Question

A satellite is revolving around the earth in a circular orbit of radius 'a' with velocity $v_0$. A particle is projected from the statellite in forward direction wiht relative velocity $V=\left[\right(\frac{\sqrt{5}}{4}\left)\right]v_0.$ During the subsequent motion of the particle, match the following:
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{I. Total energy of particle}& \text{P.}\frac{(-3G{M}_{e}m)}{8a}\\ \text{II. Minimum distance of}& \text{Q.}\frac{(-5G{M}_{e}m)}{8a}\\ \text{particle from the earth}& \\ \text{III. Maximum distance of}& \text{R.}\frac{5a}{3}\\ \text{particle from the earth}& \\ & \text{S. 2a}\\ & \text{T.}\frac{2a}{3}\\ & \text{U. a}\end{array}$

• A. $I\to Q;II\to S;III\to U$
• B. $I\to P;II\to U;III\to S$
• C. $I\to Q;II\to S;III\to R$
• D. $I\to P;II\to U;III\to R$

Answer 1

The correct option is D $I\to P;II\to U;III\to R$

Angular momentum of the particle $=m(v_0+v)a$
$=\sqrt{\frac{5}{4}}m{v}_{0}a[v_0=\sqrt{\frac{G{M}_e}{a}}]$
Toal energy of the particle $=\frac{1}{2}m(v_0+v{)}^2-\frac{G{M}_{e}m}{a}$
$=\frac{1}{2}\times \frac{5}{4}m{v}_e^2-\frac{G{M}_{e}m}{a}=\frac{5}{8}\frac{G{M}_{e}m}{a}-\frac{G{M}_{e}m}{a}=\frac{-3G{M}_{e}m}{8a}$
At any distance 'r', $T.E.=\frac{1}{2}m{u}^2-\frac{G{M}_{e}m}{a}$
But angular momentum conservation gives
$mur=\sqrt{\frac{5G{m}_e}{4a}}\Rightarrow u=\sqrt{\frac{5}{4}\frac{G{M}_{e}a}{r^2}}$
T.E. at any distance ${}^{\prime }{r}^{\prime }=\frac{1}{2}m\frac{5}{4}\frac{GMea}{r^2}-\frac{G{M}_{e}m}{r}$
But through conservation of total energy, we have
$\frac{1}{2}m\frac{5}{4}\frac{G{M}_{e}a}{r^2}-\frac{G{M}_{e}m}{r}=-\frac{3G{M}_{e}m}{8a}$
On solving, we get
$3r^2-8ar+5a^2=0$
$(r-a)(3r-5a)=0$
$R=a,r=\frac{5a}{3}$
Minimum distance = a
Maximum distance $=\frac{5a}{3}$