Question

A small satellite of mass $m$ is revolving around the earth in a circular orbit of radius $r_{0}$ with speed $v_{0}$ . At a certain point in its orbit, the direction of motion of satellite is suddenly changed by angle $\theta =\left(cos\right)^{-1}\left(\frac{3}{5}\right)$ by turning its velocity vector in the same plane of motion, such that the speed remains constant. The satellite consequently goes into an elliptical orbit around earth. The ratio of speed at perigee to speed at apogee is

• A. $3$
• B. $9$
• C. $1 / 3$
• D. $1 / 9$

Answer 1

Answer: (B)

Using conservation of angular momentum about $O$.
$ mV _{ P } r _{ P }= mV _{ A } r _{ A }= mV _0 r _0 \cos \theta$
$ V _{ A } r _{ A }= V _{ P } r _{ P }=\frac{3 v _0 r _0}{5} \ldots . . . \text { (1) }$
using conservation of energy
$\frac{1}{2} mV _{ A }^2+\frac{- GMm }{ r _{ A }}=\frac{- GMm }{ r _0}+\frac{1}{2} mV _0^2$
$\Rightarrow \frac{9 V _0^2 r _0^2}{50 r _{ A }^2}-\frac{ v _0^2 r _0}{ r _{ A }}+\frac{ v _0^2}{2}=0 $
Let $\frac{ r _0}{ r _{ A }}= x$
This is a quadratic equation in $\frac{r_0}{r_a}$
$\Rightarrow 9 x^2-50 x+25=0 \Rightarrow x=5 \text { or } x=\frac{5}{9}$


$\Rightarrow r_{A}=\frac{9}{5}r_{o} \, \ldots \ldots .\left(1\right)$
$r_{p}=\frac{r_{o}}{5} \, \, \ldots \ldots .\left(\right.2\left.\right)$
using (1) & (2)
$\Rightarrow \frac{\text{V}_{\text{P}}}{\text{V}_{\text{A}}} = \frac{\text{r}_{\text{A}}}{\text{r}_{\text{P}}} = 9$