Question

A particle is projected from the earth's surface with an initial speed of $4km/s$. What will be the maximum height, in $km$, attained by the particle?

Answer 1

Since the motion is under gravitational force which is conservative force , energy should be conserved

this gives: $initialenergy=finalenergy$

$P.E_i+K.E_i=P.E_f+K.E_f$

So, $\frac{-GMm}{R}+\frac{1}{2}m{v}^2=\frac{-GMm}{(R+h)}+0$

(Here since speed is very high, we cannot consider $g$ as constant )

$\Rightarrow -\frac{GM}{R}+16/2\times {10}^6=\frac{GM}{(R+h)}\Rightarrow \frac{-6.67\times {10}^{-11}\times 6\times {10}^{24}}{64\times {10}^5}+8\times {10}^6=\frac{-GM}{R+h}$

solving we get,

$\frac{-GM}{R+h}=-54\times {10}^6$

$R+h=\frac{GM}{54\times {10}^6}=\frac{40\times {10}^{13}}{54\times {10}^6}=74\times {10}^5$

So, $h=74\times {10}^5-64\times {10}^5$

$h=10\times {10}^{5}m=1000km$