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A communication is launched at 42000 km above Earth. Find its orbital speed. $(2876 \: ms^{-1})$

Difficulty: Easy

Solution:

Height = h = 42000 km = $42000\times 1000$ m = $42 \times10^{6}$m
        Orbital velocity = $v_{o}$ =?


$√8.27*10^{6}$
        $v_{o}$=$\frac{√(〖 GM〗_{e}}{(R+h))}$
        $v_{o}$ = $\frac{√((6.673 \times 10^{-11}\times 6.0 \times 10^{24})}{(6.4 \times 10^{6}+ 42 \times 10^{6}))}$ 
            =$\frac{√((40.038 \times10^{13})}{48.4\times 10^{6}))}$
              =$\frac{√((400.38 \times 10^{12})}{(48.4 \times 10^{6}))}$
              = $√(8.27\times 10^{6})$
            = $2.876 \times 10^{3} = 2876 ms^{-1}$

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