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 4\; R_{e}
Option: 2 2.5\; R_{e}
 
Option: 3 3\; R_{e}  
Option: 4 2\; R_{e}

Answers (1)

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}

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