A planet is revolving round the sun. Its distance from the sun at Apogee is $r_{A}$ and that at Perigee is $r_{P}$ . The masses of planet and sun are $m$ and $M$ respectively, $v_{A}$ is the velocity of planet at Apogee and $v_{P}$ is at Perigee respectively and $T$ is the time period of revolution of planet round the sun, then identify the wrong answer.

T^{2} = \frac{π^{2}}{2 G m} {(r_{A} + r_{p})}_{A}^{3}

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T^{2} = \frac{π^{2}}{2 G m} {(r_{A} + r_{p})}_{A}^{2}

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v_{A} r_{A} = v_{p} r_{p}

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v_{A} < v_{p}, r_{A} > r_{p}

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Solution

The correct option is B $T^{2} = \frac{π^{2}}{2 G m} {(r_{A} + r_{p})}_{A}^{2}$
Note that if a is the semi major axis then we have the following relation according to geometry: $a = \frac{r_{A} + r_{P}}{2}$
also if $v_{A}$ and $v_{p}$ are the velocity at apogee and perigee respectively, then by the law of conservation of angular momentum at both these points, we have: $v_{A} r_{A} = v_{P} r_{P}$
also we know by definition of apogee and perigee, $r_{A} > r_{P}$ , which implies now time period T is given by: $T^{2} = \frac{4 π^{2} a^{3}}{G m}$
$T^{2} = \frac{4 π^{2} {(\frac{r_{A} + r_{P}}{2})}^{3}}{G m} = T^{2} = \frac{π^{2} {(r_{A} + r_{P})}_{A}^{3}}{2 G m}$
(NOTE: this particular question can also be solved by dimensional analysis of equation given in option B, where the dimensions of LHS and RHS do not match)

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Column-I	Column-II
(a) speed of planet	(p) remains same
(b) distance of planet from centre of sun	(q) decreases
(c) potential energy	(r) increases
(d) angular momentum about centre of sun	(s) can not say

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