Explain how the formula for escape velocity is derived?

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Answer

$(i)$ Space launch vehicles are used for the launching of artificial satellite which overcome the earth's gravitational force and more into the space.
$(ii)$ The velocity of an object should be more than the escape velocity, then only the object can be launched in the space.
$V _{ esc }=\sqrt{\frac{2 GM }{ R }}$
$G -$ Gravitational constant $= 6.67 \times 10^{-11} \ Nm^2/kg^2$
$M -$ Mass of the planet (earth) $= 6 \times 10^{24} \ kg$
$R -$ Radius of the earth $= 6.4 \times 10^6 \ m$
$ V =\sqrt{\frac{2 GM }{ R }}$
$V =\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^6}}$
$V=11.18\times10^3 \ m/s \ \ 11.18 \ Km/s.$
Hence, the escape velocity for earth is $11.18 \ km/s.$
The satellite launch machine should have velocity more than or at least $11.2 \ km/s.$

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