# Geometry of Orbits

This note will be a solution to the following problem :

Any orbiting object with negligible size which is launched with a velocity $\vec{v_1},$ at a distance of $r_1$ from the centre a planet and that we know (only) the following values :

• Universal Gravitational Constant

• Mass of the planet ($M$)

• $r_1,|\vec{v_1}|$

• And angle between $\vec{v}$ and the line joining the center of the planet and the orbiting object $(\phi)$

Also assume that mass of the orbiting object is very - very smaller than the mass of the planet.

From Kepler's First law we know that the orbit is going to be a conic section, which can be described by the following polar equation : $r=\dfrac{l}{1+e\cos\theta}..........[A]$ Let there be some position in this orbit with distance $r_2$ from the centre of the planet and velocity $\vec{v_2}$ such that angle between $\vec{v}$ and the line joining the center of the planet and the orbiting object is $90\deg$ $\therefore r_2v_2=r_1v_1\sin\phi\space(\blue{conservation \space of\space angular\space momentum})..........[1]$ $\dfrac{1}{2}v^2_2-\dfrac{GM}{r_2}=\dfrac{1}{2}v^2_1-\dfrac{GM}{r_1}...........[2]$ Using $[1]$ and $[2]$ we can find $r_2,v_2$ $In\space [A], \space if \space l=\dfrac{q^2}{p},e=\sqrt{1-\dfrac{q^2}{p^2}}$ $Then\space from \space here\space we\space know\space that :$ $p=\dfrac{\sqrt{r_1GM}}{2GM-r_1v^2_1}$ $From \space [A]\space we \space can\space say\space that$ $r_2=\dfrac{l}{1+e}=\dfrac{q^2}{p+\sqrt{p^2-q^2}}$ $\Rightarrow q^4 - q^2 (2pr_2-r^2_2)=0..........[3]$

$Now \space before \space going \space further \space we \space need \space to \space see \space something \space important \space :$$( \blue{proof \space of \space which \space is \space left \space to \space the \space reader})$

$If \space 0\leq e <1 \space then \space the \space orbit \space is \space going \space to \space be \space an \space ellipse \space or \space a \space circle$ $\therefore 0\leq e<1\iff v_1 $where\space v_{escape} \space is \space the \space escape \space velocity$

$If\space e=1 \space then \space the \space trajectory \space will \space be \space a \space parabola$ $In \space that \space case \space v_1=v_{escape}$

$If \space e>1 \space then \space the \space trajectory \space will \space be \space a \space hyperbola$ $then \space v_1>v_{escape}$

$Now\space when \space we \space shall \space be \space solving \space for \space r_2 \space using$ $[1] \space and \space [2] \space we \space will \space get \space 2 \space poosible \space values \space of \space it$

$If\space the \space orbit \space is \space an \space ellipse \space or \space circle, \space both \space of \space them \space must \space work$ $If\space the\space trajectory\space is\space a \space parabola\space then\space there\space must\space be\space only\space one\space value\space of\space r_2$ $If\space the \space trajectory \space is \space a \space hyperbola, \space only \space one \space of \space those \space value \space will \space work$

$Now \space coming \space back \space to \space [3], \space if \space v_1=v_ {escape}\Rightarrow \space q=0$ $else, q^2=2ar_2-r_2^2$ $Now\space we \space can \space find \space e,\space l \space and \space subsequently \space the \space conic \space section \space described \space by \space it$

Note by Zakir Husain
2 weeks, 1 day ago

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