Question

The ratio of kinetic energy of a planet at perihelion and aphelion during its motion around the sun in elliptical orbit of eccentricity $e$ is

• A. $1:e$
• B. $\frac{1+e}{1-e}$
• C. ${\left(\frac{1+e}{1-e}\right)}^2$
• D. ${\left(\frac{1-e}{1+e}\right)}^2$

Answer 1

The correct option is C ${\left(\frac{1+e}{1-e}\right)}^2$

As we know that kinetic energy is given by

K.E of a planet $=\frac{1}{2}m{v}^2$

So, K.E at perihelion, $K{E}_p=\frac{1}{2}m{v}_p^2$

K.E at aphelion, $K{E}_a=\frac{1}{2}m{v}_a^2$

Using conservation of angular momentum at $\text{P}$ and $\text{A}$, we get

$\Rightarrow m{v}_p{r}_p=m{v}_A{r}_a$


$\Rightarrow \frac{v_p}{v_a}=\frac{r_a}{r_p}$

From diagram,

$r_p=a(1-e);r_a=a(1+e)$

$\therefore \frac{v_p}{v_a}=\frac{a(1+e)}{a(1-e)}$

$\Rightarrow \frac{{KE}_p}{{KE}_a}=\frac{v_p^2}{v_a^2}={\left(\frac{1+e}{1-e}\right)}^2$

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

Why this question: To make students familiar with the mathematics of the elliptical orbit of planetary motion and the application of conservation of angular momentum. Key Concept: The angular momentum of a planet is conserved about the star as the force passes through the star itself and hence external torque is zero.