# Trajectory of a Celestial Body

An interstellar body approaches the solar system. When it is first detected at a distance of $100 AU$ from the sun, it has a speed of $10 km/s$. Will its orbit past the sun be elliptical, parabolic, or hyperbolic? Its orbit takes it to less than $1 AU$ from the sun. Use energy considerations to determine its speed when it is $1 AU$ from the sun and compare this to the speed needed to escape the solar system at $1 AU$.

Characteristic energy of a body moving in a central gravitational field is by definition

$\epsilon=v^2/2-\mu/R$

$V$ is speed of body, $R$ is distance to star, $\mu=GM$ is the standard gravitational parameter ($M$ is the star mass, $G$ is the gravitational constant).

The specific trajectory is given by the value of $\epsilon$.

If $\epsilon<0$ the trajectory is elliptic (unless the body falls directly on the star).

If $\epsilon=0$ the trajectory is parabolic.

If $\epsilon>0$ the trajectory is hyperbolic.

With the given data $M_{sun}=1.989*10^30 kg$ ,$R=100 AU=1.496*10^13 m$ , $\mu=1.327*10^20 (m^3/s^2)$

the characteristic energy is

$\epsilon=(10^4)^2/2-(1.327*10^20)/(1.496*10^13)=+4.11*10^7 (J/kg)$

Therefore the trajectory is hyperbolic.

For $R_0=1 AU=1.496*10^11 m$ one has

$(v_0^2)/2-\mu/R_0 =4.11*10^(7) (J/m)$

$v_0^2=2*(4.11*10^7+(1.327*10^20)/(1.496*10^11 ))=1.856*10^9 (m^2/s^2)$

$v_0=43.08 (km/s)$

Escape velocity is by definition when =0 , that is when $v_e=\sqrt{(2GM/R0)}=\sqrt{(2\mu/R0)}$

$v_e=42.12 (km/s) < v_0$

In other words, the trajectory is

-elliptic for velocity < escape velocity,

-parabolic for velocity = escape velocity and

-hyperbolic for velocity > escape velocity