2} = \frac{mv_o^2}{r}$, which simplifies to $v_o = \sqrt{\frac{GM}{r}}$. For an orbit grazing the surface $r \approx R$, and using $GM = gR^2$ gives $v_o = \sqrt{gR}$. Substituting, $v_o = \sqrt{9.8	imes 6.4	imes10^6} = \sqrt{6.27	imes10^7} \approx 7.92	imes10^3$ 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 $\sqrt2$. 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 $v_o = v_e/\sqrt2$, so the orbital speed sits sensibly below the escape speed, exactly as expected for a bound circular orbit.

Orbital Velocity

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