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Offline Centres

A satellites is revolving round the earth in a circular orbit of radius ‘a’ with velocity $v_{0} .$ A particle of mass m is projected from the satellite in forward direction with relative velocity $v = [\sqrt{\frac{5}{4}} - 1] v_{0}$ During subsequent motion of particle, match the following

List – I

List – II
A
Total energy of particle
P
$- \frac{3 G M_{e} m}{8 a}$
B
Minimum distance of particle from the earth
Q
   $\frac{5}{8} \frac{G M_{e} m}{a}$
C
Maximum distance of particle from the earth
R
   $5a/3$
D
The kinetic energy
S
$a$

T
     $a/2$

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$A \to p B \to t C \to r p \to q$

$A \to p B \to s C \to r D \to q$

$A \to q B \to s C \to r D \to p$

$A \to q B \to t C \to r D \to q$

answer is B.

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Detailed Solution

Angular momentum of particle $= m (v_{0} + v) a = \sqrt{\frac{5}{4}} m v_{0} a ..... v_{0} = \sqrt{\frac{G M_{e}}{a}}$
Total energy of particle
$= \frac{1}{2} m {(v_{0} + v)}^{2} - \frac{G M_{e} m}{a} = \frac{1}{2} \times \frac{5}{4} m v_{0}^{2} - \frac{G M_{e} m}{a}$
$\frac{5}{8} \frac{G M_{e} m}{a} - \frac{G M_{e} m}{a} = \frac{3 G M_{e} m}{8 a}$
At any distance ‘r’ $T . E = \frac{1}{2} m u^{2} - \frac{G M_{e} m}{r}$
But angular momentum is conserved,
$m u r = m \sqrt{\frac{5 G M_{e}}{4 a}} . a \Rightarrow u = \sqrt{\frac{5}{4} \frac{G M_{e} a}{r^{2}}}$
T.E at any distance ‘r’ $\frac{1}{2} m \frac{5}{4} \frac{G M_{e} a}{r^{2}} - \frac{G M_{e} m}{r}$
But through conservation of energy, total energy $\frac{1}{2} m \frac{5}{4} \frac{G M_{e} a}{r^{2}} - \frac{G M_{e} m}{r} = - \frac{3 G m_{e} m}{8 a}$

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