Question

explain the derivation of escape velocity

Answer 1

Check out Escape Velocity of Earth for a detailed understanding of the minimum velocity required to overcome earth's gravitational pull. Escape Velocity Formula:Derivation: Assume a perfect sphere shaped planet of radius R and mass M. Now, if a body of mass m is projected from a point A on the surface of the planet...

Answer 2

$\textbf{Answer :}$


Gravitational potential energy + kinetic energy = 0


$\frac{-GMm}{R}$ + $\frac{m v_{e} ^{2} }{2}$  = $0$

$\frac{m v_{e} ^{2} }{2}$ = $\frac{GMm}{R}$

$v_{e} ^{2}$  = $\frac{2GM}{R}$ 

$v_{e}$  =  $\sqrt{ \frac{2GM}{R} }$

$v_{e}$ = $11.2 km/s$


Also we know that


$g$ = $\frac{GM}{ R^{2} }$

$R^{2}$ = $GM$

$v_{e}$  = $\sqrt{ \frac{2 R^{2}g }{R} }$

$v_{e}$ = $\sqrt{2gR}$

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