To solve this problem it is necessary to take into account the concepts of Gravitational Force and Kinetic Energy.
The kinetic energy is given by the equation:
[tex]F= \frac{mv^2}2[/tex]
La energía gravitacional por,
[tex]F=\frac{GM_cm}{d}[/tex]
Where m is the mass, v is the velocity, G the gravitational constant [tex]M_e[/tex] the mass of the earth, m the mass of the sun and d the distance ..
The sum of the energies, we must be a total energy
[tex]E= \frac{mv^2}2+\frac{GM_em}{d}[/tex]
By the type of orbit we know that
E> 0 is a hyperbolic orbit
E = 0 is a parabolic orbit
E <0 is a closed orbit.
In the case of hyperbolic orbit
E>0
[tex]\frac{mq^2}{2}-\frac{GM_em}{d}>0\\\frac{qv^2_e}{2}>\frac{GM_em}{d}\\q^2d>2\frac{GM_e}{v^2_e}\\q^2d>2[/tex]
The case of the comet is a closed orbit, so,
E<0
[tex]\frac{mv^2}2+\frac{GM_em}{d}<0\\\frac{mq^2v^2_e}{2}<\frac{GM_cm}{d}\\q^2d<2\frac{GM_e}{v^2_e}[/tex]
For parabolic orbit
E=0
[tex]\frac{mv^2_eq^2}{2}-\frac{GM_cm}{d}=0\\\frac{v^2_eq^2}{2}=\frac{GM_c}{d}\\q^2d=2\frac{GM_e}{v^2_e}\\q^2d=2[/tex]
For the sun and the earth
[tex]\frac{m_ev_e^2}{r}=\frac{GM_em_e}{r^2}[/tex]
[tex]v^2_e=\frac{GM_e}{r}\\\frac{GM_e}{v_e}=r[/tex]
where [tex]R \approx 1AU[/tex]
[tex]q^2d<2[/tex] For elliptical orbit
