A rocket is launched from the surface of Earth with an initial speed of $$10$$ $$kms-1$$. The maximum distance of rocket from the surface of Earth is n×$$104$$ km. The value of n is $$__________$$.(Radius$$ of Earth $$=$$ $$6400$$ km and escape velocity from Earth's surface $$=$$ $$11.2$$ $$km/s) Neglect atmospheric resistance.

−$$GMmR+12mv2=-GMm(R+h)$$⇒−$$GMR+12v2=$$−$$GMR+h$$⇒$$12v2=$$−$$GM(R+h)+GMR$$⇒$$12v2=GM-1R+h+1R$$ ⇒$$v2=2GMR1-11+hR$$ here, $$2GMR=ve2=11.22$$ ⇒$$1011.22=hR+h$$ ⇒$$h=2.58$$×$$104km$$

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A rocket is launched from the surface of Earth with an initial speed of 10 kms^-1. The maximum distance of rocket from the surface of Earth is n $\times$ 10⁴ km. The value of n is __________.
(Radius of Earth = 6400 km and escape velocity from Earth's surface = 11.2 km/s) Neglect atmospheric resistance.

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Solution

$\begin{array}{l} - \frac{G M m}{R} + \frac{1}{2} m v^{2} = - \frac{G M m}{(R + h)} \\ \Rightarrow \frac{- G M}{R} + \frac{1}{2} v^{2} = \frac{- G M}{R + h} \\ \Rightarrow \frac{1}{2} v^{2} = \frac{- G M}{(R + h)} + \frac{G M}{R} \\ \Rightarrow \frac{1}{2} v^{2} = G M (- \frac{1}{R + h} + \frac{1}{R}) \end{array} \Rightarrow v^{2} = \frac{2 G M}{R} (1 - \frac{1}{1 + \frac{h}{R}}) h e r e, \frac{2 G M}{R} = {v_{e}}^{2} = {(11.2)}^{2} \Rightarrow {(\frac{10}{11.2})}^{2} = \frac{h}{R + h} \Rightarrow h = 2.58 \times 10^{4} k m$

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Gravitational potential and potential energy

Remember concepts with our Masterclasses.

A rocket is launched from the surface of Earth with an initial speed of 10 kms-1. The maximum distance of rocket from the surface of Earth is n×104 km. The value of n is __________.(Radius of Earth = 6400 km and escape velocity from Earth's surface = 11.2 km/s) Neglect atmospheric resistance.

−GMmR+12mv2=-GMm(R+h)⇒−GMR+12v2=−GMR+h⇒12v2=−GM(R+h)+GMR⇒12v2=GM-1R+h+1R ⇒v2=2GMR1-11+hR here, 2GMR=ve2=11.22 ⇒1011.22=hR+h ⇒h=2.58×104km

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