Question

derivation of escape velocity.

Answer 1

dw =fdx
=fdxcos o
=fdx
dw=GMmdx/x^2
integrate
/dw= /GMmdx/x^2
w=GMm/R
GMm/R=1/2mv^2
Ve^2=2gR

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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