Question

A planet has a circular orbit around a star of mass M. However, the star just explodes, projecting its outer envelope at a much greater speed than that of the planet in orbit. Its lost mass can, therefore, be considered as having been lost instantaneously. What remains of the star has a mass M ', always greater than that of the planet. What is the eccentricity of the planet's orbit after the explosion? You can neglect the force exerted on the planet by the envelope being propelled.

The eccentricity in terms of energy E and angular momentum L is given by: e^2=1+ (2EL^2) / (MpK^2)

where Mp is the mass of the planet and where the magnitude of the gravitational force between the star and the planet is given by K/r^2

Answer 1

The energy and momentum will always be conserved, the eccentricity change will be due to the change in the K value, We know that the magnitude of force between the star of mass M' and planet mass M is,

By comparing we can say that,

Therefore the eccentricity of planet after explosion is,

We know that intially the planet was in circular orbit, which means that the initail eccentricity was 0, let the intial mass of the star be, M_s. Then we can write,

We know that Energy and Angular Momentum is always constant, so we can substitute this in the new eccentricity equation, which gives, the equation for eccentricity after explosion as,

Ms is the mass of the star initially, and M' is the mass of the star after explosion.