Why does an object in a hyperbolic orbit get a velocity kick from a planet?

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Homework Help Overview

The discussion revolves around the mechanics of an object in a hyperbolic orbit receiving a velocity "kick" when passing close to a planet. The original poster seeks to understand the underlying principles of this phenomenon, particularly in relation to gravity assists and the conditions under which this effect occurs.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the mathematical representation of hyperbolic orbits and the gravitational forces involved. Questions are raised regarding the conditions necessary for the velocity kick to occur, particularly contrasting moving versus stationary planets.

Discussion Status

Some participants provide mathematical derivations related to gravitational forces and velocity changes, while others express appreciation for the clarity of these explanations. The discussion appears to be productive, with participants engaging in both conceptual and mathematical aspects of the problem.

Contextual Notes

There is an implicit assumption that the object must be in motion at an angle to the planet for the velocity kick to occur, which is a point of discussion among participants.

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Hyperbolic Kick, Why is happens??

Homework Statement



Why does an object in a hyperbolic orbit passing close to a planet (which is in orbit about another large object like the Sun) get a velocity "kick" from it?

Why does it not work for a stationary planet?

I think it has to do with gravity assist, but I will like to know how this "kick" works.

Thanks :)
 
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Well, if you take the equation of the hyperbola with the planet at the origin, it is [tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]. This, of course, represents both the arms of the hyperbola, isolating one of the arms of the hyperbola, you get

[tex]y=b\sqrt{1+\frac{x^2}{a^2}}[/tex] (here the arms are above and below the x axis).

As the path of the planet is represented by this hyperbola, at any point on its path, say (x,y), the square of the distance would be given by

[tex]r^2=(\frac{a^2+b^2}{a^2})x^2+b^2[/tex]

(derived from [tex]r=\sqrt{x^2+y^2}[/tex] from pythagorean theorem and the equation derived above)

Now, we know, the gravitational force is given by [tex]F_g=-\frac{Gm_1m_2}{r^2}[/tex].

Substituting the above expression for [tex]r^2[/tex] to get gravitational force as a function of x, we get,

[tex]F_g=-\frac{Gm_1m_2}{(\frac{a^2+b^2}{a^2})x^2+b^2}[/tex]

[tex]F_g=m_1a[/tex] where m1 is the mass of the object and a is the acceleration.

Acceleration may be written as [tex]a=v\frac{dv}{dx}[/tex], giving you the DE,

[tex]v\frac{dv}{dx}=-\frac{Gm_2}{(\frac{a^2+b^2}{a^2})x^2+b^2}[/tex]

Upon solving, you get,

[tex]\frac{v^2-v_o^2}{2}=-Gm_2(\sqrt{\frac{a^2+b^2}{a^2b^2}})tan^{-1}x[/tex].

This shows that the square of the velocity is dependent on [tex]tan^{-1}x[/tex]. If you see the graph of arctan(x), you should be able to understand why the object gets a "kick".
 
Last edited:
If the object was not already in motion at an angle to the planet, the the line of action of force is the line joining the centers of the two bodies, therefore, the object would travel in a straight line towards the planet.
 
THANK YOU SO MUCH! This makes so much sense.
 

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