# Homework Help: Hyperbolic Kick, Why is happens?

1. Dec 5, 2007

### m0nk3y

Hyperbolic Kick, Why is happens??

1. The problem statement, all variables and given/known data

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 :)

2. Dec 5, 2007

### chaoseverlasting

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

$$y=b\sqrt{1+\frac{x^2}{a^2}}$$ (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

$$r^2=(\frac{a^2+b^2}{a^2})x^2+b^2$$

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

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

Substituting the above expression for $$r^2$$ to get gravitational force as a function of x, we get,

$$F_g=-\frac{Gm_1m_2}{(\frac{a^2+b^2}{a^2})x^2+b^2}$$

$$F_g=m_1a$$ where m1 is the mass of the object and a is the acceleration.

Acceleration may be written as $$a=v\frac{dv}{dx}$$, giving you the DE,

$$v\frac{dv}{dx}=-\frac{Gm_2}{(\frac{a^2+b^2}{a^2})x^2+b^2}$$

Upon solving, you get,

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

This shows that the square of the velocity is dependent on $$tan^{-1}x$$. If you see the graph of arctan(x), you should be able to understand why the object gets a "kick".

Last edited: Dec 5, 2007
3. Dec 5, 2007

### chaoseverlasting

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.

4. Dec 6, 2007

### m0nk3y

THANK YOU SO MUCH!! This makes so much sense.