Question

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions $\text{A, B}$ and $\text{C}$ are $K_A$, $K_B$ and $K_C$, respectively. $\text{AC}$ is the major axis and $\text{SB}$ is perpendicular to $\text{AC}$ at the position of the Sun $S$ as shown in the figure. Then

• A. $K_A<K_B<K_C$
• B. $K_A>K_B>K_C$
• C. $K_B>K_A>K_C$
• D. $K_B<K_A<K_C$

Answer 1

The correct option is B $K_A>K_B>K_C$

The magnitude of angular momentum $L$ of the planet is given by

$L=mvr$

Since the angular momentum of the planet is constant, orbital velocity will be inversely proportional to the distance of planet from the sun.

$\therefore v\propto \frac{1}{r}$

Hence, the minimum velocity will be at the farthest point and vice versa.


Point $\text{A}$ is perihelion and $\text{C}$ is aphelion.

$\therefore v_A>v_B>v_C$

So, $K_A>K_B>K_C$

Hence, option $\left(b\right)$ is correct.

Why this question: To make students check their understanding of Kepler’s second law of planetary motion. Key Concept: Kepler’s 2nd Law of planetary motion - A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.