Question

A planet of mass 'm' revolves around the sun of mass M in an elliptical orbit. The semi major axis of the orbit is a . Find out the total mechanical energy of the planet?

Answer 1

Answer:

Given the perihelion = r2 = shortest distance of a planet from Sun

                 aphelion = r1 = longest distance of planet from Sun

                 mass of planet = m.

                 mass of Sun = M

So the length of the major axis of the Elliptical orbit of the planet

    = r1 + r2.

Semimajor axis = R = (r1 + r2)/2.

According to Kepler's laws:

    The square of time period T of a planet revolving around Sun is proportional to the cube of semi major axis R of the elliptical orbit of the planet.

    T² ∞ R³

    T ∞  R³/²

    T  ∞ [r1 + r2] ³/²     Answer.

Actual value of the time period is given by:T=2pie√R³/Gm

2pie√(R1+R2 )³/8Gm