Two planets A and B of equal mass are having their period of revolutions TA and TB such that TA = 2TB. These planets are revolving in the circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?

T_A^2 - T_B^2 = \pi ^2 \over GM( r_B^3 - 4r_A^3 )

Gravitation PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q47 with Solution

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Correct answer: C

For a planet revolving around a central mass $M$ , the time period is

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

Squaring both sides,

$T^2 = \frac{4\pi^2}{GM}r^3$

So,

$T^2 \propto r^3$

This is Kepler's third law.

Given,

$T_A = 2T_B$

Squaring,

$T_A^2 = 4T_B^2$

Since $T^2 \propto r^3$ ,

$\frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3}$

Thus,

$\frac{r_A^3}{r_B^3} = 4$

So,

$r_A^3 = 4r_B^3$

From $T_A^2 = \frac{4\pi^2}{GM}r_A^3, \quad T_B^2 = \frac{4\pi^2}{GM}r_B^3$

we get

$T_A^2 - T_B^2 = \frac{4\pi^2}{GM}(r_A^3-r_B^3)$

This is not equal to

$\frac{\pi^2}{GM}(r_B^3-4r_A^3)$

So D is incorrect.

The correct relationship is

$\boxed{r_A^3 = 4r_B^3}$

Hence, option C is correct.