An artificial satellite is describing an equatorial orbit at 3600 km above the earth’s surface. Calculate its orbital speed.

6.335 km/sec

The time period of satellite is given by

$T^2=\frac{4 \pi^2}{\mathrm{GM}}(\mathrm{R}+\mathrm{h})^3=\frac{4 \pi^2}{\mathrm{~g}} \frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{R}^2}$

$\mathrm{~T}=\frac{2 \pi}{\mathrm{R}} \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{~g}}}=\frac{2 \pi}{6400 \times 10^3} \sqrt{\frac{10^{21}}{9.8}}\left[\text { as } \mathrm{R}+\mathrm{h}=(6400+3600) \times 10^3 \mathrm{~m}=10^7 \mathrm{~m}\right]$

= 31360.78 sec = 8.71 hrs

Orbital speed is

V₀ = $\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}=\sqrt{\frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}}=\frac{9.8 \times\left(6400 \times 10^3\right)^2}{10^7}$ = m/s = 6335 m/s

= 6.335 km/sec