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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