Question

derivation of escape velocity

Answer 1

Hey

For any planet, escape velocity is the minimum desired velocity that'll take you away above the planet [ against the existing gravity ]

For such a motion, the required kinetic energy must be equal to or greater than the Potential Energy :->
$K.E._e \geq P.E._e$  ---------> ( i )

We know :->
[tex]K.E._e = \frac{1}{2} mv_e^2 [ where \ v_e \ is \ the \ escape \ velocity ] \\ P.E._e = \frac{GMm}{R} [ where \ the \ letters \ have \ their \ usual \ meaning ][/tex]  -----> ( ii )

From ( i ) and ( ii ) -->

$\frac{1}{2} mv_e^2 \geq \frac{GMm}{R} \\ \\ v_e \geq \sqrt{ \frac{2GM}{R} }$

Further, if we know about 'g ( the acceleration due to gravity of a planet )' , it makes our equation more neater :->
$v_e \geq \sqrt{2gR}$ 

Now, by definition of Escape Velocity, it is the minimum such velocity, which gives us -->
$v_e = \sqrt{2gR}$

Hope it helps ^^

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

$g$ = $\frac{GM}{ R^{2} }$

$R^{2}$ = $GM$

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

$v_{e}$ = $\sqrt{2gR}$

======================================