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 centre of the earth that the satellite reaches is $R.$ The value of $R$ is-

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

Answer 1

The correct option is D $2R_e$


As we know that orbital velocity is,

$V_O=\sqrt{\frac{GM}{R}}$

And potential energy is

$PE=-\frac{GMm}{R}$

Apply energy conservation between two given situations,

$E_1=E_2$

$-\frac{GMm}{R_e}+\frac{1}{2}m{\left(\sqrt{\frac{3}{2}}{V}_O\right)}^2=-\frac{GMm}{R_{max}}+\frac{1}{2}m{V}_{min}^2$

$-\frac{GMm}{R_e}+\frac{3GMm}{4R_e}=-\frac{GMm}{R_{max}}+\frac{GMm}{2R_{max}}$

$-\frac{GMm}{4R_e}=-\frac{GMm}{2R_{max}}$

$\Rightarrow R_{max}=2R_e$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.