Question

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

Answer 1

Hint: Eccentricity: How much a conic section varies from being circular.
Perigee and apogee is the farthest and the nearest point in the elliptical path.
The path in which planets revolve is elliptical in nature and thus eccentricity is defined for the elliptical path of the planets in the question.
We will solve the above question using the concept of ellipse.

Complete step by step solution:
Let us discuss the eccentricity, apogee and perigee in detail then we will calculate the ratio of kinetic energies of the planet.
Eccentricity: Eccentricity is the measure of how much a conic section (a circle, a parabola, a hyperbola) varies from being circular. Eccentricity of the circle is zero. Bigger eccentricities are less curved.
Perigee: The moon's orbit around the earth is elliptical and the point nearest to earth on the orbit is called perigee.
Apogee: The point on the moon's orbit farthest from the earth is called Apogee.
Now we will do the calculation part:
Let ra be the distance at apogee and rp is the distance at perigee.
We will denote the distances in terms of semi major axes and eccentricity (using the concepts of ellipse).
If a is the length of semi major axis then,
${r_a} = a(1 + e)$ (for apogee)
${r_p} = a(1 - e)$ (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 va and vp.
$ \Rightarrow m{v_a}{r_a} = m{v_p}{r_p}$ (on cancelling the common terms)
$ \Rightarrow \dfrac{{{v_a}}}{{{v_p}}} = \dfrac{{{r_p}}}{{{r_a}}}$ (we have taken the ratio of velocities and distances)
$ \Rightarrow \dfrac{{{v_a}}}{{{v_p}}} = \dfrac{{(1 - e)}}{{(1 + e)}}$............(1)
We know kinetic energy is given by:
$\dfrac{1}{2}m{v^2}$ (m is the mass and v is the velocity of the moving planet)
Ratio of the kinetic energies from apogee and perigee is:
$ \Rightarrow \dfrac{{{K_p}}}{{{K_a}}} = \dfrac{{\dfrac{1}{2}m{v^2}_p}}{{\dfrac{1}{2}m{v^2}_a}}$
$ \Rightarrow \dfrac{{{K_p}}}{{{K_a}}} = \dfrac{{{{(1 + e)}^2}}}{{{{(1 - e)}^2}}}$(on substituting the values of velocities from equation one)

Ratio of kinetic energies is: $\dfrac{{{K_p}}}{{{K_a}}} = \dfrac{{{{(1 + e)}^2}}}{{{{(1 - e)}^2}}}$.

Note: Elliptical path of the planets is responsible for the seasons we witness on our planet earth. When the earth is nearest to the sun summer season takes place and when the Earth goes far away from the sun while the winter season takes place, less heat is experienced by the Earth.