Question

A planet is revolving around the Sun as shown in the figure. The radius vectors joining the Sun and the planet at points $A$ and $B$ are $90 \times 10^{6} km$ and $60 \times 10^{6} km$, respectively. The ratio of velocities of the planet at the points $A$ and $B$ when its velocities make angle $30^{\circ}$ and $60^{\circ}$ with major-axis of the orbit is

• A. $\frac{3}{2\sqrt{3}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

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

Answer 2

According to the law of conservation of angular momentum, angular momentum $(J)$ of a planet is constant.

$\Rightarrow m u_{A} r_{A} \sin \theta_{A}=m u_{B} r_{B} \sin \theta_{B}$

or $\frac{u_{A}}{v_{B}}=\frac{r_{B}}{r_{A}} \frac{\sin \theta_{B}}{\sin \theta_{ A }}$

Given, $r_{A}=90 \times 10^{6} km , r_{B}=60 \times 10^{6} \,km$

$\theta_{A}=30^{\circ}, \theta_{B}=60^{\circ}$

or $\frac{u_{A}}{u_{B}}=\frac{60 \times 10^{6}}{90 \times 10^{6}} \times \frac{\sin 60^{\circ}}{\sin 30^{\circ}}$

$4=\frac{2}{3} \times \frac{\sqrt{3} / 2}{1 / 2}$

or $\frac{u_{A}}{u_{B}}=\frac{2}{\sqrt{3}}$

Hence, the ratio of velocities of the planet is $2 \sqrt{3}$