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(B) According to the law of conservation of angular momentum,the angular momentum $(L)$ of a planet is constant.$L = mvr \sin 	heta$,where $	heta$ i

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

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(B) According to the law of conservation of angular momentum,the angular momentum $(L)$ of a planet is constant.$L = mvr \sin 	heta$,where $	heta$ is the angle between the velocity vector and the radius vector.However,the problem specifies the angle between the velocity vector and the major axis. Let $\phi$ be the angle between the velocity vector and the radius vector. The angular momentum is $L = mvr \sin \phi$.From the geometry of an ellipse,the angle between the radius vector and the tangent (velocity vector) is related to the angle with the major axis. Given the standard approach for such problems where the angular momentum component perpendicular to the radius vector is $v \sin \phi$,the conservation law is $m v_A r_A \sin \phi_A = m v_B r_B \sin \phi_B$.Given $r_A = 90 	imes 10^6 	ext{ km}$,$r_B = 60 	imes 10^6 	ext{ km}$,$\phi_A = 30^{\circ}$,and $\phi_B = 60^{\circ}$:$\frac{v_A}{v_B} = \frac{r_B}{r_A} 	imes \frac{\sin \phi_B}{\sin \phi_A}$$\frac{v_A}{v_B} = \frac{60 	imes 10^6}{90 	imes 10^6} 	imes \frac{\sin 60^{\circ}}{\sin 30^{\circ}}$$\frac{v_A}{v_B} = \frac{2}{3} 	imes \frac{\sqrt{3}/2}{1/2} = \frac{2}{3} 	imes \sqrt{3} = \frac{2}{\sqrt{3}}$.

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