Question

The ratio of kinetic energy of a planet at perigee and apogee during its motion around the sun in an elliptical orbit of eccentricity $e$is?

Answer 1

Step 1: Given data

$\text{eccentricity}=e$

Step 2: To Find

We have to determine the ratio of the kinetic energy of a planet at perigee and apogee.

Step 3: Calculate the ratio of the kinetic energy of a planet at perigee and apogee

Let,

$r_a$ be the distance at apogee.

$r_p$ is the distance at perigee.

Denote the distances in terms of semi-major axes and eccentricity (using the concepts of the ellipse).

If $a$ is the length of the semi-major axis then,

$r_a=a(1+e)$ (for apogee), and $r_p=a(1-e)$ (for perigee)

By the law of conservation of momentum, we will equate the angular momentum of the planet which has mass $m$ and velocities of apogee and perigee as $v_a$ and $v_p$.

$\Rightarrow m{v}_a{r}_a=m{v}_p{r}_p$ (on canceling the common terms)
$\Rightarrow {\displaystyle \frac{v_a}{v_p}}={\displaystyle \frac{r_p}{r_a}}$ (we have taken the ratio of velocities and distances)
$\frac{v_a}{v_p}=\frac{1-e}{1+e}$ ...(1)

We know kinetic energy is given by:
${\displaystyle \frac{1}{2}}m{v}^2$ (m is the mass and v is the velocity of the moving planet)

The ratio of the kinetic energies from apogee and perigee is:
$\Rightarrow {\displaystyle \frac{K_p}{K_a}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}m{v^2}_p}{{\displaystyle \frac{1}{2}}m{v^2}_a}}$
$\Rightarrow {\displaystyle \frac{K_p}{K_a}}={\displaystyle \frac{{(1+e)}^2}{{(1-e)}^2}}$(on substituting the values of velocities from equation one)
Hence, the ratio of kinetic energies is ${\displaystyle \frac{K_p}{K_a}}={\displaystyle \frac{{(1+e)}^2}{{(1-e)}^2}}$.