Question

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 $\sqrt{\frac{3}{2}}$times larger. Due to this the farthest distance from the center of the earth that the satellite reaches is $R$. Value of $R$is:

• A. $2R_e$
• B. $3R_e$
• C. $4R_e$
• D. $2.5R_e$

Answer 1

The correct option is B

$3R_e$

Step 1. Given data :

Earth's radius $=R_e$

Orbit's radius $=R$

Mass of the earth $=M$

Mass of the satellite $=m$

The velocity of the satellite $=V_{\circ }$

Step 2. Formula

${(K.E)}_A+{(P.E)}_A={(K.E)}_B+{(P.E)}_B$

$m{V}_{\circ }{R}_e=mVR$

Step 3. Find the value of $R$:

$V_{\circ }\to \sqrt{\frac{3}{2}}{V}_{\circ }$

From the conservation of angular momentum about the center of earth $m{V}_{\circ }{R}_e=mVR$

$\left[\frac{m{v}^2}{r}=\frac{GMm}{r^2}\Rightarrow V_{\circ }=\sqrt{\frac{GM}{R_e}}\right]$

$\sqrt{\frac{3}{2}}\times \sqrt{\frac{GM}{R_e}}\times R_e=V\times R$ $\to \left(1\right)$

Applying conservation of total energy for satellite :

${(K.E)}_A+{(P.E)}_A={(K.E)}_B+{(P.E)}_B$ (where, $A$is for minimum and $B$ is for maximum)

$\Rightarrow$$\frac{1}{2}m{\left(\sqrt{\frac{3}{2}}{v}_0\right)}^2-\frac{GMm}{R_e}=\frac{1}{2}m{v_0}^2-\frac{GMm}{R}$

$\Rightarrow$ $\frac{3}{4}\frac{GM}{R_e}-\frac{GM}{R_e}=\frac{V^2}{2}-\frac{GM}{R}$

substitute $V$ value from equation $\left(1\right)$

$\Rightarrow$ $-\frac{GM}{4R_e}=\frac{3}{2}\frac{{R_e}^2}{R^2}\times \frac{GM}{R_e}\times \frac{1}{2}-\frac{GM}{R}$

$\Rightarrow$ $-\frac{1}{4R_e}=\frac{3}{4}\frac{R_e}{R^2}-\frac{1}{R}$

$\Rightarrow$ $-\frac{1}{4R_e}=\frac{3R_e-4R}{4R^2}$

$\Rightarrow$ $-R^2=3{R_e}^2-4R{R}_e$

$\Rightarrow$$R^2-4R{R}_e+3{R_e}^2=0$

$\therefore$$\mathit{R}\mathbf{=}{\mathit{R}}_{\mathbf{e}\mathbf{}}$and $\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}{\mathit{R}}_{\mathbf{e}}$

Hence, option (B) is correct.