Question 188 of 223

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius ${R}_{e}.$ By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become $\sqrt{\dfrac{3}{2}}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $R$. Value of $R$ is :

A$4{R}_{e}$B$2.5{R}_{e}$C$3{R}_{e}$D$2{R}_{e}$

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Answer: (C) $3{R}_{e}$

$V=\sqrt{\dfrac{GM}{{R}_{e}}}$
From angular momentum conversation
$-\dfrac{GMm}{{R}_{e}}+\dfrac{1}{2}m{\left(\sqrt{\dfrac{3}{2}}V\right)}^{2}=-\dfrac{GMm}{{R}_{\max }}+\dfrac{1}{2}m{V}_{\min }^{2}$
From angular momentum conversation
$\sqrt{\dfrac{3}{2}}{VR}_{e}={V}_{\min }{R}_{\max }$
Eliminating ${V}_{\text{min}}$ from equation (i) and (ii) we get
${R}_{\max }=3{R}_{e}$