A rocket vertically from the surface of the earth with an initil velocity(v_o ) show that its velocity v at height h is given by v_o ^2 −v^2 =((2gh)/(1+(h/R))) where R is radius of earth and g is the acceleration due to gravity at the earth surface


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Question Number 74411 by peter frank last updated on 24/Nov/19

$A r o c k e t v e r t i c a l l y f r o m$ $t h e s u r f a c e o f t h e e a r t h$ $w i t h a n i n i t i l v e l o c i t y (v_{o})$ $s h o w t h a t i t s v e l o c i t y v$ $a t h e i g h t h i s g i v e n b y$ $v_{o}^{2} - v^{2} = \frac{2 g h}{1 + \frac{h}{R}}$ $w h e r e R i s r a d i u s o f e a r t h$ $a n d g i s t h e a c c e l e r a t i o n$ $d u e t o g r a v i t y a t t h e e a r t h$ $s u r f a c e$
	Answered by ajfour last updated on 24/Nov/19

	$- \frac{G M m}{R} + \frac{{m v}_{0}^{2}}{2} = - \frac{G M m}{(R + h)} + \frac{{m v}^{2}}{2}$ $\Rightarrow v_{0}^{2} - v^{2} = 2 (\frac{G M}{R^{2}}) R (1 - \frac{R}{R + h})$ $\Rightarrow v_{0}^{2} - v^{2} = \frac{2 g h}{1 + h / R} .$
	Commented by peter frank last updated on 24/Nov/19

	$t h a n k y o u$
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