A satellite is revolving round the earth with orbital speed v0. If it stops suddenly, the speed withwhich it will strike the surface of earth would be ve2−kv02(ve= escape velocity of a particle onearth’s surface). Find the value of K.

# A satellite is revolving round the earth with orbital speed ${v}_{0}$. If it stops suddenly, the speed withwhich it will strike the surface of earth would be $\sqrt{{\mathrm{v}}_{\mathrm{e}}^{2}-{\mathrm{kv}}_{0}^{2}}$$\left({v}_{e}=$ escape velocity of a particle onearth’s surface). Find the value of K.

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### Solution:

$\frac{-\mathrm{GMm}}{\mathrm{r}}+\mathrm{O}=\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2}{\mathrm{mV}}^{2}$

$⇒\frac{-\mathrm{GM}}{\mathrm{r}}=\frac{-\mathrm{GM}}{\mathrm{R}}+\frac{{\mathrm{V}}^{2}}{2}⇒{\mathrm{V}}^{2}=\frac{2\mathrm{GM}}{\mathrm{R}}-\frac{2\mathrm{GM}}{\mathrm{r}}$

$⇒\mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{e}}^{2}-2{\mathrm{V}}_{0}^{2}}$

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