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#### A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly to that of earth's radius $R_{e}$. By firing rockets attached to it its speed is instanteneously increased in the direction of its motion so that it becomes $\sqrt{3/2}$ times larger. Due to this the farthest distance frim the centre of the earth that the satellite reaches is R. value of R is:      Option: 1 Option: 2   Option: 3   Option: 4

Initially at A

$\frac{GM_em}{R_e^2}=\frac{mV_0^2}{R_e}\\ V_0=\sqrt{\frac{GM_e}{R_e}}$

Finally at A,

$V_1=\sqrt{\frac{ 3}{2}}V_0=\sqrt{\frac{3GM_e}{2R_e}}$

From energy conversation
$\small \dpi{120} -\frac{\mathrm{GM_em}}{\mathrm{R}_{\mathrm{e}}}+\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{3}{2}} \mathrm{~V}\right)^{2}=-\frac{\mathrm{GM_em}}{\mathrm{R}_{\max }}+\frac{1}{2} \mathrm{mV}_{\min }^{2}$
From angular momentum conversation
$\sqrt{\frac{3}{2}} \mathrm{VR}_{\mathrm{e}}=\mathrm{V}_{\mathrm{min}} \mathrm{R}_{\max }$
Eliminating $V_{\text {min }}$  from equation (i) and (ii) we get

$R_{\max }=3 R_{e}$