The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is '…
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is ' $\mathrm{V}$ '. For the satellite orbiting at an altitude of half the earth's radius, the orbital velocity is
$\frac{3}{2} \mathrm{~V}$
$\sqrt{\frac{3}{2}} \mathrm{~V}$
$\sqrt{\frac{2}{3}} \mathrm{~V}$
$\frac{2}{3} \mathrm{~V}$
Solution
The orbital velocity near the surface of the earth
$\mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$
At an altitude $\frac{R}{2}$, the orbital velocity
$V^{\prime}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\frac{\mathrm{R}}{2}}}=\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}}=\sqrt{\frac{2}{3}} \mathrm{~V}$