Picture of Nova Cygni 1992 from APOD taken 2 years after it exploded

__Trajectory of ejecta__

Consider a particle on the surface of a rotating star of radius r_{o} having an angle θ
with the polar axis. Based on observation of sunspots, we can assume that prior
to the star exploding, the particle is confined to a fixed latitude. In figure 1, plane
AA' represents this fixed latitude. If the star rotates with an angular velocity ω_{o},
the particle at the surface at a distance of r_{o} sin θ
from the polar axis will have an
initial velocity prior to the explosion, V_{p} given by:

V_{p}= ω_{o } r_{o } sin θ e_{Φ } |
(i) |

**e**_{Φ} is the azimuthal unit vector. In Figure 1, the particle denoted by **P** can be
imagined to be moving out of the paper (or out of your screen if you are reading
this from your PC!)

Fig 1: Planes to which the particle **P** is confined.

Both AA' and BB' are planes normal to this figure.

Let's assume that after the star explodes, the trajectory of a particle from the
surface depends only on its momentum and the star's gravitational field. In other
words, after the explosion, it becomes a 'ballistic particle'. As the surface
explodes outward, the particle from the surface acquires a velocity component in
the radial direction. For the particle **P**, if we sum up the the vectors representing
the initial velocity prior to the explosion (V_{p}), and a radial component (V_{o})
outward, the resultant vector for velocity of ejecta (V_{E}) would be:

V_{E}= ω_{o } r_{o } sin θ e + V_{Φ }_{o} e _{r } |
(ii) |

**e _{r}** is the radial unit vector. After being ejected from the surface of the star, this
ballistic particle would be attraced toward the centre of gravity, which is the
centre of the star. The plane BB’ consists of both the centre of gravity and the
vector V

Fig 2: Mirrored image particle

The line intersecting planes BB' and CC' passes through the origin and is on equatorial plane of the star. If the centre of the star is denoted as O, the line OP makes 90° with the intersecting line between planes BB' and CC'. This proves that P and P' might collide at the equatorial plane.

*Will a collision actually take place at the equator?*

To answer this question, we need to look at the trajectory of the particle as seen on plane BB'. It is known that using Newton’s laws, the path of a ballistic object would either be a circle, ellipse, parabola or hyberbola. If the path is a parabola or hyperbola, the particle will never reach the equatorial plane.

Fig 3: Elliptic path on the BB' plane

After the explosion, the angular momentum of the particle at the surface having a polar angle θ , (per unit mass with respect to centre of the star), h is given by

h ( θ ) = r_{o }^{2} ω_{o } sin θ |
(iii) |

The energy of the particle per unit mass, E is given by the sum of its kinetic and potential energy.

E ( θ ) = (v_{o }^{2} + r_{o }^{2} ω_{o }^{2} sin ^{2} θ)/2 - GM/r_{o } |
(iv) |

Let’s use (ρ,φ) to denote the position of the particle on plane B-B' in polar coordinates. The equations conserving energy (E) and angular momentum (h) per unit mass are:

E = ( (dρ/dt)^{2} + ρ^{2} (dφ/dt)^{2} )/2 - GM/ρ |
(v) |

h = ρ ^{2} (dφ/dt) |
(vi) |

E and h would depend on the angle that the plane B-B' is inclined to the rotation axis of the star (θ) as given in equations (iii) and (iv). Eliminating t from (v) and (vi), the trajectory is

ρ = (h^{2}/GM) / (1 - e sin (φ + α)) |
(vii) |

where α is an arbitrary constant of integration that is equal to zero if the apogee is chosen to be at φ=π/2 as shown in Fig 3 and e, the eccentricity is given by

e = (1 + 2Eh^{2}/G^{2}M^{2}) ^{1/2} |
(viii) |

For the trajectory to be an ellipse, the eccentricity must be between 0 and 1. From equation viii, this is achieved when E satisfies the following:

-G^{2}M^{2}/2h^{2} < E < 0 |
(ix) |

*How do we explain the triple ring nebula of SN1987A?*

If a single exploding star is capable of forming a single ring, how many exploding stars does it take to form 3 rings? Click here for the answer.