Two planets A and B having masses m₁ and m₂ move around the sun in circular orbits of r₁ and r₂ radii respectively. If angular momentum of A is L and that of B is 3 L, the ratio of time period ((TA)/(TB)) is: (a) ((r₂)/(r₁))3/2 (b) 27((m₁)/(m₂))³ (c) ((r₁)/(r₂))³ (d) 1/27((m₂)/(m₁))³

Answer: (d). L = m√(GMr), so (LB)/(LA) = (m₂)/(m₁)√((r₂)/(r₁)) = 3. √((r₁)/(r₂)) = (m₂)/(3m₁) (TA)/(TB) = ((r₁)/(r₂))3/2 = ((m₂)/(3m₁))³ = 1/27((m₂)/(m₁))³ So the answer is (d).

Two planets A and B having masses m₁ and m₂ move around the sun in… | JEE Main 2024 (8 Apr, Shift 1)

Q27JEE Main 2024 · 8 APR · SHIFT 1Medium

Satellites - Orbital Velocity & Period

Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of $r_1$ and $r_2$ radii respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3$ $L$ , the ratio of time period $\left(\frac{T_A}{T_B}\right)$ is:

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Answer: (d)

$L = m\sqrt{GMr}$ , so $\frac{L_B}{L_A} = \frac{m_2}{m_1}\sqrt{\frac{r_2}{r_1}} = 3$ .
$\sqrt{\frac{r_1}{r_2}} = \frac{m_2}{3m_1}$
$\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{m_2}{3m_1}\right)^3 = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^3$
So the answer is (d).

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