Question

Two planets of equal masses orbit a much more massive star. Planet \(m_1\) moves in a circular orbit of radius \(1 \times 10^8~\text{ km}\) with a period of \(2~\text{ years}\) and planet \(m_2\) moves in an elliptical orbit with closest distance \(r_1 = 1 \times 10^8~\text{km}\) and farthest distance \(r_2 = 1.8 \times 10^8~\text{km}\), as shown:


Compare the speed of planet \(m_2\) at \(\text{P}\)\((V_P)\) with that of at \(\text{A}\) \((V_A)\).

• A. $V_P=2.4V_A$
• B. $V_P=1.8V_A$
• C. $V_P=4.2V_A$
• D. $V_P=3.6V_A$

Answer 1

The correct option is B $V_P=1.8V_A$

$\text{A}$ is at the farthest distance and $\text{P}$ is at the closest distance.

Since, there are no external torques about the sun, angular momentum of $m_2$ about the Sun will the conserved.

$m_2{V}_A{r}_2=m_2{V}_P{r}_1$

Since, mass of the planet is same at the both points, $m_2=m_1$

$\Rightarrow V_P=\frac{r_2}{r_1}{V}_A$

$\Rightarrow V_P=\frac{1.8\times {10}^8}{1\times {10}^8}\times V_A$

$\Rightarrow V_P=1.8V_A$

Hence, option (d) is the correct answer.