A rocket is accelerated to speed $v = 2 \sqrt{g R}$ near the earth's surface ( $R =$ radius of earth). Show that very far from earth its speed will be $v = 2 \sqrt{g R}$ .

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Solution

$K_{i} + U_{i} = K_{f} + U_{f}$
$∴ \frac{1}{2} m (2 \sqrt{g R})^{2} - \frac{G M m}{R}$
$= \frac{1}{2} m v^{2} + 0$
$∴ v^{2} = 4 g R - \frac{2 G M}{R}$
Putting $G M = g R^{2}$ ,we get
$v^{2} = 4 g R - 2 g R$
$∴ v = \sqrt{2 g R}$

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A rocket is accelerated to speed v=2√gR near the earth's surface (R= radius of earth). Show that very far from earth its speed will be v=2√gR.

Ki+Ui=Kf+Uf∴12m(2√gR)2−GMmR=12mv2+0∴v2=4gR−2GMRPutting GM=gR2,we get v2=4gR−2gR∴v=√2gR

A rocket is accelerated to speed $v = 2 \sqrt{g R}$ near the earth's surface ( $R =$ radius of earth). Show that very far from earth its speed will be $v = 2 \sqrt{g R}$ .

$K_{i} + U_{i} = K_{f} + U_{f}$
$∴ \frac{1}{2} m (2 \sqrt{g R})^{2} - \frac{G M m}{R}$
$= \frac{1}{2} m v^{2} + 0$
$∴ v^{2} = 4 g R - \frac{2 G M}{R}$
Putting $G M = g R^{2}$ ,we get
$v^{2} = 4 g R - 2 g R$
$∴ v = \sqrt{2 g R}$