Question

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.
A. ${T^2} = \dfrac{{{\pi ^2}}}{{2Gm}}{\left( {{r_A} + {r_P}} \right)^3}$
B. ${T^2} = \dfrac{{{\pi ^2}}}{{2Gm}}{\left( {{r_A} + {r_P}} \right)^2}$
C. ${v_A}{r_A} = {v_P}{r_P}$
D. ${v_A} < {v_P},{r_A} > {r_P}$

Answer 1

In order to solve this question, we are going to first see the relation for the semi major axis of the ellipse and then analyze the conditions for the apogee and perigee and also that for time period by putting the value of the semi major axis in it as well and all the correct options are found.

Complete step by step answer:
If we consider the semi major axis of the ellipse to be $a$, then according to geometry, we write the following relation.
$a = \dfrac{{\left( {{r_A} + {r_P}} \right)}}{2}$
Now as it is given that the ${v_A}$ is the velocity of the planet at Apogee and ${v_P}$ is at Perigee. Now, by the law of conservation of angular momentum at both these points is:
${v_A}{r_A} = {v_P}{r_P}$
Also, by definition of apogee and perigee, we have
${r_A} > {r_P}$
This implies that time period is given by
${T^2} = \dfrac{{4{\pi ^2}{a^3}}}{{Gm}}$
Putting the value of $a$ in here, we get
${T^2} = \dfrac{{4{\pi ^2}}}{{Gm}}{\left( {\dfrac{{{r_A} + {r_P}}}{2}} \right)^3}$
Simplifying this,
$\therefore {T^2} = \dfrac{{{\pi ^2}}}{{2Gm}}{\left( {{r_A} + {r_P}} \right)^3}$

Hence, the correct answer is option B.

Note: Alternatively, 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 which clearly indicates that this option is wrong while all the other options are correct dimensionally and physically as well.