$\begin{bmatrix}\text{where M}_{\text{p}}=\text{Mass of planet}\\\ \ \ \ \ \ \ \ \ \ \text{R}_{\text{p}}=\text{Radius of planet}\end{bmatrix}$
If the object is thrown from surface of a planet h = 0, we get
$(\text{v}_{\text{i}})_{\text{e}}=\sqrt{\frac{2\text{GM}_{\text{p}}}{\text{R}_{\text{p}}}}$
$\text{but }\text{g}=\frac{\text{GM}_{\text{p}}}{\text{R}_{\text{p}}}\text{ we get}$
$(\text{v}_{\text{i}})_{\text{e}}=\sqrt{2\text{gR}_{\text{p}}}$
Depends on location as 'g' varies with location, as most of the celestial bodies are not perfectly spherical.
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