A planet is revolving around the Sun as shown in the figure. The radius vectors joining the Sun and the…
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 \(90 \times 10^6 \mathrm{~km}\) and \(60 \times 10^6 \mathrm{~km}\), 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
\(\frac{3}{2 \sqrt{3}}\)
\(\frac{2}{\sqrt{3}}\)
\(\frac{1}{\sqrt{3}}\)
\(\frac{\sqrt{3}}{2}\)
Solution
According to the law of conservation of angular momentum, angular momentum \((J)\) of a planet is constant.
\(\begin{aligned}
& \Rightarrow \quad m u_A r_A \sin \theta_A=m u_B r_B \sin \theta_B \\
& \text {or } \quad \frac{u_A}{v_B}=\frac{r_B}{r_A} \frac{\sin \theta_B}{\sin \theta_{\mathrm{A}}} \\
& \text {Given, } r_A=90 \times 10^6 \mathrm{~km}, r_B=60 \times 10^6 \mathrm{~km} \\
& \theta_A=30^{\circ}, \theta_B=60^{\circ}
\end{aligned}\)
\(\begin{aligned}
\text{or } \frac{u_A}{u_B} & =\frac{60 \times 10^6}{90 \times 10^6} \times \frac{\sin 60^{\circ}}{\sin 30^{\circ}} \\
& =\frac{2}{3} \times \frac{\sqrt{3} / 2}{1 / 2} \\
\text{or } \frac{u_A}{u_B} & =\frac{2}{\sqrt{3}}
\end{aligned}\)
Hence, the ratio of velocities of the planet is \(2 \sqrt{3}\).
