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 v1,\vec{v_1}, at a distance of r1r_1 from the centre a planet and that we know (only) the following values :

  • Universal Gravitational Constant

  • Mass of the planet (MM)

  • r1,v1r_1,|\vec{v_1}|

  • And angle between v\vec{v} and the line joining the center of the planet and the orbiting object (ϕ)(\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=l1+ecosθ..........[A]r=\dfrac{l}{1+e\cos\theta}..........[A] Let there be some position in this orbit with distance r2r_2 from the centre of the planet and velocity v2\vec{v_2} such that angle between v\vec{v} and the line joining the center of the planet and the orbiting object is 90deg90\deg r2v2=r1v1sinϕ (conservation of angular momentum)..........[1]\therefore r_2v_2=r_1v_1\sin\phi\space(\blue{conservation \space of\space angular\space momentum})..........[1] 12v22GMr2=12v12GMr1...........[2]\dfrac{1}{2}v^2_2-\dfrac{GM}{r_2}=\dfrac{1}{2}v^2_1-\dfrac{GM}{r_1}...........[2] Using [1][1] and [2][2] we can find r2,v2r_2,v_2 In [A], if l=q2p,e=1q2p2In\space [A], \space if \space l=\dfrac{q^2}{p},e=\sqrt{1-\dfrac{q^2}{p^2}} Then from here we know that:Then\space from \space here\space we\space know\space that : p=r1GM2GMr1v12p=\dfrac{\sqrt{r_1GM}}{2GM-r_1v^2_1} From [A] we can say thatFrom \space [A]\space we \space can\space say\space that r2=l1+e=q2p+p2q2r_2=\dfrac{l}{1+e}=\dfrac{q^2}{p+\sqrt{p^2-q^2}} q4q2(2pr2r22)=0..........[3]\Rightarrow q^4 - q^2 (2pr_2-r^2_2)=0..........[3]

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

If 0e<1 then the orbit is going to be an ellipse or a circleIf \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 0e<1    v1<vescape\therefore 0\leq e<1\iff v_1<v_{escape} where vescape is the escape velocitywhere\space v_{escape} \space is \space the \space escape \space velocity

If e=1 then the trajectory will be a parabolaIf\space e=1 \space then \space the \space trajectory \space will \space be \space a \space parabola In that case v1=vescapeIn \space that \space case \space v_1=v_{escape}

If e>1 then the trajectory will be a hyperbolaIf \space e>1 \space then \space the \space trajectory \space will \space be \space a \space hyperbola then v1>vescapethen \space v_1>v_{escape}

Now when we shall be solving for r2 usingNow\space when \space we \space shall \space be \space solving \space for \space r_2 \space using [1] and [2] we will get 2 poosible values of it [1] \space and \space [2] \space we \space will \space get \space 2 \space poosible \space values \space of \space it

If the orbit is an ellipse or circle, both of them must workIf\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 the trajectory is a parabola then there must be only one value of r2If\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 the trajectory is a hyperbola, only one of those value will workIf\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 coming back to [3], if v1=vescape q=0Now \space coming \space back \space to \space [3], \space if \space v_1=v_ {escape}\Rightarrow \space q=0 else,q2=2ar2r22else, q^2=2ar_2-r_2^2 Now we can find e, l and subsequently the conic section described by itNow\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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