A satellite rotating around earth is orbiting with a speed $V_o$ if it is suddenly stopped then it reaches to earth surface with a velocity $\left[V_e\to \right.$escape velocity on surface)

$-KE=\frac{1}{2}PE$

$PE=-2KE$

$=-2\frac{1}{2}m{v}_o^2$

$PE=-m{v}_o^2$

When it is stopped, only it has $PE=-m{v}_o^2$

Apply law of conservation of energy

when it is on the surface of the earth its PE is $-\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_{\mathrm{e}}\right)}^2$

loss in PE=KE

$$\begin{gathered} \frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_{\mathrm{e}}\right)}^2-m{v}_o^2=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{v}=\sqrt{{\left({\mathrm{v}}_{\mathrm{e}}\right)}^2-2{\mathrm{v}}_{\mathrm{o}}^2} \end{gathered}$$