A particle of mass m is thrown in a parabolic path towards earth having mass M and radius R. It comes to greatest approach distance R/2 from surface of earth. Magnitude of total energy change required to make the particle moving in an elliptical path having eccentricity ½, with the greatest approach point either apogee or perigee are:

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$\frac{1}{6} \frac{G M m}{R}$

$\frac{G M m}{R}$

$\frac{1}{3} \frac{G M m}{R}$

$\frac{1}{2} \frac{G M m}{R}$

answer is B, C.

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Detailed Solution

Initially parabolic path means escape velocity & hence $(T E = 0)$
For perihelion
$a (1 - e) = R + h = \frac{3 R}{2} \Rightarrow a = 3 R$
Also, for elliptical path
$T . E . = - \frac{G M m}{2 a} = \frac{- G M m}{6 R}$
$\Rightarrow$ change in $T E = \frac{G M m}{6 R}$
For aphelion:
$a (1 + e) = R + h = \frac{3 R}{2} \Rightarrow a = R & T . E . = - \frac{G M m}{2 a} = - \frac{G M m}{2 R}$
$\Rightarrow$ change in $T E = \frac{G M m}{2 R}$

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