Question

A satellite is orbiting around the earth in a circular orbit of radius $r$. A particle of mass $m$ is projected from the satellite in a forward direction with a velocity $v=2/3$ times the orbital velocity (this velocity is given w.r.t earth). During subsequent motion of the particle, its minimum distance from the centre of earth is

• A. $\frac{2r}{7}$
• B. $r$
• C. $\frac{2r}{3}$
• D. $\frac{4r}{5}$

Answer 1

The correct option is A $\frac{2r}{7}$

Since in this case, the particle will move in elleptical orbit, the minimum distance will be at a position opposite to the initial position.

Conserving angular momentum,

$mr\left(\frac{2}{3}\sqrt{\frac{GM}{r}}\right)=mvx$

Also, conserving total energy,

$\frac{1}{2}m\left(\frac{4}{9}\frac{GM}{r}\right)-\frac{GMm}{r}=\frac{1}{2}m{v}^2-\frac{GMm}{x}$

$\frac{GM}{x}-\frac{1}{2}{v}^2=\frac{7}{9}\frac{GM}{r}$

$GMx-\frac{1}{2}{\left(vx\right)}^2=\frac{7}{9}\frac{GM}{r}{x}^2$

$GMx-\frac{2}{9}GMr=\frac{7}{9}\frac{GM}{r}{x}^2$

$x=\frac{2r}{7}$